Find all real valued $ntimes n$ matrices $A$ such that for any $vinmathbbR^n$ there exists $yinmathbbR^n$ with $Av=v+y$

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












I've started with the case $n=2$ to help myself gain a general idea, but I don't really understand where to continue. Consider $A=beginbmatrix a_1 & a_2 \ a_3 & a_4 endbmatrix$, $v^T=(v_1,v_2)$ and $y^T=(y_1,y_2)$. Then if $Av=v+y$, solving for the components of $y$ we have $y_1=a_1v_1+a_4v_2-v_2$ and $y_2=a_3v_1+a_4v_2-v_2$. So, is this not true for any $2times 2$ matrix $A$, if I define $y$ in this way then I am done?



Thanks in advance.










share|cite|improve this question

























    up vote
    0
    down vote

    favorite












    I've started with the case $n=2$ to help myself gain a general idea, but I don't really understand where to continue. Consider $A=beginbmatrix a_1 & a_2 \ a_3 & a_4 endbmatrix$, $v^T=(v_1,v_2)$ and $y^T=(y_1,y_2)$. Then if $Av=v+y$, solving for the components of $y$ we have $y_1=a_1v_1+a_4v_2-v_2$ and $y_2=a_3v_1+a_4v_2-v_2$. So, is this not true for any $2times 2$ matrix $A$, if I define $y$ in this way then I am done?



    Thanks in advance.










    share|cite|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I've started with the case $n=2$ to help myself gain a general idea, but I don't really understand where to continue. Consider $A=beginbmatrix a_1 & a_2 \ a_3 & a_4 endbmatrix$, $v^T=(v_1,v_2)$ and $y^T=(y_1,y_2)$. Then if $Av=v+y$, solving for the components of $y$ we have $y_1=a_1v_1+a_4v_2-v_2$ and $y_2=a_3v_1+a_4v_2-v_2$. So, is this not true for any $2times 2$ matrix $A$, if I define $y$ in this way then I am done?



      Thanks in advance.










      share|cite|improve this question













      I've started with the case $n=2$ to help myself gain a general idea, but I don't really understand where to continue. Consider $A=beginbmatrix a_1 & a_2 \ a_3 & a_4 endbmatrix$, $v^T=(v_1,v_2)$ and $y^T=(y_1,y_2)$. Then if $Av=v+y$, solving for the components of $y$ we have $y_1=a_1v_1+a_4v_2-v_2$ and $y_2=a_3v_1+a_4v_2-v_2$. So, is this not true for any $2times 2$ matrix $A$, if I define $y$ in this way then I am done?



      Thanks in advance.







      linear-algebra matrices






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Sep 10 at 18:55









      Jeff

      1449




      1449




















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          2
          down vote



          accepted










          Isn't there something missing from the question? Some condition, maybe?
          Every real valued matrix is good: let $y= Av-v$.






          share|cite|improve this answer




















          • No, I took it straight off of an old qualifying exam. I guess it is just that simple?
            – Jeff
            Sep 10 at 19:06










          • It is possible. Exam problems are often not tricky, they just test if you understand some basic notion or theorem. So yes, I guess it is just that simple.
            – A. Pongrácz
            Sep 10 at 19:11










          • Or maybe there was a typo on the exam and they didn't release the corrected version. Oh well, thanks.
            – Jeff
            Sep 10 at 19:12










          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: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          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%2f2912230%2ffind-all-real-valued-n-times-n-matrices-a-such-that-for-any-v-in-mathbbr%23new-answer', 'question_page');

          );

          Post as a guest






























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          Isn't there something missing from the question? Some condition, maybe?
          Every real valued matrix is good: let $y= Av-v$.






          share|cite|improve this answer




















          • No, I took it straight off of an old qualifying exam. I guess it is just that simple?
            – Jeff
            Sep 10 at 19:06










          • It is possible. Exam problems are often not tricky, they just test if you understand some basic notion or theorem. So yes, I guess it is just that simple.
            – A. Pongrácz
            Sep 10 at 19:11










          • Or maybe there was a typo on the exam and they didn't release the corrected version. Oh well, thanks.
            – Jeff
            Sep 10 at 19:12














          up vote
          2
          down vote



          accepted










          Isn't there something missing from the question? Some condition, maybe?
          Every real valued matrix is good: let $y= Av-v$.






          share|cite|improve this answer




















          • No, I took it straight off of an old qualifying exam. I guess it is just that simple?
            – Jeff
            Sep 10 at 19:06










          • It is possible. Exam problems are often not tricky, they just test if you understand some basic notion or theorem. So yes, I guess it is just that simple.
            – A. Pongrácz
            Sep 10 at 19:11










          • Or maybe there was a typo on the exam and they didn't release the corrected version. Oh well, thanks.
            – Jeff
            Sep 10 at 19:12












          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          Isn't there something missing from the question? Some condition, maybe?
          Every real valued matrix is good: let $y= Av-v$.






          share|cite|improve this answer












          Isn't there something missing from the question? Some condition, maybe?
          Every real valued matrix is good: let $y= Av-v$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 10 at 19:04









          A. Pongrácz

          4,465725




          4,465725











          • No, I took it straight off of an old qualifying exam. I guess it is just that simple?
            – Jeff
            Sep 10 at 19:06










          • It is possible. Exam problems are often not tricky, they just test if you understand some basic notion or theorem. So yes, I guess it is just that simple.
            – A. Pongrácz
            Sep 10 at 19:11










          • Or maybe there was a typo on the exam and they didn't release the corrected version. Oh well, thanks.
            – Jeff
            Sep 10 at 19:12
















          • No, I took it straight off of an old qualifying exam. I guess it is just that simple?
            – Jeff
            Sep 10 at 19:06










          • It is possible. Exam problems are often not tricky, they just test if you understand some basic notion or theorem. So yes, I guess it is just that simple.
            – A. Pongrácz
            Sep 10 at 19:11










          • Or maybe there was a typo on the exam and they didn't release the corrected version. Oh well, thanks.
            – Jeff
            Sep 10 at 19:12















          No, I took it straight off of an old qualifying exam. I guess it is just that simple?
          – Jeff
          Sep 10 at 19:06




          No, I took it straight off of an old qualifying exam. I guess it is just that simple?
          – Jeff
          Sep 10 at 19:06












          It is possible. Exam problems are often not tricky, they just test if you understand some basic notion or theorem. So yes, I guess it is just that simple.
          – A. Pongrácz
          Sep 10 at 19:11




          It is possible. Exam problems are often not tricky, they just test if you understand some basic notion or theorem. So yes, I guess it is just that simple.
          – A. Pongrácz
          Sep 10 at 19:11












          Or maybe there was a typo on the exam and they didn't release the corrected version. Oh well, thanks.
          – Jeff
          Sep 10 at 19:12




          Or maybe there was a typo on the exam and they didn't release the corrected version. Oh well, thanks.
          – Jeff
          Sep 10 at 19:12

















           

          draft saved


          draft discarded















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2912230%2ffind-all-real-valued-n-times-n-matrices-a-such-that-for-any-v-in-mathbbr%23new-answer', 'question_page');

          );

          Post as a guest













































































          這個網誌中的熱門文章

          How to combine Bézier curves to a surface?

          Why am i infinitely getting the same tweet with the Twitter Search API?

          Carbon dioxide