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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.

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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.

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  • $\begingroup$ Because $\phi(P)$ is invertible since $P$ is invertible and it is assumed that $\phi(M)$ is invertible. $\endgroup$ – Tsemo Aristide Mar 23 at 14:07
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(2) seems false: $det(P-\lambda M)$ is a polynomial in $\lambda$, so it has a root.

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    $\begingroup$ Maybe it can be a nonzero constant polynomial. $\endgroup$ – mechanodroid Mar 23 at 14:39
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(2) (the hint) Suppose $r=rank(M)<n$, 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, $(P-\lambda M)\in GL_n(\mathbb{C})$.

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