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Let $M$ and $N$ be square matrices such that $M^2+N^2=MN$. Then prove that $MN-NM$ is singular.

So basically I have to prove: $\det(MN-NM)=0$. I tried to prove this by multiplying the given condition by the inverse of matrices $M$ and $N$, but could not come to the answer. Could anyone please give a hint?

user26857
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Schrodinger
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1 Answers1

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We work in $M_n(\mathbb{C})$.

We put $M=uX+vY,N=wX+tY$ and we seek the complex numbers $u,v,w,t$ so that $ut-vw\not= 0$ and the coefficients of $X^2,Y^2$ are $0$.

We obtain, for example, $M=X+Y,N=-jX-j^2Y$, where $j=\exp(2i\pi/3)$, and $XY=j^2YX$.

Therefore $MN-NM=(j-j^2)(XY-YX)$ and $\det(MN-NM)=0$ iff $\det(XY-YX)=0$ iff $\det(YX)=0$.

Since $\det(XY)=j^{2n}\det(YX)$, there are $2$ cases

i) $n$ is not a multiple of $3$. Then $\det(XY)=0$ and we are done.

ii) $n$ is a multiple of $3$. Then, over $\mathbb{C}$, the required result is false

Example for $n=3$: $X=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix},Y=\begin{pmatrix}0&1&0\\0&0&j\\j^2 &0&0\end{pmatrix}$.