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Let $X$ ,$Y$ be two $n\times n$ real matrices such that $XY=X^2+X+I$.
Which of the following statements are necessarily true?

1.$X$ is invertible.
2.$X+I$ is invertible.
3.$XY=YX$.
4.$Y$ is invertible.

Since $XY=X^2+X+I\implies X(Y-X-I)=I$.Therefore $X$ is invertible.
Now as $X$ is invertible therefore $Y=X+X^{-1}+I$.
So, $YX=(X+X^{-1}+I)X=X^2+X+I=XY$. Thus option (1) and (3) are correct.
For option(2) I have a matrix $$A=\begin{bmatrix} -1&0&\\ 0&-1 \end{bmatrix}_{2\times 2}$$ Which satisfies the given condition but then $X+I$ is a zero matrix.So option (2) is incorrect.
I am unable to solve option(4).I tried to show that $Y$ is not invertible always by trying counterexamples but did not find any counterexample.(may be $Y$ is invertible).

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    $\begingroup$ Can you find a $2\times 2$ matrix whose characteristic polynomial is $\lambda^{2}+\lambda +1$? $\endgroup$
    – geetha290krm
    Commented Apr 28 at 11:47
  • $\begingroup$ @geetha290krm Yes.Let the first row of this matrix is [-2,3] and second row is [-1,1] then it will give the desired characteristic polynomial. $\endgroup$
    – Gggg
    Commented Apr 28 at 11:52
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    $\begingroup$ For such a matrix $X$, $X^{2}+X+1=0$. Take $Y=0$. $\endgroup$
    – geetha290krm
    Commented Apr 28 at 11:54
  • $\begingroup$ @geetha290krm using Cayley Hamilton theorem ?? $\endgroup$
    – Gggg
    Commented Apr 28 at 11:59
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    $\begingroup$ Yes, that is what I meant. $\endgroup$
    – geetha290krm
    Commented Apr 28 at 12:05

1 Answer 1

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Consider $X=\left(\begin{matrix}-1&1\\-1&0\end{matrix}\right)$. It's easy to verify that $XY=X^2+X+I=0$. Since $X$ is invertible, then $Y$ cannot be invertible.

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    $\begingroup$ What is $Y$ in your case? Why not just take $Y=0$? $\endgroup$
    – Dietrich Burde
    Commented Apr 28 at 11:51
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    $\begingroup$ @DietrichBurde If $X$ and $Y$ are both invertible then $XY$ is also invertible.But in this case $XY=0$ and $X$ is invertible so $Y$ must not be invertible. $\endgroup$
    – Gggg
    Commented Apr 28 at 12:05
  • $\begingroup$ @DietrichBurde Infact $Y$ does not other choices except being a zero matrix.As $XY=0$ and $X$ is invertible therefore $Y$ must be zero. $\endgroup$
    – Gggg
    Commented Apr 28 at 12:22
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    $\begingroup$ @SandeepTiwari So $4.$ is false, because we just take $Y=0$ and this matrix $X$. $\endgroup$
    – Dietrich Burde
    Commented Apr 28 at 15:33
  • $\begingroup$ @DietrichBurde Yes sir.Thank you for your valuable comments. $\endgroup$
    – Gggg
    Commented Apr 28 at 16:17

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