2
$\begingroup$

I have two square matrices $X$ and $Y$ of the order of $n \times n$. It is given that $X = I - XY$. I have to show that $X$ is invertible and that $XY = YX$.

The first part is fairly easy, moving stuff around gets me to the point that $X(I + Y) = I$ and therefore $X$ multiplied by another matrix yields the identity matrix and therefore $X$ is Invertible.

I am stuck at the second part which requires showing that $XY = YX$.

I tried multiplying $(I-XY)B$ but it got me nowhere.

$\endgroup$
4
$\begingroup$

Remark: Suppose that $AB=I$; then $BA=I=AB$.



As you have mentioned $X(I+Y)=I$.

Considering the above remark we can write

$$X(I+Y)=I=(I+Y)X;$$

now expanding both sides, one gets that:

$$X+XY=X+YX \qquad \Longrightarrow \qquad XY=YX.$$

$\endgroup$
2
$\begingroup$

Since\begin{align}X&=\operatorname{Id}-XY\\&=XX^{-1}-XY\\&=X(X^{-1}-Y)\end{align}and since $X$ is invertible, $X^{-1}-Y=\operatorname{Id}$, and this means that $Y=X^{-1}-\operatorname{Id}$. Therefore, $XY=YX$, since obviously $X$ and $X^{-1}-\operatorname{Id}$ commute (both $X(X^{-1}-\operatorname{Id})$ and $(X^{-1}-\operatorname{Id})X$ are equal to $\operatorname{Id}-X$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.