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

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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?



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    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.










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      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






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      asked Sep 10 at 18:55









      Jeff

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          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










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          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

















           

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