Let $X$ ,$Y$ be two $n\times n$ real matrices such that $XY=X^2+X+I$.

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).

• Can you find a $2\times 2$ matrix whose characteristic polynomial is $\lambda^{2}+\lambda +1$? Commented Apr 28 at 11:47
• @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.
– Gggg
Commented Apr 28 at 11:52
• For such a matrix $X$, $X^{2}+X+1=0$. Take $Y=0$. Commented Apr 28 at 11:54
• @geetha290krm using Cayley Hamilton theorem ??
– Gggg
Commented Apr 28 at 11:59
• Yes, that is what I meant. Commented Apr 28 at 12:05

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.

• What is $Y$ in your case? Why not just take $Y=0$? Commented Apr 28 at 11:51
• @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.
– Gggg
Commented Apr 28 at 12:05
• @DietrichBurde Infact $Y$ does not other choices except being a zero matrix.As $XY=0$ and $X$ is invertible therefore $Y$ must be zero.
– Gggg
Commented Apr 28 at 12:22
• @SandeepTiwari So $4.$ is false, because we just take $Y=0$ and this matrix $X$. Commented Apr 28 at 15:33