Let $M$ be an $n\times n$ real matrix such that $M^2+M^T=I_n$. Prove that $M$ is invertible
Here is my progress:
- Playing with determinant:
one has $\det(M^2)=\det(I_n-M^T)$ hence $\det(M)^2=\det(I_n-M)$
and $\det(M^T)=\det(I_n-M^2)$, hence $\det(M)=\det(I_n-M)\det(I_n+M)$
Combining both equalities yield $$\det(I_n-M)(\det(I_n-M)\det(I_n+M)-1)=0$$
- Playing with the original assumption:
transposing yields $(M^T)^2+M=I_n$, and combining gives $(M^2-I_n)^2=I_n-M$
that is $M^4-2M^2+M=0$.
$M$ is therefore diagonalizable and its eigenvalues lie in the set $\{0,1,-\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2}\}$
- Misc
Multiplying $M^2+M^T=I_n$ by $M$ in two different ways, one has $MM^T=M^TM$
- Looking for a contradiction ?
Supposing $M$ is not invertible, there is some $X$ such that $MX=0$. This in turn implies $M^TX=X$... So what ?