Let $M\in M(n,\Bbb{R})$ such that $M^3=I$ and $Mv\ne v$ for any non-zero vector $v$. Prove that $n$ is even.

Question

I have tried the problem in the following manner but I haven't got any solution.

Let $\lambda$ be an eigenvalue of $M$, then $\exists v\ne 0$ such that $Mv=\lambda v\implies M^3 v=\lambda^3v\implies Iv=\lambda^3v$
$\implies\lambda^3$ is an eigenvalue of the $n\times n$ identity matrix $I\implies |I-\lambda^3I|=0\implies(1-\lambda^3)^n=0$ $\implies(\lambda^2+\lambda+1)^n=0$ since $\lambda\ne 1$, given that $Mv\ne v$.

Now, from this I can conclude nothing?
Can anybody derive a solution for this problem?

Answer 1

If $n$ is odd, then $M$ has some real eigenvalue $\lambda$. Since $M^3=\operatorname{Id}_3$, $\lambda^3=1$, which is equivalent to the assertion that $\lambda=1$. So, if $v$ is an eigenvector of $M$ with eigenvalue $1$, $M.v=\lambda v=v$.

Answer 2

By the hypotheses,

• $M$ is invertible with inverse $M^{-1}=M^2$,
• $M-I\,$ is invertible as $1$ is not an eigenvalue.

This yields $$0\:=\:M^3-I \:=\: (M-I)(M^2+M+I)\:\: \implies M^{-1}+M+I = 0$$ thus eigenvalues of $M$ must satisfy $$\frac 1\lambda + \lambda\:=\: -1\quad\Longleftrightarrow\quad\lambda\:=\: \frac{-1\pm\sqrt 3i}2 \:=\: \exp\left(\pm\, i\frac{2\pi} 3\right)\,. $$ Hence $M\,$ has no real eigenvalues, and because for real matrices non-real eigenvalues necessarily show up as conjugate-complex pairs, the matrix size $n\,$ cannot be odd.