Proof that $X = left{ A in mathbb R^{6,5} : V subset ker(A) right}$ is linear subspace and find its $dim$












3












$begingroup$


Proof that if $dim V = 3 $ then $X = left{ A in mathbb R^{6,5} : V subset ker(A) right}$ is linear subspace and find its $dim$



What I've done



I think that I have done proof.
$ forall vec{v} in V. Avec{v} = 0 $ and $Bvec{v}=0$

but from property of the matrix product it follows that:
$$ Avec{v} + Bvec{v} = (A+B)vec{v} $$



Moreover if $Avec{x}=0$ then $(alpha A)vec{x} =0$ and in to other side too. So ok.



My problem



But how in use of given informations find its dimension? My scribbles do not lead to anything.






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you denote $Bbb R^{6,5}$?
    $endgroup$
    – Bernard
    Dec 31 '18 at 17:23












  • $begingroup$
    Sorry, I used to write RR but there it does not work I didn't know how to write it. But now I will remember, thanks
    $endgroup$
    – VirtualUser
    Dec 31 '18 at 17:33
















3












$begingroup$


Proof that if $dim V = 3 $ then $X = left{ A in mathbb R^{6,5} : V subset ker(A) right}$ is linear subspace and find its $dim$



What I've done



I think that I have done proof.
$ forall vec{v} in V. Avec{v} = 0 $ and $Bvec{v}=0$

but from property of the matrix product it follows that:
$$ Avec{v} + Bvec{v} = (A+B)vec{v} $$



Moreover if $Avec{x}=0$ then $(alpha A)vec{x} =0$ and in to other side too. So ok.



My problem



But how in use of given informations find its dimension? My scribbles do not lead to anything.






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you denote $Bbb R^{6,5}$?
    $endgroup$
    – Bernard
    Dec 31 '18 at 17:23












  • $begingroup$
    Sorry, I used to write RR but there it does not work I didn't know how to write it. But now I will remember, thanks
    $endgroup$
    – VirtualUser
    Dec 31 '18 at 17:33














3












3








3





$begingroup$


Proof that if $dim V = 3 $ then $X = left{ A in mathbb R^{6,5} : V subset ker(A) right}$ is linear subspace and find its $dim$



What I've done



I think that I have done proof.
$ forall vec{v} in V. Avec{v} = 0 $ and $Bvec{v}=0$

but from property of the matrix product it follows that:
$$ Avec{v} + Bvec{v} = (A+B)vec{v} $$



Moreover if $Avec{x}=0$ then $(alpha A)vec{x} =0$ and in to other side too. So ok.



My problem



But how in use of given informations find its dimension? My scribbles do not lead to anything.






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$




Proof that if $dim V = 3 $ then $X = left{ A in mathbb R^{6,5} : V subset ker(A) right}$ is linear subspace and find its $dim$



What I've done



I think that I have done proof.
$ forall vec{v} in V. Avec{v} = 0 $ and $Bvec{v}=0$

but from property of the matrix product it follows that:
$$ Avec{v} + Bvec{v} = (A+B)vec{v} $$



Moreover if $Avec{x}=0$ then $(alpha A)vec{x} =0$ and in to other side too. So ok.



My problem



But how in use of given informations find its dimension? My scribbles do not lead to anything.






linear-algebra vector-spaces




linear-algebra vector-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 17:21









Bernard

121k740116




121k740116










asked Dec 31 '18 at 17:12









VirtualUserVirtualUser

899114




899114












  • $begingroup$
    What do you denote $Bbb R^{6,5}$?
    $endgroup$
    – Bernard
    Dec 31 '18 at 17:23












  • $begingroup$
    Sorry, I used to write RR but there it does not work I didn't know how to write it. But now I will remember, thanks
    $endgroup$
    – VirtualUser
    Dec 31 '18 at 17:33


















  • $begingroup$
    What do you denote $Bbb R^{6,5}$?
    $endgroup$
    – Bernard
    Dec 31 '18 at 17:23












  • $begingroup$
    Sorry, I used to write RR but there it does not work I didn't know how to write it. But now I will remember, thanks
    $endgroup$
    – VirtualUser
    Dec 31 '18 at 17:33
















$begingroup$
What do you denote $Bbb R^{6,5}$?
$endgroup$
– Bernard
Dec 31 '18 at 17:23






$begingroup$
What do you denote $Bbb R^{6,5}$?
$endgroup$
– Bernard
Dec 31 '18 at 17:23














$begingroup$
Sorry, I used to write RR but there it does not work I didn't know how to write it. But now I will remember, thanks
$endgroup$
– VirtualUser
Dec 31 '18 at 17:33




$begingroup$
Sorry, I used to write RR but there it does not work I didn't know how to write it. But now I will remember, thanks
$endgroup$
– VirtualUser
Dec 31 '18 at 17:33










2 Answers
2






active

oldest

votes


















1












$begingroup$

$mathbb R^{6,5}$ is a linear space of dimension $6 times 5=30$.



If $(v_1, v_2, v_3)$ (with $v_i in mathbb R^5$) is a basis of $V$. You have



$$X = {A in mathbb R^{6,5} ; A.v_1=A.v_2=A.v_3=0}.$$



It means that you're imposing $3 times 6 = 18$ independent conditions in the dual space $left(mathbb R^{6,5}right)^vee$. Hence $dim X = 30 -18=12$.



Another proof not using the dual space



Without loss of generality, you can suppose that $(v_1, dots, v_5)$ is a basis of $mathcal V = mathbb R^5$ where $(v_1, v_2,v_3)$ is a basis of $V$. And that $(v_1^prime, dots, v_6^prime)$ is a basis of $mathcal V^prime=mathbb R^6$.



For $A =(a_{ij}) in X$ expressed in those basis, you must have $$a_{11} =dots =a_{61}=a_{12}=dots=a_{62}=a_{13}=dots=a_{63}=0$$ as $V subseteq ker A$.
If we denote $E_{ij}$ the matrix with coefficient $i,j$ equal to $1$ and the others vanishing, $(E_{14}, dots, E_{64}, E_{15}, dots, E_{65})$ is a basis of $X$. Proving that the dimension of that linear subspace is equal to $12$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I don't understand this notation: $A.v_1=A.v_2=A.v_3=0$. Does $A.v_{i} $ means $i$ column? I suspect that these are not the same vectors as the vectors from base $V$?
    $endgroup$
    – VirtualUser
    Dec 31 '18 at 17:47










  • $begingroup$
    For a vector $v in V$ (with 5 coordinates) $A.v$ means the image by $A$ of vector $v$. $A.v$ is a vector with 6 coordinates.
    $endgroup$
    – mathcounterexamples.net
    Dec 31 '18 at 17:49










  • $begingroup$
    Can it be done without dual space? I haven't got this on lecture and probably I can't use that
    $endgroup$
    – VirtualUser
    Jan 2 at 19:12












  • $begingroup$
    What are the key theorems of your lecture?
    $endgroup$
    – mathcounterexamples.net
    Jan 2 at 20:35










  • $begingroup$
    image, kernel, rank, basis of linear space, sum of linear spaces linear independence and linear combinations
    $endgroup$
    – VirtualUser
    Jan 2 at 20:39



















0












$begingroup$

Since $Vsubsetker(A)forall Ain X$, the rows of $A$ are orthogonal to all vectors in $V$ and thus belong to $V^perp$, which has dimension $2$. Each row vector of $A$ is the linear combination of the two basis vectors $v_1,v_2$ of $V^perp$.



$$A=begin{bmatrix}a_{11}v_1^T+a_{12}v_2^T\a_{21}v_1^T+a_{22}v_2^T\vdots\a_{61}v_1^T+a_{62}v_2^Tend{bmatrix}=a_{11}begin{bmatrix}v_1^T\0\vdots\0end{bmatrix}+a_{12}begin{bmatrix}v_2^T\0\vdots\0end{bmatrix}+a_{21}begin{bmatrix}0\v_1^T\vdots\0end{bmatrix}+...+a_{62}begin{bmatrix}0\0\vdots\v_2^Tend{bmatrix}$$



Thus, $X$ has $12$ basis vectors.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






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    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    $mathbb R^{6,5}$ is a linear space of dimension $6 times 5=30$.



    If $(v_1, v_2, v_3)$ (with $v_i in mathbb R^5$) is a basis of $V$. You have



    $$X = {A in mathbb R^{6,5} ; A.v_1=A.v_2=A.v_3=0}.$$



    It means that you're imposing $3 times 6 = 18$ independent conditions in the dual space $left(mathbb R^{6,5}right)^vee$. Hence $dim X = 30 -18=12$.



    Another proof not using the dual space



    Without loss of generality, you can suppose that $(v_1, dots, v_5)$ is a basis of $mathcal V = mathbb R^5$ where $(v_1, v_2,v_3)$ is a basis of $V$. And that $(v_1^prime, dots, v_6^prime)$ is a basis of $mathcal V^prime=mathbb R^6$.



    For $A =(a_{ij}) in X$ expressed in those basis, you must have $$a_{11} =dots =a_{61}=a_{12}=dots=a_{62}=a_{13}=dots=a_{63}=0$$ as $V subseteq ker A$.
    If we denote $E_{ij}$ the matrix with coefficient $i,j$ equal to $1$ and the others vanishing, $(E_{14}, dots, E_{64}, E_{15}, dots, E_{65})$ is a basis of $X$. Proving that the dimension of that linear subspace is equal to $12$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I don't understand this notation: $A.v_1=A.v_2=A.v_3=0$. Does $A.v_{i} $ means $i$ column? I suspect that these are not the same vectors as the vectors from base $V$?
      $endgroup$
      – VirtualUser
      Dec 31 '18 at 17:47










    • $begingroup$
      For a vector $v in V$ (with 5 coordinates) $A.v$ means the image by $A$ of vector $v$. $A.v$ is a vector with 6 coordinates.
      $endgroup$
      – mathcounterexamples.net
      Dec 31 '18 at 17:49










    • $begingroup$
      Can it be done without dual space? I haven't got this on lecture and probably I can't use that
      $endgroup$
      – VirtualUser
      Jan 2 at 19:12












    • $begingroup$
      What are the key theorems of your lecture?
      $endgroup$
      – mathcounterexamples.net
      Jan 2 at 20:35










    • $begingroup$
      image, kernel, rank, basis of linear space, sum of linear spaces linear independence and linear combinations
      $endgroup$
      – VirtualUser
      Jan 2 at 20:39
















    1












    $begingroup$

    $mathbb R^{6,5}$ is a linear space of dimension $6 times 5=30$.



    If $(v_1, v_2, v_3)$ (with $v_i in mathbb R^5$) is a basis of $V$. You have



    $$X = {A in mathbb R^{6,5} ; A.v_1=A.v_2=A.v_3=0}.$$



    It means that you're imposing $3 times 6 = 18$ independent conditions in the dual space $left(mathbb R^{6,5}right)^vee$. Hence $dim X = 30 -18=12$.



    Another proof not using the dual space



    Without loss of generality, you can suppose that $(v_1, dots, v_5)$ is a basis of $mathcal V = mathbb R^5$ where $(v_1, v_2,v_3)$ is a basis of $V$. And that $(v_1^prime, dots, v_6^prime)$ is a basis of $mathcal V^prime=mathbb R^6$.



    For $A =(a_{ij}) in X$ expressed in those basis, you must have $$a_{11} =dots =a_{61}=a_{12}=dots=a_{62}=a_{13}=dots=a_{63}=0$$ as $V subseteq ker A$.
    If we denote $E_{ij}$ the matrix with coefficient $i,j$ equal to $1$ and the others vanishing, $(E_{14}, dots, E_{64}, E_{15}, dots, E_{65})$ is a basis of $X$. Proving that the dimension of that linear subspace is equal to $12$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I don't understand this notation: $A.v_1=A.v_2=A.v_3=0$. Does $A.v_{i} $ means $i$ column? I suspect that these are not the same vectors as the vectors from base $V$?
      $endgroup$
      – VirtualUser
      Dec 31 '18 at 17:47










    • $begingroup$
      For a vector $v in V$ (with 5 coordinates) $A.v$ means the image by $A$ of vector $v$. $A.v$ is a vector with 6 coordinates.
      $endgroup$
      – mathcounterexamples.net
      Dec 31 '18 at 17:49










    • $begingroup$
      Can it be done without dual space? I haven't got this on lecture and probably I can't use that
      $endgroup$
      – VirtualUser
      Jan 2 at 19:12












    • $begingroup$
      What are the key theorems of your lecture?
      $endgroup$
      – mathcounterexamples.net
      Jan 2 at 20:35










    • $begingroup$
      image, kernel, rank, basis of linear space, sum of linear spaces linear independence and linear combinations
      $endgroup$
      – VirtualUser
      Jan 2 at 20:39














    1












    1








    1





    $begingroup$

    $mathbb R^{6,5}$ is a linear space of dimension $6 times 5=30$.



    If $(v_1, v_2, v_3)$ (with $v_i in mathbb R^5$) is a basis of $V$. You have



    $$X = {A in mathbb R^{6,5} ; A.v_1=A.v_2=A.v_3=0}.$$



    It means that you're imposing $3 times 6 = 18$ independent conditions in the dual space $left(mathbb R^{6,5}right)^vee$. Hence $dim X = 30 -18=12$.



    Another proof not using the dual space



    Without loss of generality, you can suppose that $(v_1, dots, v_5)$ is a basis of $mathcal V = mathbb R^5$ where $(v_1, v_2,v_3)$ is a basis of $V$. And that $(v_1^prime, dots, v_6^prime)$ is a basis of $mathcal V^prime=mathbb R^6$.



    For $A =(a_{ij}) in X$ expressed in those basis, you must have $$a_{11} =dots =a_{61}=a_{12}=dots=a_{62}=a_{13}=dots=a_{63}=0$$ as $V subseteq ker A$.
    If we denote $E_{ij}$ the matrix with coefficient $i,j$ equal to $1$ and the others vanishing, $(E_{14}, dots, E_{64}, E_{15}, dots, E_{65})$ is a basis of $X$. Proving that the dimension of that linear subspace is equal to $12$.






    share|cite|improve this answer











    $endgroup$



    $mathbb R^{6,5}$ is a linear space of dimension $6 times 5=30$.



    If $(v_1, v_2, v_3)$ (with $v_i in mathbb R^5$) is a basis of $V$. You have



    $$X = {A in mathbb R^{6,5} ; A.v_1=A.v_2=A.v_3=0}.$$



    It means that you're imposing $3 times 6 = 18$ independent conditions in the dual space $left(mathbb R^{6,5}right)^vee$. Hence $dim X = 30 -18=12$.



    Another proof not using the dual space



    Without loss of generality, you can suppose that $(v_1, dots, v_5)$ is a basis of $mathcal V = mathbb R^5$ where $(v_1, v_2,v_3)$ is a basis of $V$. And that $(v_1^prime, dots, v_6^prime)$ is a basis of $mathcal V^prime=mathbb R^6$.



    For $A =(a_{ij}) in X$ expressed in those basis, you must have $$a_{11} =dots =a_{61}=a_{12}=dots=a_{62}=a_{13}=dots=a_{63}=0$$ as $V subseteq ker A$.
    If we denote $E_{ij}$ the matrix with coefficient $i,j$ equal to $1$ and the others vanishing, $(E_{14}, dots, E_{64}, E_{15}, dots, E_{65})$ is a basis of $X$. Proving that the dimension of that linear subspace is equal to $12$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 3 at 5:51

























    answered Dec 31 '18 at 17:28









    mathcounterexamples.netmathcounterexamples.net

    26.9k22157




    26.9k22157












    • $begingroup$
      I don't understand this notation: $A.v_1=A.v_2=A.v_3=0$. Does $A.v_{i} $ means $i$ column? I suspect that these are not the same vectors as the vectors from base $V$?
      $endgroup$
      – VirtualUser
      Dec 31 '18 at 17:47










    • $begingroup$
      For a vector $v in V$ (with 5 coordinates) $A.v$ means the image by $A$ of vector $v$. $A.v$ is a vector with 6 coordinates.
      $endgroup$
      – mathcounterexamples.net
      Dec 31 '18 at 17:49










    • $begingroup$
      Can it be done without dual space? I haven't got this on lecture and probably I can't use that
      $endgroup$
      – VirtualUser
      Jan 2 at 19:12












    • $begingroup$
      What are the key theorems of your lecture?
      $endgroup$
      – mathcounterexamples.net
      Jan 2 at 20:35










    • $begingroup$
      image, kernel, rank, basis of linear space, sum of linear spaces linear independence and linear combinations
      $endgroup$
      – VirtualUser
      Jan 2 at 20:39


















    • $begingroup$
      I don't understand this notation: $A.v_1=A.v_2=A.v_3=0$. Does $A.v_{i} $ means $i$ column? I suspect that these are not the same vectors as the vectors from base $V$?
      $endgroup$
      – VirtualUser
      Dec 31 '18 at 17:47










    • $begingroup$
      For a vector $v in V$ (with 5 coordinates) $A.v$ means the image by $A$ of vector $v$. $A.v$ is a vector with 6 coordinates.
      $endgroup$
      – mathcounterexamples.net
      Dec 31 '18 at 17:49










    • $begingroup$
      Can it be done without dual space? I haven't got this on lecture and probably I can't use that
      $endgroup$
      – VirtualUser
      Jan 2 at 19:12












    • $begingroup$
      What are the key theorems of your lecture?
      $endgroup$
      – mathcounterexamples.net
      Jan 2 at 20:35










    • $begingroup$
      image, kernel, rank, basis of linear space, sum of linear spaces linear independence and linear combinations
      $endgroup$
      – VirtualUser
      Jan 2 at 20:39
















    $begingroup$
    I don't understand this notation: $A.v_1=A.v_2=A.v_3=0$. Does $A.v_{i} $ means $i$ column? I suspect that these are not the same vectors as the vectors from base $V$?
    $endgroup$
    – VirtualUser
    Dec 31 '18 at 17:47




    $begingroup$
    I don't understand this notation: $A.v_1=A.v_2=A.v_3=0$. Does $A.v_{i} $ means $i$ column? I suspect that these are not the same vectors as the vectors from base $V$?
    $endgroup$
    – VirtualUser
    Dec 31 '18 at 17:47












    $begingroup$
    For a vector $v in V$ (with 5 coordinates) $A.v$ means the image by $A$ of vector $v$. $A.v$ is a vector with 6 coordinates.
    $endgroup$
    – mathcounterexamples.net
    Dec 31 '18 at 17:49




    $begingroup$
    For a vector $v in V$ (with 5 coordinates) $A.v$ means the image by $A$ of vector $v$. $A.v$ is a vector with 6 coordinates.
    $endgroup$
    – mathcounterexamples.net
    Dec 31 '18 at 17:49












    $begingroup$
    Can it be done without dual space? I haven't got this on lecture and probably I can't use that
    $endgroup$
    – VirtualUser
    Jan 2 at 19:12






    $begingroup$
    Can it be done without dual space? I haven't got this on lecture and probably I can't use that
    $endgroup$
    – VirtualUser
    Jan 2 at 19:12














    $begingroup$
    What are the key theorems of your lecture?
    $endgroup$
    – mathcounterexamples.net
    Jan 2 at 20:35




    $begingroup$
    What are the key theorems of your lecture?
    $endgroup$
    – mathcounterexamples.net
    Jan 2 at 20:35












    $begingroup$
    image, kernel, rank, basis of linear space, sum of linear spaces linear independence and linear combinations
    $endgroup$
    – VirtualUser
    Jan 2 at 20:39




    $begingroup$
    image, kernel, rank, basis of linear space, sum of linear spaces linear independence and linear combinations
    $endgroup$
    – VirtualUser
    Jan 2 at 20:39











    0












    $begingroup$

    Since $Vsubsetker(A)forall Ain X$, the rows of $A$ are orthogonal to all vectors in $V$ and thus belong to $V^perp$, which has dimension $2$. Each row vector of $A$ is the linear combination of the two basis vectors $v_1,v_2$ of $V^perp$.



    $$A=begin{bmatrix}a_{11}v_1^T+a_{12}v_2^T\a_{21}v_1^T+a_{22}v_2^T\vdots\a_{61}v_1^T+a_{62}v_2^Tend{bmatrix}=a_{11}begin{bmatrix}v_1^T\0\vdots\0end{bmatrix}+a_{12}begin{bmatrix}v_2^T\0\vdots\0end{bmatrix}+a_{21}begin{bmatrix}0\v_1^T\vdots\0end{bmatrix}+...+a_{62}begin{bmatrix}0\0\vdots\v_2^Tend{bmatrix}$$



    Thus, $X$ has $12$ basis vectors.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Since $Vsubsetker(A)forall Ain X$, the rows of $A$ are orthogonal to all vectors in $V$ and thus belong to $V^perp$, which has dimension $2$. Each row vector of $A$ is the linear combination of the two basis vectors $v_1,v_2$ of $V^perp$.



      $$A=begin{bmatrix}a_{11}v_1^T+a_{12}v_2^T\a_{21}v_1^T+a_{22}v_2^T\vdots\a_{61}v_1^T+a_{62}v_2^Tend{bmatrix}=a_{11}begin{bmatrix}v_1^T\0\vdots\0end{bmatrix}+a_{12}begin{bmatrix}v_2^T\0\vdots\0end{bmatrix}+a_{21}begin{bmatrix}0\v_1^T\vdots\0end{bmatrix}+...+a_{62}begin{bmatrix}0\0\vdots\v_2^Tend{bmatrix}$$



      Thus, $X$ has $12$ basis vectors.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Since $Vsubsetker(A)forall Ain X$, the rows of $A$ are orthogonal to all vectors in $V$ and thus belong to $V^perp$, which has dimension $2$. Each row vector of $A$ is the linear combination of the two basis vectors $v_1,v_2$ of $V^perp$.



        $$A=begin{bmatrix}a_{11}v_1^T+a_{12}v_2^T\a_{21}v_1^T+a_{22}v_2^T\vdots\a_{61}v_1^T+a_{62}v_2^Tend{bmatrix}=a_{11}begin{bmatrix}v_1^T\0\vdots\0end{bmatrix}+a_{12}begin{bmatrix}v_2^T\0\vdots\0end{bmatrix}+a_{21}begin{bmatrix}0\v_1^T\vdots\0end{bmatrix}+...+a_{62}begin{bmatrix}0\0\vdots\v_2^Tend{bmatrix}$$



        Thus, $X$ has $12$ basis vectors.






        share|cite|improve this answer









        $endgroup$



        Since $Vsubsetker(A)forall Ain X$, the rows of $A$ are orthogonal to all vectors in $V$ and thus belong to $V^perp$, which has dimension $2$. Each row vector of $A$ is the linear combination of the two basis vectors $v_1,v_2$ of $V^perp$.



        $$A=begin{bmatrix}a_{11}v_1^T+a_{12}v_2^T\a_{21}v_1^T+a_{22}v_2^T\vdots\a_{61}v_1^T+a_{62}v_2^Tend{bmatrix}=a_{11}begin{bmatrix}v_1^T\0\vdots\0end{bmatrix}+a_{12}begin{bmatrix}v_2^T\0\vdots\0end{bmatrix}+a_{21}begin{bmatrix}0\v_1^T\vdots\0end{bmatrix}+...+a_{62}begin{bmatrix}0\0\vdots\v_2^Tend{bmatrix}$$



        Thus, $X$ has $12$ basis vectors.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 12:40









        Shubham JohriShubham Johri

        5,189718




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