Proof that $exists x in C$ that $A - xcdot I_n$ is invertible












2












$begingroup$


We know that $A in mathbb C^{n,n}$

Proof that $exists x in C$ that $A - xcdot I_n$ is invertible. How many such x exists?

I am thinking about this problem and generally I have only few little concepts...
We should consider two cases:

1. $A$ is reversible.

Then we can put $ x = 0 $ and $$ A - xcdot I_n = A $$ so it works.


2. $A$ is not invertible - and there I have problem. If A is invertible It means that $det_n A neq 0$. Ok, but does it give me something? We can take such $x$ to make at least one element to 0 on diagonal - But if we have $ 0 $ on diagonal, It does not guarantee that $A - xcdot I_n = A$ will be invertible... thanks for your time






linear-algebra matrices inverse







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What is the determinant of $A - xcdot I_n$?
    $endgroup$
    – user3482749
    Dec 31 '18 at 17:01










  • $begingroup$
    On my lecture, I was only proof that Cauchy theorem about determinants - I have nothing about sum of matrices - unless you mean something trivial, which I do not see at the moment
    $endgroup$
    – VirtualUser
    Dec 31 '18 at 17:03










  • $begingroup$
    If $A$ is not invertible (i.e., singular) then its determinant is zero.
    $endgroup$
    – Sean Roberson
    Dec 31 '18 at 17:08










  • $begingroup$
    Yes, I mean something very simple. Just write out the determinant formula in full, if you need to do so to see it.
    $endgroup$
    – user3482749
    Dec 31 '18 at 17:39
















2












$begingroup$


We know that $A in mathbb C^{n,n}$

Proof that $exists x in C$ that $A - xcdot I_n$ is invertible. How many such x exists?

I am thinking about this problem and generally I have only few little concepts...
We should consider two cases:

1. $A$ is reversible.

Then we can put $ x = 0 $ and $$ A - xcdot I_n = A $$ so it works.


2. $A$ is not invertible - and there I have problem. If A is invertible It means that $det_n A neq 0$. Ok, but does it give me something? We can take such $x$ to make at least one element to 0 on diagonal - But if we have $ 0 $ on diagonal, It does not guarantee that $A - xcdot I_n = A$ will be invertible... thanks for your time






linear-algebra matrices inverse







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What is the determinant of $A - xcdot I_n$?
    $endgroup$
    – user3482749
    Dec 31 '18 at 17:01










  • $begingroup$
    On my lecture, I was only proof that Cauchy theorem about determinants - I have nothing about sum of matrices - unless you mean something trivial, which I do not see at the moment
    $endgroup$
    – VirtualUser
    Dec 31 '18 at 17:03










  • $begingroup$
    If $A$ is not invertible (i.e., singular) then its determinant is zero.
    $endgroup$
    – Sean Roberson
    Dec 31 '18 at 17:08










  • $begingroup$
    Yes, I mean something very simple. Just write out the determinant formula in full, if you need to do so to see it.
    $endgroup$
    – user3482749
    Dec 31 '18 at 17:39














2












2








2





$begingroup$


We know that $A in mathbb C^{n,n}$

Proof that $exists x in C$ that $A - xcdot I_n$ is invertible. How many such x exists?

I am thinking about this problem and generally I have only few little concepts...
We should consider two cases:

1. $A$ is reversible.

Then we can put $ x = 0 $ and $$ A - xcdot I_n = A $$ so it works.


2. $A$ is not invertible - and there I have problem. If A is invertible It means that $det_n A neq 0$. Ok, but does it give me something? We can take such $x$ to make at least one element to 0 on diagonal - But if we have $ 0 $ on diagonal, It does not guarantee that $A - xcdot I_n = A$ will be invertible... thanks for your time






linear-algebra matrices inverse







share|cite|improve this question











$endgroup$




We know that $A in mathbb C^{n,n}$

Proof that $exists x in C$ that $A - xcdot I_n$ is invertible. How many such x exists?

I am thinking about this problem and generally I have only few little concepts...
We should consider two cases:

1. $A$ is reversible.

Then we can put $ x = 0 $ and $$ A - xcdot I_n = A $$ so it works.


2. $A$ is not invertible - and there I have problem. If A is invertible It means that $det_n A neq 0$. Ok, but does it give me something? We can take such $x$ to make at least one element to 0 on diagonal - But if we have $ 0 $ on diagonal, It does not guarantee that $A - xcdot I_n = A$ will be invertible... thanks for your time






linear-algebra matrices inverse




linear-algebra matrices inverse






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 17:08









José Carlos Santos

164k22131234




164k22131234










asked Dec 31 '18 at 17:00









VirtualUserVirtualUser

899114




899114








  • 1




    $begingroup$
    What is the determinant of $A - xcdot I_n$?
    $endgroup$
    – user3482749
    Dec 31 '18 at 17:01










  • $begingroup$
    On my lecture, I was only proof that Cauchy theorem about determinants - I have nothing about sum of matrices - unless you mean something trivial, which I do not see at the moment
    $endgroup$
    – VirtualUser
    Dec 31 '18 at 17:03










  • $begingroup$
    If $A$ is not invertible (i.e., singular) then its determinant is zero.
    $endgroup$
    – Sean Roberson
    Dec 31 '18 at 17:08










  • $begingroup$
    Yes, I mean something very simple. Just write out the determinant formula in full, if you need to do so to see it.
    $endgroup$
    – user3482749
    Dec 31 '18 at 17:39














  • 1




    $begingroup$
    What is the determinant of $A - xcdot I_n$?
    $endgroup$
    – user3482749
    Dec 31 '18 at 17:01










  • $begingroup$
    On my lecture, I was only proof that Cauchy theorem about determinants - I have nothing about sum of matrices - unless you mean something trivial, which I do not see at the moment
    $endgroup$
    – VirtualUser
    Dec 31 '18 at 17:03










  • $begingroup$
    If $A$ is not invertible (i.e., singular) then its determinant is zero.
    $endgroup$
    – Sean Roberson
    Dec 31 '18 at 17:08










  • $begingroup$
    Yes, I mean something very simple. Just write out the determinant formula in full, if you need to do so to see it.
    $endgroup$
    – user3482749
    Dec 31 '18 at 17:39








1




1




$begingroup$
What is the determinant of $A - xcdot I_n$?
$endgroup$
– user3482749
Dec 31 '18 at 17:01




$begingroup$
What is the determinant of $A - xcdot I_n$?
$endgroup$
– user3482749
Dec 31 '18 at 17:01












$begingroup$
On my lecture, I was only proof that Cauchy theorem about determinants - I have nothing about sum of matrices - unless you mean something trivial, which I do not see at the moment
$endgroup$
– VirtualUser
Dec 31 '18 at 17:03




$begingroup$
On my lecture, I was only proof that Cauchy theorem about determinants - I have nothing about sum of matrices - unless you mean something trivial, which I do not see at the moment
$endgroup$
– VirtualUser
Dec 31 '18 at 17:03












$begingroup$
If $A$ is not invertible (i.e., singular) then its determinant is zero.
$endgroup$
– Sean Roberson
Dec 31 '18 at 17:08




$begingroup$
If $A$ is not invertible (i.e., singular) then its determinant is zero.
$endgroup$
– Sean Roberson
Dec 31 '18 at 17:08












$begingroup$
Yes, I mean something very simple. Just write out the determinant formula in full, if you need to do so to see it.
$endgroup$
– user3482749
Dec 31 '18 at 17:39




$begingroup$
Yes, I mean something very simple. Just write out the determinant formula in full, if you need to do so to see it.
$endgroup$
– user3482749
Dec 31 '18 at 17:39










2 Answers
2






active

oldest

votes


















2












$begingroup$

Suppose $A-xI_n$ is not invertible. Then there exists $v_xne0$ such that $(A-xI_n)v_x=0$, which means $Av_x=xv_x$.



Suppose $xne y$ and that both $A-xI_n$ and $A-yI_n$ are not invertible. Suppose also
$$
av_x+bv_y=0
$$
Then, multiplying by $A$, we get $aAv_x+bAv_y=0$, that is, $axv_x+byv_y=0$. If we instead multiply by $y$, we obtain $ayv_x+byv_y=0$. Subtracting the two relations, we obtain
$$
axv_x-ayv_x=0
$$
and since $xne y$, we conclude $a=0$ and therefore $b=0$. In other words, $v_x$ and $v_y$ are linearly independent.



This can be generalized: suppose we have $x_1,x_2,dots,x_m$ pairwise distinct such that $A-x_kI_n$ is not invertible. With induction based on the same technique as before we conclude that the set
$$
{v_{x_1},v_{x_2},dots,v_{x_m}}
$$
is linearly independent. Since the vectors belong to $mathbb{C}^n$, we see that $mle n$.



Hence $A-xI_n$ can be non invertible for at most $n$ distinct values of $x$.






share|cite|improve this answer









$endgroup$





















    6












    $begingroup$

    The matrix $A-xoperatorname{Id}$ is invertible if and only if $det(A-xoperatorname{Id})neq0$. But $det(A-xoperatorname{Id})$ is a polynomial function of degree $n$ (in fact, it's the characteristic polynomial of $A$) and therefore it has at most $n$ zeros (and at least one). For all other values of $x$, the matrix is invertible.






    share|cite|improve this answer











    $endgroup$













      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',
      autoActivateHeartbeat: false,
      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%2f3057865%2fproof-that-exists-x-in-c-that-a-x-cdot-i-n-is-invertible%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Suppose $A-xI_n$ is not invertible. Then there exists $v_xne0$ such that $(A-xI_n)v_x=0$, which means $Av_x=xv_x$.



      Suppose $xne y$ and that both $A-xI_n$ and $A-yI_n$ are not invertible. Suppose also
      $$
      av_x+bv_y=0
      $$
      Then, multiplying by $A$, we get $aAv_x+bAv_y=0$, that is, $axv_x+byv_y=0$. If we instead multiply by $y$, we obtain $ayv_x+byv_y=0$. Subtracting the two relations, we obtain
      $$
      axv_x-ayv_x=0
      $$
      and since $xne y$, we conclude $a=0$ and therefore $b=0$. In other words, $v_x$ and $v_y$ are linearly independent.



      This can be generalized: suppose we have $x_1,x_2,dots,x_m$ pairwise distinct such that $A-x_kI_n$ is not invertible. With induction based on the same technique as before we conclude that the set
      $$
      {v_{x_1},v_{x_2},dots,v_{x_m}}
      $$
      is linearly independent. Since the vectors belong to $mathbb{C}^n$, we see that $mle n$.



      Hence $A-xI_n$ can be non invertible for at most $n$ distinct values of $x$.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Suppose $A-xI_n$ is not invertible. Then there exists $v_xne0$ such that $(A-xI_n)v_x=0$, which means $Av_x=xv_x$.



        Suppose $xne y$ and that both $A-xI_n$ and $A-yI_n$ are not invertible. Suppose also
        $$
        av_x+bv_y=0
        $$
        Then, multiplying by $A$, we get $aAv_x+bAv_y=0$, that is, $axv_x+byv_y=0$. If we instead multiply by $y$, we obtain $ayv_x+byv_y=0$. Subtracting the two relations, we obtain
        $$
        axv_x-ayv_x=0
        $$
        and since $xne y$, we conclude $a=0$ and therefore $b=0$. In other words, $v_x$ and $v_y$ are linearly independent.



        This can be generalized: suppose we have $x_1,x_2,dots,x_m$ pairwise distinct such that $A-x_kI_n$ is not invertible. With induction based on the same technique as before we conclude that the set
        $$
        {v_{x_1},v_{x_2},dots,v_{x_m}}
        $$
        is linearly independent. Since the vectors belong to $mathbb{C}^n$, we see that $mle n$.



        Hence $A-xI_n$ can be non invertible for at most $n$ distinct values of $x$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Suppose $A-xI_n$ is not invertible. Then there exists $v_xne0$ such that $(A-xI_n)v_x=0$, which means $Av_x=xv_x$.



          Suppose $xne y$ and that both $A-xI_n$ and $A-yI_n$ are not invertible. Suppose also
          $$
          av_x+bv_y=0
          $$
          Then, multiplying by $A$, we get $aAv_x+bAv_y=0$, that is, $axv_x+byv_y=0$. If we instead multiply by $y$, we obtain $ayv_x+byv_y=0$. Subtracting the two relations, we obtain
          $$
          axv_x-ayv_x=0
          $$
          and since $xne y$, we conclude $a=0$ and therefore $b=0$. In other words, $v_x$ and $v_y$ are linearly independent.



          This can be generalized: suppose we have $x_1,x_2,dots,x_m$ pairwise distinct such that $A-x_kI_n$ is not invertible. With induction based on the same technique as before we conclude that the set
          $$
          {v_{x_1},v_{x_2},dots,v_{x_m}}
          $$
          is linearly independent. Since the vectors belong to $mathbb{C}^n$, we see that $mle n$.



          Hence $A-xI_n$ can be non invertible for at most $n$ distinct values of $x$.






          share|cite|improve this answer









          $endgroup$



          Suppose $A-xI_n$ is not invertible. Then there exists $v_xne0$ such that $(A-xI_n)v_x=0$, which means $Av_x=xv_x$.



          Suppose $xne y$ and that both $A-xI_n$ and $A-yI_n$ are not invertible. Suppose also
          $$
          av_x+bv_y=0
          $$
          Then, multiplying by $A$, we get $aAv_x+bAv_y=0$, that is, $axv_x+byv_y=0$. If we instead multiply by $y$, we obtain $ayv_x+byv_y=0$. Subtracting the two relations, we obtain
          $$
          axv_x-ayv_x=0
          $$
          and since $xne y$, we conclude $a=0$ and therefore $b=0$. In other words, $v_x$ and $v_y$ are linearly independent.



          This can be generalized: suppose we have $x_1,x_2,dots,x_m$ pairwise distinct such that $A-x_kI_n$ is not invertible. With induction based on the same technique as before we conclude that the set
          $$
          {v_{x_1},v_{x_2},dots,v_{x_m}}
          $$
          is linearly independent. Since the vectors belong to $mathbb{C}^n$, we see that $mle n$.



          Hence $A-xI_n$ can be non invertible for at most $n$ distinct values of $x$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 31 '18 at 17:17









          egregegreg

          183k1486205




          183k1486205























              6












              $begingroup$

              The matrix $A-xoperatorname{Id}$ is invertible if and only if $det(A-xoperatorname{Id})neq0$. But $det(A-xoperatorname{Id})$ is a polynomial function of degree $n$ (in fact, it's the characteristic polynomial of $A$) and therefore it has at most $n$ zeros (and at least one). For all other values of $x$, the matrix is invertible.






              share|cite|improve this answer











              $endgroup$


















                6












                $begingroup$

                The matrix $A-xoperatorname{Id}$ is invertible if and only if $det(A-xoperatorname{Id})neq0$. But $det(A-xoperatorname{Id})$ is a polynomial function of degree $n$ (in fact, it's the characteristic polynomial of $A$) and therefore it has at most $n$ zeros (and at least one). For all other values of $x$, the matrix is invertible.






                share|cite|improve this answer











                $endgroup$
















                  6












                  6








                  6





                  $begingroup$

                  The matrix $A-xoperatorname{Id}$ is invertible if and only if $det(A-xoperatorname{Id})neq0$. But $det(A-xoperatorname{Id})$ is a polynomial function of degree $n$ (in fact, it's the characteristic polynomial of $A$) and therefore it has at most $n$ zeros (and at least one). For all other values of $x$, the matrix is invertible.






                  share|cite|improve this answer











                  $endgroup$



                  The matrix $A-xoperatorname{Id}$ is invertible if and only if $det(A-xoperatorname{Id})neq0$. But $det(A-xoperatorname{Id})$ is a polynomial function of degree $n$ (in fact, it's the characteristic polynomial of $A$) and therefore it has at most $n$ zeros (and at least one). For all other values of $x$, the matrix is invertible.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 31 '18 at 17:10









                  Jakobian

                  2,650721




                  2,650721










                  answered Dec 31 '18 at 17:06









                  José Carlos SantosJosé Carlos Santos

                  164k22131234




                  164k22131234






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057865%2fproof-that-exists-x-in-c-that-a-x-cdot-i-n-is-invertible%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)

                      Image
                      Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                      Read more

                      Quarter-circle Tiles

                      Image
                      up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                      Read more

                      Mont Emei

                      Image
                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                      Read more