Prove that function is bijective Let $n \in \mathbb{N} \setminus \{ 0 \}  $ and $A \in M_n(\mathbb{R})$ with $m \in \mathbb{N} \setminus \{ 0 \}$ as $A^m= \alpha \times I_n$, with $ \alpha \in \mathbb{R} \setminus \{ -1,1 \}$. 
Prove that $f: M_n(\mathbb{R})\to M_n(\mathbb{R}), f(X)=X+AXA$ is bijective.
I honestly don't have any idea how should I prove this, I'd be grateful for any help. I found the problem in the archives of a contest, but no hints or solving is provided.
 A: Suppose $f(X)=0$. Then $AXA=-X$. Hence show that $A^mXA^m=(-1)^mX$. Yet by the property of $A$, we have $A^mXA^m=\alpha^2X$. Argue that $X=0$ and $f$ is injective. Now, why does it follow that $f$ is bijective?
A: Oops, missed the homework tag first time around.
First find a formula for $\det f$ in terms of the eigenvalues of $A$.

First, suppose that $A$ has a full set of eigenvectors. Let $A u_k = \lambda u_k $ and let $v_k^* A = \lambda_k v_k^*$. Then it is straightforward to show that $f(u_j v_j^*) = (1+ \lambda_i {\lambda_j})u_j v_j^*$, from which we obtain $\det f = \prod_{i,j} (1+\lambda_i {\lambda_j})$. It follows by a continuity argument that this formula holds for arbitrary $A$.

Second, show that $\det f \neq 0$.

Since $A^m = \alpha I$, it follows that for any eigenvalue, we have $\lambda^m = \alpha$, hence $|\lambda| = \sqrt[m]{|\alpha|}$. Hence $|\lambda_i \lambda_k| = (\sqrt[m]{|\alpha|})^2 \neq 1$, and hence $(1+\lambda_i {\lambda_j}) \neq 0$ for all $i,j$. So we have $\det f \neq 0$. Hence $f$ is bijective.

