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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 $A=\begin{bmatrix} a_1 & a_2 \\ a_3 & a_4 \end{bmatrix}$, $v^T=(v_1,v_2)$ 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 $2\times 2$ matrix $A$, if I define $y$ in this way then I am done?

Thanks in advance.

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

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  • $\begingroup$ No, I took it straight off of an old qualifying exam. I guess it is just that simple? $\endgroup$ – Gengar Sep 10 '18 at 19:06
  • $\begingroup$ 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. $\endgroup$ – A. Pongrácz Sep 10 '18 at 19:11
  • $\begingroup$ Or maybe there was a typo on the exam and they didn't release the corrected version. Oh well, thanks. $\endgroup$ – Gengar Sep 10 '18 at 19:12

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