Let ${^ts}:Eto F$ be the transpose of $s:Fto E$. Show that $text{Im}(s)cap ker (,^ts)={0_E}$.











up vote
3
down vote

favorite
1













Let $sin mathcal{L}(F,E)$



$$displaystyle F overset{s}{longrightarrow} Eoverset{^ts}{longrightarrow} F$$
I spent two days to show that :$$text{Im}(s)cap ker (,^ts)={0_E}qquad tag{1}$$




I'm not sure that is right, I tried with specific matrix (3x3), and it works.



But when I want to provide a general proof, I struggle.



I used the Annihilator but no way to find a solution, may be this statement is wrong...?



I add more information about $E$ and $F$ regarding the comments



$F=mathbb{R}^p$ and $E=mathbb{R}^n$ with $ple n$



Or $F=(mathbb{R}^p,langlecdot,cdotrangle)$ and $E=(mathbb{R}^n,langlecdot,cdotrangle)$



The matrix associated at $u$ is $M$ and $Min mathcal{M}_{n,p}(mathbb{R})$










share|cite|improve this question




















  • 2




    I think the problem is not clear. What is the base field here? Presumably, you have nondegenerate bilinear forms on $E$ and $F$. However, not all bilinear forms work. If the base field is of a positive characteristic, then this is false. Even when the characteristic of the base field is $0$, you can come up with a nondegenerate symmetric bilinear form that contradicts the claim.
    – Batominovski
    Nov 16 at 17:03






  • 1




    So, I would like the OP to confirm whether: (1) the base field is $mathbb{R}$, (2) $E$ and $F$ are finite-dimensional vector spaces, and (3) the vector spaces are equipped with positive-definite symmetric bilinear forms.
    – Batominovski
    Nov 16 at 17:05












  • I added more informations, but I didn't want to use the inner product at first...but it seems I must do.
    – Stu
    Nov 16 at 17:23















up vote
3
down vote

favorite
1













Let $sin mathcal{L}(F,E)$



$$displaystyle F overset{s}{longrightarrow} Eoverset{^ts}{longrightarrow} F$$
I spent two days to show that :$$text{Im}(s)cap ker (,^ts)={0_E}qquad tag{1}$$




I'm not sure that is right, I tried with specific matrix (3x3), and it works.



But when I want to provide a general proof, I struggle.



I used the Annihilator but no way to find a solution, may be this statement is wrong...?



I add more information about $E$ and $F$ regarding the comments



$F=mathbb{R}^p$ and $E=mathbb{R}^n$ with $ple n$



Or $F=(mathbb{R}^p,langlecdot,cdotrangle)$ and $E=(mathbb{R}^n,langlecdot,cdotrangle)$



The matrix associated at $u$ is $M$ and $Min mathcal{M}_{n,p}(mathbb{R})$










share|cite|improve this question




















  • 2




    I think the problem is not clear. What is the base field here? Presumably, you have nondegenerate bilinear forms on $E$ and $F$. However, not all bilinear forms work. If the base field is of a positive characteristic, then this is false. Even when the characteristic of the base field is $0$, you can come up with a nondegenerate symmetric bilinear form that contradicts the claim.
    – Batominovski
    Nov 16 at 17:03






  • 1




    So, I would like the OP to confirm whether: (1) the base field is $mathbb{R}$, (2) $E$ and $F$ are finite-dimensional vector spaces, and (3) the vector spaces are equipped with positive-definite symmetric bilinear forms.
    – Batominovski
    Nov 16 at 17:05












  • I added more informations, but I didn't want to use the inner product at first...but it seems I must do.
    – Stu
    Nov 16 at 17:23













up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1






Let $sin mathcal{L}(F,E)$



$$displaystyle F overset{s}{longrightarrow} Eoverset{^ts}{longrightarrow} F$$
I spent two days to show that :$$text{Im}(s)cap ker (,^ts)={0_E}qquad tag{1}$$




I'm not sure that is right, I tried with specific matrix (3x3), and it works.



But when I want to provide a general proof, I struggle.



I used the Annihilator but no way to find a solution, may be this statement is wrong...?



I add more information about $E$ and $F$ regarding the comments



$F=mathbb{R}^p$ and $E=mathbb{R}^n$ with $ple n$



Or $F=(mathbb{R}^p,langlecdot,cdotrangle)$ and $E=(mathbb{R}^n,langlecdot,cdotrangle)$



The matrix associated at $u$ is $M$ and $Min mathcal{M}_{n,p}(mathbb{R})$










share|cite|improve this question
















Let $sin mathcal{L}(F,E)$



$$displaystyle F overset{s}{longrightarrow} Eoverset{^ts}{longrightarrow} F$$
I spent two days to show that :$$text{Im}(s)cap ker (,^ts)={0_E}qquad tag{1}$$




I'm not sure that is right, I tried with specific matrix (3x3), and it works.



But when I want to provide a general proof, I struggle.



I used the Annihilator but no way to find a solution, may be this statement is wrong...?



I add more information about $E$ and $F$ regarding the comments



$F=mathbb{R}^p$ and $E=mathbb{R}^n$ with $ple n$



Or $F=(mathbb{R}^p,langlecdot,cdotrangle)$ and $E=(mathbb{R}^n,langlecdot,cdotrangle)$



The matrix associated at $u$ is $M$ and $Min mathcal{M}_{n,p}(mathbb{R})$







linear-algebra vector-spaces linear-transformations bilinear-form transpose






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 16 at 18:31









Batominovski

31.6k23188




31.6k23188










asked Nov 16 at 15:57









Stu

1,1011313




1,1011313








  • 2




    I think the problem is not clear. What is the base field here? Presumably, you have nondegenerate bilinear forms on $E$ and $F$. However, not all bilinear forms work. If the base field is of a positive characteristic, then this is false. Even when the characteristic of the base field is $0$, you can come up with a nondegenerate symmetric bilinear form that contradicts the claim.
    – Batominovski
    Nov 16 at 17:03






  • 1




    So, I would like the OP to confirm whether: (1) the base field is $mathbb{R}$, (2) $E$ and $F$ are finite-dimensional vector spaces, and (3) the vector spaces are equipped with positive-definite symmetric bilinear forms.
    – Batominovski
    Nov 16 at 17:05












  • I added more informations, but I didn't want to use the inner product at first...but it seems I must do.
    – Stu
    Nov 16 at 17:23














  • 2




    I think the problem is not clear. What is the base field here? Presumably, you have nondegenerate bilinear forms on $E$ and $F$. However, not all bilinear forms work. If the base field is of a positive characteristic, then this is false. Even when the characteristic of the base field is $0$, you can come up with a nondegenerate symmetric bilinear form that contradicts the claim.
    – Batominovski
    Nov 16 at 17:03






  • 1




    So, I would like the OP to confirm whether: (1) the base field is $mathbb{R}$, (2) $E$ and $F$ are finite-dimensional vector spaces, and (3) the vector spaces are equipped with positive-definite symmetric bilinear forms.
    – Batominovski
    Nov 16 at 17:05












  • I added more informations, but I didn't want to use the inner product at first...but it seems I must do.
    – Stu
    Nov 16 at 17:23








2




2




I think the problem is not clear. What is the base field here? Presumably, you have nondegenerate bilinear forms on $E$ and $F$. However, not all bilinear forms work. If the base field is of a positive characteristic, then this is false. Even when the characteristic of the base field is $0$, you can come up with a nondegenerate symmetric bilinear form that contradicts the claim.
– Batominovski
Nov 16 at 17:03




I think the problem is not clear. What is the base field here? Presumably, you have nondegenerate bilinear forms on $E$ and $F$. However, not all bilinear forms work. If the base field is of a positive characteristic, then this is false. Even when the characteristic of the base field is $0$, you can come up with a nondegenerate symmetric bilinear form that contradicts the claim.
– Batominovski
Nov 16 at 17:03




1




1




So, I would like the OP to confirm whether: (1) the base field is $mathbb{R}$, (2) $E$ and $F$ are finite-dimensional vector spaces, and (3) the vector spaces are equipped with positive-definite symmetric bilinear forms.
– Batominovski
Nov 16 at 17:05






So, I would like the OP to confirm whether: (1) the base field is $mathbb{R}$, (2) $E$ and $F$ are finite-dimensional vector spaces, and (3) the vector spaces are equipped with positive-definite symmetric bilinear forms.
– Batominovski
Nov 16 at 17:05














I added more informations, but I didn't want to use the inner product at first...but it seems I must do.
– Stu
Nov 16 at 17:23




I added more informations, but I didn't want to use the inner product at first...but it seems I must do.
– Stu
Nov 16 at 17:23










4 Answers
4






active

oldest

votes

















up vote
2
down vote













Let $s:Fto E$ be a linear transformation from a vector space $F$ to another vector space $E$ over the base field $mathbb{R}$. Both $E$ and $F$ are equipped with nondegenerate symmetric bilinear forms $langle_,_rangle_E$ and $langle_,_rangle_F$, respectively. Suppose further that $langle_,_rangle_E$ is positive-definite. Write $s^top:Eto F$ for the transpose of $s$. We claim that
$$text{im}(s)capker(s^top)={0_E},.$$



Recall that $s^top$ is the unique linear transformation from $E$ to $F$ such that $$biglangle s(x),ybigrangle_E=biglangle x,s^top(y)rangle_F$$
for all $xin F$ and $yin E$. Suppose that $yintext{im}(s)capker(s^top)$. Then, $yintext{im}(s)$, so $y=s(x)$ for some $xin F$. Sinc $yinker(s^top)$, we get $$s^topbig(s(x)big)=s^top(y)=0_F,.$$
Therefore,
$$Biglangle x,s^topbig(s(x)big)Bigrangle_F=biglangle x,0_Fbigrangle_F=0,.$$
By the definition of $s^top$, we have
$$langle y,yrangle_E=biglangle s(x),s(x)bigrangle_E=Biglangle x,s^topbig(s(x)big)Bigrangle_F=0,.$$
Since $langle_,_rangle_E$ is positive-definite, we have $y=0_E$. This proves the claim.



P.S.: If the ground field is $mathbb{C}$, we instead assume that $langle_,_rangle_E$ and $langle_,_rangle_F$ are nondegenerate Hermitian forms. The transpose map has to be replaced by the Hermitian conjugate. For the claim to hold, we need to assume further that $langle_,_rangle_E$ is positive-definite.





If the ground field $mathbb{K}$ is a field of characteristic $0$, but there is no positive-definiteness assumption on $langle_,_rangle_E$ (which makes no sense when $mathbb{K}$ is not a subfied of $mathbb{R}$ anyhow), then the claim is not true. For example, let $mathbb{K}:=mathbb{R}$, $E:=mathbb{R}^2$, and $F:=mathbb{R}^2$. Then we set $e_1:=(1,0)$, $e_2:=(0,1)$. Let $langle_,_rangle$ be the nondegenerate symmetric bilinear form on both $E$ and $F$ given by $$langle a_1e_1+a_2e_2,b_1e_1+b_2e_2rangle:=a_1b_1-a_2b_2$$
for all $a_1,a_2,b_1,b_2inmathbb{R}$. Take $s:Fto E$ to be the linear map sending both $e_1$ and $e_2$ to $e_1+e_2$. That is, $s$ is given by the matrix
$$begin{bmatrix}1&1\1&1end{bmatrix}$$
with respect to the ordered basis $(e_1,e_2)$ of both $F$ and $E$. Then, $s^top:Eto F$ is given by the matrix
$$begin{bmatrix}1&-1\-1&1end{bmatrix}$$
with respect to the ordered basis $(e_1,e_2)$ of both $E$ and $F$. This shows that $$text{im}(s)=text{span}_mathbb{R}{e_1+e_2}=ker(s^top),.$$





The claim fails in positive characteristics. If the ground field $mathbb{K}$ is of a positive characteristric $p$, then take $E$ and $F$ to be both $mathbb{K}^p$. Fix a basis ${e_1,e_2,ldots,e_p}$ of both $E$ and $F$. Equip $E$ and $F$ with the nondegenerate symmetric bilinear form $langle_,_rangle$ given by
$$leftlanglesum_{i=1}^p,a_ie_i,sum_{i=1}^p,b_ie_irightrangle:=sum_{i=1}^p,a_ib_i$$
for all $a_1,a_2,ldots,a_p,b_1,b_2,ldots,b_pinmathbb{K}$. If the map $s:Fto E$ is the $mathbb{K}$-linear map sending $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$, then its transpose $s^top:Eto F$ also sends $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$. In particular, this shows that $$text{span}_mathbb{K}{e_1+e_2+ldots+e_p}=text{im}(s)capker(s^top),.$$






share|cite|improve this answer






























    up vote
    2
    down vote













    It is sufficient to show that
    $$
    ker s^tsubset(text{Im } s)^perp
    $$

    because we can then conclude by:
    $$
    ker s^t cap text{Im} s subset (text{Im} s)^perp cap text{Im} s ={0_E}
    $$





    The claim $ker s^tsubset(text{Im } s)^perp$ can be proven as follows:



    begin{align*}
    einker s^t &Rightarrow s^t(e)=0_F \
    &Rightarrow forall fin F, langle f,s^t(e) rangle_F=0 \ &Rightarrow forall fin F, langle s(f),erangle_E=0 \
    &Rightarrow forall fin F, s(f) perp e \
    &Rightarrow e in (text{Im} s)^perp
    end{align*}






    share|cite|improve this answer



















    • 1




      I would like to note that the equality $ker (s^top)=big(text{im}(s)big)^perp$ is true for any nondegenerate symmetric bilinear form $langle_,_rangle_E$. However, the last equality $big(text{im}(s)big)^perpcaptext{im}(s)={0_E}$ must rely on some strong conditions on $langle_,_rangle_E$ such as positive-definiteness.
      – Batominovski
      Nov 16 at 18:27












    • thanks a lot I need some time to understand everything, and I'll be back
      – Stu
      Nov 16 at 18:28










    • @Batominovski you are right, and that is the reason why I simply wrote positive definite right at the beginning. Also a more general result would also hold only using duality bracket $<. ,.>_{Etimes E^*}$ and $<. ,.>_{Ftimes F^*}$ (without introducing scalar products $<. ,.>_E$ and $<. ,.>_F$
      – Picaud Vincent
      Nov 16 at 18:30












    • Shorter proof (sorry it is nearly a complete rewrite, but the idea is the same)
      – Picaud Vincent
      Nov 16 at 21:39




















    up vote
    1
    down vote













    In order that the transpose is a map $Fto E$ you need some assumptions: both vector spaces need to be equipped with a nondegenerate bilinear form, so that we can define isomorphisms $Eto E^*$ and $Fto F^*$ (the dual spaces), assuming finite dimensionality.



    In particular, if the base field is $mathbb{R}$, the two space might be inner product spaces.



    If $langle{cdot},{cdot}rangle_E$ is the form on $E$, then for each $vin E$, the map
    $$
    e_vcolon Eto K,qquad e_v(x)=langle v,xrangle_E
    $$

    is an element of $E^*$. If $vne0$, then nondegeneracy implies there exists $xin E$ with $langle v,xrangle_Ene0$, so $ecolon Eto E^*$, $vmapsto e_v$ is an isomorphism (it is clear from bilinearity that $e_v$ is linear, for every $vin E$, and $e$ is linear as well).



    Then the transpose ${}^{t!}s$ is the dual map composed with these isomorphisms:
    $$
    {}^{t!}s=e^{-1}circ s^*circ e
    $$

    (I denote by $e$ both maps $Eto E^*$ and $Fto F^*$, no confusion should arise).



    If $w=s(v)$ and ${}^{t!}s(w)=0$, then also $s^*circ e_w=0$, which means $e_wcirc s=0$, that is,
    $$
    e_w(s(x))=0quadtext{for all $xin V$}
    $$

    hence
    $$
    langle w,s(x)rangle_F=0
    $$

    In particular, $langle s(v),s(v)rangle_F=0$, so by nondegeneracy, $s(v)=0$ and $w=0$.






    share|cite|improve this answer




























      up vote
      0
      down vote













      If you represent a linear map by a matrix then the image of the map is the sub-space spanned of the columns of the matrix. And the kernel of the map is the sub-space that is perpendicular to the rows of the matrix (because the inner product of each row with any vector in the kernel must be zero).



      So the image of $s$ is the sub-space that is spanned by the columns of the matrix representing $s$ (relative to specific bases in $E$ and $F$).



      And the kernel of $^ts$ is the sub-space that is perpendicular to the sub-space spanned by the rows of the matrix representing $^ts$.



      But the columns of the first matrix are the rows of the second matrix. The result follows.






      share|cite|improve this answer





















        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














         

        draft saved


        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3001306%2flet-tse-to-f-be-the-transpose-of-sf-to-e-show-that-textims-cap%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        2
        down vote













        Let $s:Fto E$ be a linear transformation from a vector space $F$ to another vector space $E$ over the base field $mathbb{R}$. Both $E$ and $F$ are equipped with nondegenerate symmetric bilinear forms $langle_,_rangle_E$ and $langle_,_rangle_F$, respectively. Suppose further that $langle_,_rangle_E$ is positive-definite. Write $s^top:Eto F$ for the transpose of $s$. We claim that
        $$text{im}(s)capker(s^top)={0_E},.$$



        Recall that $s^top$ is the unique linear transformation from $E$ to $F$ such that $$biglangle s(x),ybigrangle_E=biglangle x,s^top(y)rangle_F$$
        for all $xin F$ and $yin E$. Suppose that $yintext{im}(s)capker(s^top)$. Then, $yintext{im}(s)$, so $y=s(x)$ for some $xin F$. Sinc $yinker(s^top)$, we get $$s^topbig(s(x)big)=s^top(y)=0_F,.$$
        Therefore,
        $$Biglangle x,s^topbig(s(x)big)Bigrangle_F=biglangle x,0_Fbigrangle_F=0,.$$
        By the definition of $s^top$, we have
        $$langle y,yrangle_E=biglangle s(x),s(x)bigrangle_E=Biglangle x,s^topbig(s(x)big)Bigrangle_F=0,.$$
        Since $langle_,_rangle_E$ is positive-definite, we have $y=0_E$. This proves the claim.



        P.S.: If the ground field is $mathbb{C}$, we instead assume that $langle_,_rangle_E$ and $langle_,_rangle_F$ are nondegenerate Hermitian forms. The transpose map has to be replaced by the Hermitian conjugate. For the claim to hold, we need to assume further that $langle_,_rangle_E$ is positive-definite.





        If the ground field $mathbb{K}$ is a field of characteristic $0$, but there is no positive-definiteness assumption on $langle_,_rangle_E$ (which makes no sense when $mathbb{K}$ is not a subfied of $mathbb{R}$ anyhow), then the claim is not true. For example, let $mathbb{K}:=mathbb{R}$, $E:=mathbb{R}^2$, and $F:=mathbb{R}^2$. Then we set $e_1:=(1,0)$, $e_2:=(0,1)$. Let $langle_,_rangle$ be the nondegenerate symmetric bilinear form on both $E$ and $F$ given by $$langle a_1e_1+a_2e_2,b_1e_1+b_2e_2rangle:=a_1b_1-a_2b_2$$
        for all $a_1,a_2,b_1,b_2inmathbb{R}$. Take $s:Fto E$ to be the linear map sending both $e_1$ and $e_2$ to $e_1+e_2$. That is, $s$ is given by the matrix
        $$begin{bmatrix}1&1\1&1end{bmatrix}$$
        with respect to the ordered basis $(e_1,e_2)$ of both $F$ and $E$. Then, $s^top:Eto F$ is given by the matrix
        $$begin{bmatrix}1&-1\-1&1end{bmatrix}$$
        with respect to the ordered basis $(e_1,e_2)$ of both $E$ and $F$. This shows that $$text{im}(s)=text{span}_mathbb{R}{e_1+e_2}=ker(s^top),.$$





        The claim fails in positive characteristics. If the ground field $mathbb{K}$ is of a positive characteristric $p$, then take $E$ and $F$ to be both $mathbb{K}^p$. Fix a basis ${e_1,e_2,ldots,e_p}$ of both $E$ and $F$. Equip $E$ and $F$ with the nondegenerate symmetric bilinear form $langle_,_rangle$ given by
        $$leftlanglesum_{i=1}^p,a_ie_i,sum_{i=1}^p,b_ie_irightrangle:=sum_{i=1}^p,a_ib_i$$
        for all $a_1,a_2,ldots,a_p,b_1,b_2,ldots,b_pinmathbb{K}$. If the map $s:Fto E$ is the $mathbb{K}$-linear map sending $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$, then its transpose $s^top:Eto F$ also sends $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$. In particular, this shows that $$text{span}_mathbb{K}{e_1+e_2+ldots+e_p}=text{im}(s)capker(s^top),.$$






        share|cite|improve this answer



























          up vote
          2
          down vote













          Let $s:Fto E$ be a linear transformation from a vector space $F$ to another vector space $E$ over the base field $mathbb{R}$. Both $E$ and $F$ are equipped with nondegenerate symmetric bilinear forms $langle_,_rangle_E$ and $langle_,_rangle_F$, respectively. Suppose further that $langle_,_rangle_E$ is positive-definite. Write $s^top:Eto F$ for the transpose of $s$. We claim that
          $$text{im}(s)capker(s^top)={0_E},.$$



          Recall that $s^top$ is the unique linear transformation from $E$ to $F$ such that $$biglangle s(x),ybigrangle_E=biglangle x,s^top(y)rangle_F$$
          for all $xin F$ and $yin E$. Suppose that $yintext{im}(s)capker(s^top)$. Then, $yintext{im}(s)$, so $y=s(x)$ for some $xin F$. Sinc $yinker(s^top)$, we get $$s^topbig(s(x)big)=s^top(y)=0_F,.$$
          Therefore,
          $$Biglangle x,s^topbig(s(x)big)Bigrangle_F=biglangle x,0_Fbigrangle_F=0,.$$
          By the definition of $s^top$, we have
          $$langle y,yrangle_E=biglangle s(x),s(x)bigrangle_E=Biglangle x,s^topbig(s(x)big)Bigrangle_F=0,.$$
          Since $langle_,_rangle_E$ is positive-definite, we have $y=0_E$. This proves the claim.



          P.S.: If the ground field is $mathbb{C}$, we instead assume that $langle_,_rangle_E$ and $langle_,_rangle_F$ are nondegenerate Hermitian forms. The transpose map has to be replaced by the Hermitian conjugate. For the claim to hold, we need to assume further that $langle_,_rangle_E$ is positive-definite.





          If the ground field $mathbb{K}$ is a field of characteristic $0$, but there is no positive-definiteness assumption on $langle_,_rangle_E$ (which makes no sense when $mathbb{K}$ is not a subfied of $mathbb{R}$ anyhow), then the claim is not true. For example, let $mathbb{K}:=mathbb{R}$, $E:=mathbb{R}^2$, and $F:=mathbb{R}^2$. Then we set $e_1:=(1,0)$, $e_2:=(0,1)$. Let $langle_,_rangle$ be the nondegenerate symmetric bilinear form on both $E$ and $F$ given by $$langle a_1e_1+a_2e_2,b_1e_1+b_2e_2rangle:=a_1b_1-a_2b_2$$
          for all $a_1,a_2,b_1,b_2inmathbb{R}$. Take $s:Fto E$ to be the linear map sending both $e_1$ and $e_2$ to $e_1+e_2$. That is, $s$ is given by the matrix
          $$begin{bmatrix}1&1\1&1end{bmatrix}$$
          with respect to the ordered basis $(e_1,e_2)$ of both $F$ and $E$. Then, $s^top:Eto F$ is given by the matrix
          $$begin{bmatrix}1&-1\-1&1end{bmatrix}$$
          with respect to the ordered basis $(e_1,e_2)$ of both $E$ and $F$. This shows that $$text{im}(s)=text{span}_mathbb{R}{e_1+e_2}=ker(s^top),.$$





          The claim fails in positive characteristics. If the ground field $mathbb{K}$ is of a positive characteristric $p$, then take $E$ and $F$ to be both $mathbb{K}^p$. Fix a basis ${e_1,e_2,ldots,e_p}$ of both $E$ and $F$. Equip $E$ and $F$ with the nondegenerate symmetric bilinear form $langle_,_rangle$ given by
          $$leftlanglesum_{i=1}^p,a_ie_i,sum_{i=1}^p,b_ie_irightrangle:=sum_{i=1}^p,a_ib_i$$
          for all $a_1,a_2,ldots,a_p,b_1,b_2,ldots,b_pinmathbb{K}$. If the map $s:Fto E$ is the $mathbb{K}$-linear map sending $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$, then its transpose $s^top:Eto F$ also sends $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$. In particular, this shows that $$text{span}_mathbb{K}{e_1+e_2+ldots+e_p}=text{im}(s)capker(s^top),.$$






          share|cite|improve this answer

























            up vote
            2
            down vote










            up vote
            2
            down vote









            Let $s:Fto E$ be a linear transformation from a vector space $F$ to another vector space $E$ over the base field $mathbb{R}$. Both $E$ and $F$ are equipped with nondegenerate symmetric bilinear forms $langle_,_rangle_E$ and $langle_,_rangle_F$, respectively. Suppose further that $langle_,_rangle_E$ is positive-definite. Write $s^top:Eto F$ for the transpose of $s$. We claim that
            $$text{im}(s)capker(s^top)={0_E},.$$



            Recall that $s^top$ is the unique linear transformation from $E$ to $F$ such that $$biglangle s(x),ybigrangle_E=biglangle x,s^top(y)rangle_F$$
            for all $xin F$ and $yin E$. Suppose that $yintext{im}(s)capker(s^top)$. Then, $yintext{im}(s)$, so $y=s(x)$ for some $xin F$. Sinc $yinker(s^top)$, we get $$s^topbig(s(x)big)=s^top(y)=0_F,.$$
            Therefore,
            $$Biglangle x,s^topbig(s(x)big)Bigrangle_F=biglangle x,0_Fbigrangle_F=0,.$$
            By the definition of $s^top$, we have
            $$langle y,yrangle_E=biglangle s(x),s(x)bigrangle_E=Biglangle x,s^topbig(s(x)big)Bigrangle_F=0,.$$
            Since $langle_,_rangle_E$ is positive-definite, we have $y=0_E$. This proves the claim.



            P.S.: If the ground field is $mathbb{C}$, we instead assume that $langle_,_rangle_E$ and $langle_,_rangle_F$ are nondegenerate Hermitian forms. The transpose map has to be replaced by the Hermitian conjugate. For the claim to hold, we need to assume further that $langle_,_rangle_E$ is positive-definite.





            If the ground field $mathbb{K}$ is a field of characteristic $0$, but there is no positive-definiteness assumption on $langle_,_rangle_E$ (which makes no sense when $mathbb{K}$ is not a subfied of $mathbb{R}$ anyhow), then the claim is not true. For example, let $mathbb{K}:=mathbb{R}$, $E:=mathbb{R}^2$, and $F:=mathbb{R}^2$. Then we set $e_1:=(1,0)$, $e_2:=(0,1)$. Let $langle_,_rangle$ be the nondegenerate symmetric bilinear form on both $E$ and $F$ given by $$langle a_1e_1+a_2e_2,b_1e_1+b_2e_2rangle:=a_1b_1-a_2b_2$$
            for all $a_1,a_2,b_1,b_2inmathbb{R}$. Take $s:Fto E$ to be the linear map sending both $e_1$ and $e_2$ to $e_1+e_2$. That is, $s$ is given by the matrix
            $$begin{bmatrix}1&1\1&1end{bmatrix}$$
            with respect to the ordered basis $(e_1,e_2)$ of both $F$ and $E$. Then, $s^top:Eto F$ is given by the matrix
            $$begin{bmatrix}1&-1\-1&1end{bmatrix}$$
            with respect to the ordered basis $(e_1,e_2)$ of both $E$ and $F$. This shows that $$text{im}(s)=text{span}_mathbb{R}{e_1+e_2}=ker(s^top),.$$





            The claim fails in positive characteristics. If the ground field $mathbb{K}$ is of a positive characteristric $p$, then take $E$ and $F$ to be both $mathbb{K}^p$. Fix a basis ${e_1,e_2,ldots,e_p}$ of both $E$ and $F$. Equip $E$ and $F$ with the nondegenerate symmetric bilinear form $langle_,_rangle$ given by
            $$leftlanglesum_{i=1}^p,a_ie_i,sum_{i=1}^p,b_ie_irightrangle:=sum_{i=1}^p,a_ib_i$$
            for all $a_1,a_2,ldots,a_p,b_1,b_2,ldots,b_pinmathbb{K}$. If the map $s:Fto E$ is the $mathbb{K}$-linear map sending $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$, then its transpose $s^top:Eto F$ also sends $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$. In particular, this shows that $$text{span}_mathbb{K}{e_1+e_2+ldots+e_p}=text{im}(s)capker(s^top),.$$






            share|cite|improve this answer














            Let $s:Fto E$ be a linear transformation from a vector space $F$ to another vector space $E$ over the base field $mathbb{R}$. Both $E$ and $F$ are equipped with nondegenerate symmetric bilinear forms $langle_,_rangle_E$ and $langle_,_rangle_F$, respectively. Suppose further that $langle_,_rangle_E$ is positive-definite. Write $s^top:Eto F$ for the transpose of $s$. We claim that
            $$text{im}(s)capker(s^top)={0_E},.$$



            Recall that $s^top$ is the unique linear transformation from $E$ to $F$ such that $$biglangle s(x),ybigrangle_E=biglangle x,s^top(y)rangle_F$$
            for all $xin F$ and $yin E$. Suppose that $yintext{im}(s)capker(s^top)$. Then, $yintext{im}(s)$, so $y=s(x)$ for some $xin F$. Sinc $yinker(s^top)$, we get $$s^topbig(s(x)big)=s^top(y)=0_F,.$$
            Therefore,
            $$Biglangle x,s^topbig(s(x)big)Bigrangle_F=biglangle x,0_Fbigrangle_F=0,.$$
            By the definition of $s^top$, we have
            $$langle y,yrangle_E=biglangle s(x),s(x)bigrangle_E=Biglangle x,s^topbig(s(x)big)Bigrangle_F=0,.$$
            Since $langle_,_rangle_E$ is positive-definite, we have $y=0_E$. This proves the claim.



            P.S.: If the ground field is $mathbb{C}$, we instead assume that $langle_,_rangle_E$ and $langle_,_rangle_F$ are nondegenerate Hermitian forms. The transpose map has to be replaced by the Hermitian conjugate. For the claim to hold, we need to assume further that $langle_,_rangle_E$ is positive-definite.





            If the ground field $mathbb{K}$ is a field of characteristic $0$, but there is no positive-definiteness assumption on $langle_,_rangle_E$ (which makes no sense when $mathbb{K}$ is not a subfied of $mathbb{R}$ anyhow), then the claim is not true. For example, let $mathbb{K}:=mathbb{R}$, $E:=mathbb{R}^2$, and $F:=mathbb{R}^2$. Then we set $e_1:=(1,0)$, $e_2:=(0,1)$. Let $langle_,_rangle$ be the nondegenerate symmetric bilinear form on both $E$ and $F$ given by $$langle a_1e_1+a_2e_2,b_1e_1+b_2e_2rangle:=a_1b_1-a_2b_2$$
            for all $a_1,a_2,b_1,b_2inmathbb{R}$. Take $s:Fto E$ to be the linear map sending both $e_1$ and $e_2$ to $e_1+e_2$. That is, $s$ is given by the matrix
            $$begin{bmatrix}1&1\1&1end{bmatrix}$$
            with respect to the ordered basis $(e_1,e_2)$ of both $F$ and $E$. Then, $s^top:Eto F$ is given by the matrix
            $$begin{bmatrix}1&-1\-1&1end{bmatrix}$$
            with respect to the ordered basis $(e_1,e_2)$ of both $E$ and $F$. This shows that $$text{im}(s)=text{span}_mathbb{R}{e_1+e_2}=ker(s^top),.$$





            The claim fails in positive characteristics. If the ground field $mathbb{K}$ is of a positive characteristric $p$, then take $E$ and $F$ to be both $mathbb{K}^p$. Fix a basis ${e_1,e_2,ldots,e_p}$ of both $E$ and $F$. Equip $E$ and $F$ with the nondegenerate symmetric bilinear form $langle_,_rangle$ given by
            $$leftlanglesum_{i=1}^p,a_ie_i,sum_{i=1}^p,b_ie_irightrangle:=sum_{i=1}^p,a_ib_i$$
            for all $a_1,a_2,ldots,a_p,b_1,b_2,ldots,b_pinmathbb{K}$. If the map $s:Fto E$ is the $mathbb{K}$-linear map sending $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$, then its transpose $s^top:Eto F$ also sends $e_imapsto e_1+e_2+ldots+e_p$ for all $i=1,2,ldots,p$. In particular, this shows that $$text{span}_mathbb{K}{e_1+e_2+ldots+e_p}=text{im}(s)capker(s^top),.$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 16 at 18:40

























            answered Nov 16 at 17:49









            Batominovski

            31.6k23188




            31.6k23188






















                up vote
                2
                down vote













                It is sufficient to show that
                $$
                ker s^tsubset(text{Im } s)^perp
                $$

                because we can then conclude by:
                $$
                ker s^t cap text{Im} s subset (text{Im} s)^perp cap text{Im} s ={0_E}
                $$





                The claim $ker s^tsubset(text{Im } s)^perp$ can be proven as follows:



                begin{align*}
                einker s^t &Rightarrow s^t(e)=0_F \
                &Rightarrow forall fin F, langle f,s^t(e) rangle_F=0 \ &Rightarrow forall fin F, langle s(f),erangle_E=0 \
                &Rightarrow forall fin F, s(f) perp e \
                &Rightarrow e in (text{Im} s)^perp
                end{align*}






                share|cite|improve this answer



















                • 1




                  I would like to note that the equality $ker (s^top)=big(text{im}(s)big)^perp$ is true for any nondegenerate symmetric bilinear form $langle_,_rangle_E$. However, the last equality $big(text{im}(s)big)^perpcaptext{im}(s)={0_E}$ must rely on some strong conditions on $langle_,_rangle_E$ such as positive-definiteness.
                  – Batominovski
                  Nov 16 at 18:27












                • thanks a lot I need some time to understand everything, and I'll be back
                  – Stu
                  Nov 16 at 18:28










                • @Batominovski you are right, and that is the reason why I simply wrote positive definite right at the beginning. Also a more general result would also hold only using duality bracket $<. ,.>_{Etimes E^*}$ and $<. ,.>_{Ftimes F^*}$ (without introducing scalar products $<. ,.>_E$ and $<. ,.>_F$
                  – Picaud Vincent
                  Nov 16 at 18:30












                • Shorter proof (sorry it is nearly a complete rewrite, but the idea is the same)
                  – Picaud Vincent
                  Nov 16 at 21:39

















                up vote
                2
                down vote













                It is sufficient to show that
                $$
                ker s^tsubset(text{Im } s)^perp
                $$

                because we can then conclude by:
                $$
                ker s^t cap text{Im} s subset (text{Im} s)^perp cap text{Im} s ={0_E}
                $$





                The claim $ker s^tsubset(text{Im } s)^perp$ can be proven as follows:



                begin{align*}
                einker s^t &Rightarrow s^t(e)=0_F \
                &Rightarrow forall fin F, langle f,s^t(e) rangle_F=0 \ &Rightarrow forall fin F, langle s(f),erangle_E=0 \
                &Rightarrow forall fin F, s(f) perp e \
                &Rightarrow e in (text{Im} s)^perp
                end{align*}






                share|cite|improve this answer



















                • 1




                  I would like to note that the equality $ker (s^top)=big(text{im}(s)big)^perp$ is true for any nondegenerate symmetric bilinear form $langle_,_rangle_E$. However, the last equality $big(text{im}(s)big)^perpcaptext{im}(s)={0_E}$ must rely on some strong conditions on $langle_,_rangle_E$ such as positive-definiteness.
                  – Batominovski
                  Nov 16 at 18:27












                • thanks a lot I need some time to understand everything, and I'll be back
                  – Stu
                  Nov 16 at 18:28










                • @Batominovski you are right, and that is the reason why I simply wrote positive definite right at the beginning. Also a more general result would also hold only using duality bracket $<. ,.>_{Etimes E^*}$ and $<. ,.>_{Ftimes F^*}$ (without introducing scalar products $<. ,.>_E$ and $<. ,.>_F$
                  – Picaud Vincent
                  Nov 16 at 18:30












                • Shorter proof (sorry it is nearly a complete rewrite, but the idea is the same)
                  – Picaud Vincent
                  Nov 16 at 21:39















                up vote
                2
                down vote










                up vote
                2
                down vote









                It is sufficient to show that
                $$
                ker s^tsubset(text{Im } s)^perp
                $$

                because we can then conclude by:
                $$
                ker s^t cap text{Im} s subset (text{Im} s)^perp cap text{Im} s ={0_E}
                $$





                The claim $ker s^tsubset(text{Im } s)^perp$ can be proven as follows:



                begin{align*}
                einker s^t &Rightarrow s^t(e)=0_F \
                &Rightarrow forall fin F, langle f,s^t(e) rangle_F=0 \ &Rightarrow forall fin F, langle s(f),erangle_E=0 \
                &Rightarrow forall fin F, s(f) perp e \
                &Rightarrow e in (text{Im} s)^perp
                end{align*}






                share|cite|improve this answer














                It is sufficient to show that
                $$
                ker s^tsubset(text{Im } s)^perp
                $$

                because we can then conclude by:
                $$
                ker s^t cap text{Im} s subset (text{Im} s)^perp cap text{Im} s ={0_E}
                $$





                The claim $ker s^tsubset(text{Im } s)^perp$ can be proven as follows:



                begin{align*}
                einker s^t &Rightarrow s^t(e)=0_F \
                &Rightarrow forall fin F, langle f,s^t(e) rangle_F=0 \ &Rightarrow forall fin F, langle s(f),erangle_E=0 \
                &Rightarrow forall fin F, s(f) perp e \
                &Rightarrow e in (text{Im} s)^perp
                end{align*}







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 16 at 22:13

























                answered Nov 16 at 18:20









                Picaud Vincent

                71815




                71815








                • 1




                  I would like to note that the equality $ker (s^top)=big(text{im}(s)big)^perp$ is true for any nondegenerate symmetric bilinear form $langle_,_rangle_E$. However, the last equality $big(text{im}(s)big)^perpcaptext{im}(s)={0_E}$ must rely on some strong conditions on $langle_,_rangle_E$ such as positive-definiteness.
                  – Batominovski
                  Nov 16 at 18:27












                • thanks a lot I need some time to understand everything, and I'll be back
                  – Stu
                  Nov 16 at 18:28










                • @Batominovski you are right, and that is the reason why I simply wrote positive definite right at the beginning. Also a more general result would also hold only using duality bracket $<. ,.>_{Etimes E^*}$ and $<. ,.>_{Ftimes F^*}$ (without introducing scalar products $<. ,.>_E$ and $<. ,.>_F$
                  – Picaud Vincent
                  Nov 16 at 18:30












                • Shorter proof (sorry it is nearly a complete rewrite, but the idea is the same)
                  – Picaud Vincent
                  Nov 16 at 21:39
















                • 1




                  I would like to note that the equality $ker (s^top)=big(text{im}(s)big)^perp$ is true for any nondegenerate symmetric bilinear form $langle_,_rangle_E$. However, the last equality $big(text{im}(s)big)^perpcaptext{im}(s)={0_E}$ must rely on some strong conditions on $langle_,_rangle_E$ such as positive-definiteness.
                  – Batominovski
                  Nov 16 at 18:27












                • thanks a lot I need some time to understand everything, and I'll be back
                  – Stu
                  Nov 16 at 18:28










                • @Batominovski you are right, and that is the reason why I simply wrote positive definite right at the beginning. Also a more general result would also hold only using duality bracket $<. ,.>_{Etimes E^*}$ and $<. ,.>_{Ftimes F^*}$ (without introducing scalar products $<. ,.>_E$ and $<. ,.>_F$
                  – Picaud Vincent
                  Nov 16 at 18:30












                • Shorter proof (sorry it is nearly a complete rewrite, but the idea is the same)
                  – Picaud Vincent
                  Nov 16 at 21:39










                1




                1




                I would like to note that the equality $ker (s^top)=big(text{im}(s)big)^perp$ is true for any nondegenerate symmetric bilinear form $langle_,_rangle_E$. However, the last equality $big(text{im}(s)big)^perpcaptext{im}(s)={0_E}$ must rely on some strong conditions on $langle_,_rangle_E$ such as positive-definiteness.
                – Batominovski
                Nov 16 at 18:27






                I would like to note that the equality $ker (s^top)=big(text{im}(s)big)^perp$ is true for any nondegenerate symmetric bilinear form $langle_,_rangle_E$. However, the last equality $big(text{im}(s)big)^perpcaptext{im}(s)={0_E}$ must rely on some strong conditions on $langle_,_rangle_E$ such as positive-definiteness.
                – Batominovski
                Nov 16 at 18:27














                thanks a lot I need some time to understand everything, and I'll be back
                – Stu
                Nov 16 at 18:28




                thanks a lot I need some time to understand everything, and I'll be back
                – Stu
                Nov 16 at 18:28












                @Batominovski you are right, and that is the reason why I simply wrote positive definite right at the beginning. Also a more general result would also hold only using duality bracket $<. ,.>_{Etimes E^*}$ and $<. ,.>_{Ftimes F^*}$ (without introducing scalar products $<. ,.>_E$ and $<. ,.>_F$
                – Picaud Vincent
                Nov 16 at 18:30






                @Batominovski you are right, and that is the reason why I simply wrote positive definite right at the beginning. Also a more general result would also hold only using duality bracket $<. ,.>_{Etimes E^*}$ and $<. ,.>_{Ftimes F^*}$ (without introducing scalar products $<. ,.>_E$ and $<. ,.>_F$
                – Picaud Vincent
                Nov 16 at 18:30














                Shorter proof (sorry it is nearly a complete rewrite, but the idea is the same)
                – Picaud Vincent
                Nov 16 at 21:39






                Shorter proof (sorry it is nearly a complete rewrite, but the idea is the same)
                – Picaud Vincent
                Nov 16 at 21:39












                up vote
                1
                down vote













                In order that the transpose is a map $Fto E$ you need some assumptions: both vector spaces need to be equipped with a nondegenerate bilinear form, so that we can define isomorphisms $Eto E^*$ and $Fto F^*$ (the dual spaces), assuming finite dimensionality.



                In particular, if the base field is $mathbb{R}$, the two space might be inner product spaces.



                If $langle{cdot},{cdot}rangle_E$ is the form on $E$, then for each $vin E$, the map
                $$
                e_vcolon Eto K,qquad e_v(x)=langle v,xrangle_E
                $$

                is an element of $E^*$. If $vne0$, then nondegeneracy implies there exists $xin E$ with $langle v,xrangle_Ene0$, so $ecolon Eto E^*$, $vmapsto e_v$ is an isomorphism (it is clear from bilinearity that $e_v$ is linear, for every $vin E$, and $e$ is linear as well).



                Then the transpose ${}^{t!}s$ is the dual map composed with these isomorphisms:
                $$
                {}^{t!}s=e^{-1}circ s^*circ e
                $$

                (I denote by $e$ both maps $Eto E^*$ and $Fto F^*$, no confusion should arise).



                If $w=s(v)$ and ${}^{t!}s(w)=0$, then also $s^*circ e_w=0$, which means $e_wcirc s=0$, that is,
                $$
                e_w(s(x))=0quadtext{for all $xin V$}
                $$

                hence
                $$
                langle w,s(x)rangle_F=0
                $$

                In particular, $langle s(v),s(v)rangle_F=0$, so by nondegeneracy, $s(v)=0$ and $w=0$.






                share|cite|improve this answer

























                  up vote
                  1
                  down vote













                  In order that the transpose is a map $Fto E$ you need some assumptions: both vector spaces need to be equipped with a nondegenerate bilinear form, so that we can define isomorphisms $Eto E^*$ and $Fto F^*$ (the dual spaces), assuming finite dimensionality.



                  In particular, if the base field is $mathbb{R}$, the two space might be inner product spaces.



                  If $langle{cdot},{cdot}rangle_E$ is the form on $E$, then for each $vin E$, the map
                  $$
                  e_vcolon Eto K,qquad e_v(x)=langle v,xrangle_E
                  $$

                  is an element of $E^*$. If $vne0$, then nondegeneracy implies there exists $xin E$ with $langle v,xrangle_Ene0$, so $ecolon Eto E^*$, $vmapsto e_v$ is an isomorphism (it is clear from bilinearity that $e_v$ is linear, for every $vin E$, and $e$ is linear as well).



                  Then the transpose ${}^{t!}s$ is the dual map composed with these isomorphisms:
                  $$
                  {}^{t!}s=e^{-1}circ s^*circ e
                  $$

                  (I denote by $e$ both maps $Eto E^*$ and $Fto F^*$, no confusion should arise).



                  If $w=s(v)$ and ${}^{t!}s(w)=0$, then also $s^*circ e_w=0$, which means $e_wcirc s=0$, that is,
                  $$
                  e_w(s(x))=0quadtext{for all $xin V$}
                  $$

                  hence
                  $$
                  langle w,s(x)rangle_F=0
                  $$

                  In particular, $langle s(v),s(v)rangle_F=0$, so by nondegeneracy, $s(v)=0$ and $w=0$.






                  share|cite|improve this answer























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    In order that the transpose is a map $Fto E$ you need some assumptions: both vector spaces need to be equipped with a nondegenerate bilinear form, so that we can define isomorphisms $Eto E^*$ and $Fto F^*$ (the dual spaces), assuming finite dimensionality.



                    In particular, if the base field is $mathbb{R}$, the two space might be inner product spaces.



                    If $langle{cdot},{cdot}rangle_E$ is the form on $E$, then for each $vin E$, the map
                    $$
                    e_vcolon Eto K,qquad e_v(x)=langle v,xrangle_E
                    $$

                    is an element of $E^*$. If $vne0$, then nondegeneracy implies there exists $xin E$ with $langle v,xrangle_Ene0$, so $ecolon Eto E^*$, $vmapsto e_v$ is an isomorphism (it is clear from bilinearity that $e_v$ is linear, for every $vin E$, and $e$ is linear as well).



                    Then the transpose ${}^{t!}s$ is the dual map composed with these isomorphisms:
                    $$
                    {}^{t!}s=e^{-1}circ s^*circ e
                    $$

                    (I denote by $e$ both maps $Eto E^*$ and $Fto F^*$, no confusion should arise).



                    If $w=s(v)$ and ${}^{t!}s(w)=0$, then also $s^*circ e_w=0$, which means $e_wcirc s=0$, that is,
                    $$
                    e_w(s(x))=0quadtext{for all $xin V$}
                    $$

                    hence
                    $$
                    langle w,s(x)rangle_F=0
                    $$

                    In particular, $langle s(v),s(v)rangle_F=0$, so by nondegeneracy, $s(v)=0$ and $w=0$.






                    share|cite|improve this answer












                    In order that the transpose is a map $Fto E$ you need some assumptions: both vector spaces need to be equipped with a nondegenerate bilinear form, so that we can define isomorphisms $Eto E^*$ and $Fto F^*$ (the dual spaces), assuming finite dimensionality.



                    In particular, if the base field is $mathbb{R}$, the two space might be inner product spaces.



                    If $langle{cdot},{cdot}rangle_E$ is the form on $E$, then for each $vin E$, the map
                    $$
                    e_vcolon Eto K,qquad e_v(x)=langle v,xrangle_E
                    $$

                    is an element of $E^*$. If $vne0$, then nondegeneracy implies there exists $xin E$ with $langle v,xrangle_Ene0$, so $ecolon Eto E^*$, $vmapsto e_v$ is an isomorphism (it is clear from bilinearity that $e_v$ is linear, for every $vin E$, and $e$ is linear as well).



                    Then the transpose ${}^{t!}s$ is the dual map composed with these isomorphisms:
                    $$
                    {}^{t!}s=e^{-1}circ s^*circ e
                    $$

                    (I denote by $e$ both maps $Eto E^*$ and $Fto F^*$, no confusion should arise).



                    If $w=s(v)$ and ${}^{t!}s(w)=0$, then also $s^*circ e_w=0$, which means $e_wcirc s=0$, that is,
                    $$
                    e_w(s(x))=0quadtext{for all $xin V$}
                    $$

                    hence
                    $$
                    langle w,s(x)rangle_F=0
                    $$

                    In particular, $langle s(v),s(v)rangle_F=0$, so by nondegeneracy, $s(v)=0$ and $w=0$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 16 at 23:16









                    egreg

                    174k1383198




                    174k1383198






















                        up vote
                        0
                        down vote













                        If you represent a linear map by a matrix then the image of the map is the sub-space spanned of the columns of the matrix. And the kernel of the map is the sub-space that is perpendicular to the rows of the matrix (because the inner product of each row with any vector in the kernel must be zero).



                        So the image of $s$ is the sub-space that is spanned by the columns of the matrix representing $s$ (relative to specific bases in $E$ and $F$).



                        And the kernel of $^ts$ is the sub-space that is perpendicular to the sub-space spanned by the rows of the matrix representing $^ts$.



                        But the columns of the first matrix are the rows of the second matrix. The result follows.






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          If you represent a linear map by a matrix then the image of the map is the sub-space spanned of the columns of the matrix. And the kernel of the map is the sub-space that is perpendicular to the rows of the matrix (because the inner product of each row with any vector in the kernel must be zero).



                          So the image of $s$ is the sub-space that is spanned by the columns of the matrix representing $s$ (relative to specific bases in $E$ and $F$).



                          And the kernel of $^ts$ is the sub-space that is perpendicular to the sub-space spanned by the rows of the matrix representing $^ts$.



                          But the columns of the first matrix are the rows of the second matrix. The result follows.






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            If you represent a linear map by a matrix then the image of the map is the sub-space spanned of the columns of the matrix. And the kernel of the map is the sub-space that is perpendicular to the rows of the matrix (because the inner product of each row with any vector in the kernel must be zero).



                            So the image of $s$ is the sub-space that is spanned by the columns of the matrix representing $s$ (relative to specific bases in $E$ and $F$).



                            And the kernel of $^ts$ is the sub-space that is perpendicular to the sub-space spanned by the rows of the matrix representing $^ts$.



                            But the columns of the first matrix are the rows of the second matrix. The result follows.






                            share|cite|improve this answer












                            If you represent a linear map by a matrix then the image of the map is the sub-space spanned of the columns of the matrix. And the kernel of the map is the sub-space that is perpendicular to the rows of the matrix (because the inner product of each row with any vector in the kernel must be zero).



                            So the image of $s$ is the sub-space that is spanned by the columns of the matrix representing $s$ (relative to specific bases in $E$ and $F$).



                            And the kernel of $^ts$ is the sub-space that is perpendicular to the sub-space spanned by the rows of the matrix representing $^ts$.



                            But the columns of the first matrix are the rows of the second matrix. The result follows.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 16 at 16:48









                            gandalf61

                            7,197523




                            7,197523






























                                 

                                draft saved


                                draft discarded



















































                                 


                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3001306%2flet-tse-to-f-be-the-transpose-of-sf-to-e-show-that-textims-cap%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                Popular posts from this blog

                                Ellipse (mathématiques)

                                Quarter-circle Tiles

                                Mont Emei