# Endomorphism $\phi: M_{n}(\mathbb{C}) \rightarrow M_{n}(\mathbb{C})$ that stabilizes $GL_{n}(\mathbb{C}).$

I've been given this as a homework assignment, and have no idea how to proceed. Can anyone help? The question is:

(1) Let $$\phi: M_{n}(\mathbb{C}) \rightarrow M_{n}(\mathbb{C})$$ be an endomorphism such that $$M \in \operatorname{GL}_{n}(\mathbb{C}) \implies \phi (M) \in \operatorname{GL}_{n}(\mathbb{C}).$$ Show that, for any $$M \in \operatorname{GL}_{n}(\mathbb{C})$$, we have $$M \in \operatorname{GL}_{n}(\mathbb{C}) \iff \phi (M) \in \operatorname{GL}_{n}(\mathbb{C}).$$

For this problem, we received a hint from the professor:

(2) If rank$$(M) < n,$$ then there exists $$P \in \operatorname{GL}_{n}(\mathbb{C})$$ such that, for any $$\lambda \in \mathbb{C}$$, $$P - \lambda M$$ is invertible.

I don't know how to prove (1) nor (2). Although my main goal is to prove (1), any help with proving (2) would be appreciated.

You have to show that $$\phi(M)$$ is invertible implies that $$M$$ is invertible.
Suppose that $$M$$ is not invertible, there exists $$P$$ invertible such that for every $$\lambda, P-\lambda M$$ is invertible this implies that $$\phi(P)-\lambda\phi(M)$$ is invertible and $$I-\lambda \phi(P)^{-1}\phi(M)$$ invertible for every $$\lambda$$. The matrix $$\phi(P)^{-1}\phi(M)$$ is invertible, it has an eigenvalue $$c$$, $$I-{1\over c}\phi(P)^{-1}\phi(M)$$ is not invertible. Contradiction.
• Because $\phi(P)$ is invertible since $P$ is invertible and it is assumed that $\phi(M)$ is invertible. – Tsemo Aristide Mar 23 at 14:07
(2) seems false: $$det(P-\lambda M)$$ is a polynomial in $$\lambda$$, so it has a root.
(2) (the hint) Suppose $$r=rank(M), so there exists $$A,B\in GL_n(\mathbb{C})$$ such that $$M=AN_rB,$$ where $$N_r=\left[\begin{array}{cc}0&I_r\\0&0\end{array}\right]$$. Defines $$P=AB,$$ with $$A,B$$ as above. So, for any $$\lambda\in\mathbb{C}$$, we have $$P-\lambda M=AB-\lambda AN_rB=A(I-\lambda N_r)B.$$ Now we just need to see that $$(I-\lambda N_r)\in GL_n(\mathbb{C})$$. But this is evident, because $$(I-\lambda N_r)$$ is upper triangular matrix with all diagonal entries equal to $$1$$, so its determinant equals to $$1$$ and this gives us $$(I-\lambda N_r)\in Gl_n(\mathbb{C})$$. Finally, $$(P-\lambda M)\in GL_n(\mathbb{C})$$.