Hermitian matrix such that $4M^5+2M^3+M=7I_n$

$n$ is a positive integer. Besides the identity matrix $I_n$, does there exist other $n\times n$ Hermitian matrix $M$, such that the following equality

$$4M^5+2M^3+M=7I_n $$

hold?

I try this: Since $4M^5+2M^3+M=7I_n$, then

$$(M-I_n)(4M^4+4M^3+6M^2+6M+7I_n)=0$$

but, What should I do next? Thanks!

Solutions Collecting From Web of "Hermitian matrix such that $4M^5+2M^3+M=7I_n$"

Hermitian matrices are diagonalizable and they have only real eigenvalues. Let
$$
M=H^{-1}\mathrm{diag}(d_1,\ldots,d_n)H,
$$
where $D=\mathrm{diag}(d_1,\ldots,d_n)$ is the diagonal matrix which contains the eigenvalues of $M$. Note that
$$
4x^4+4x^3+6x^2+6x+7= (2x^2+1)^2+3(x+1)^2+2x^2+4 \ge 5,
$$
for all real $x$. This implies that
$$
L=4M^4+4M^3+6M^2+6M+7= H^{-1}\mathrm{diag}(r_1,\ldots,r_n)H,
$$
with $r_i\ge 5$, for all $i$. Thus $L$ is invertible. Hence
$$
(4M^4+4M^3+6M^2+6M+7)(M-I)=0,
$$
implies that $M=I$!

The polynomial $P(x)=4x^5+2x^3+x-7$ annihilates the matrix $A$ so the eigenvalues of $A$ belong to the set of roots of $P$ and since $A$ is hermitian then it has only real eigenvalues hence
$$\operatorname{sp}(A)=\{1\}$$
since $1$ is the only real root of $P$. Now knowing that every hermitian matrix is diagonalizable then $A=I_n$.

Because it is hermitian you can find a matrix $Q$ such that $Q^tQ=I$ and $Q^TMQ$ is diagonal. Apply $Q^t$ and $Q$ to the equation and you get equations (the same equation) on the eigenvalues.

Using the eigenvalues build other diagonal matrices $D$ different from the identity and multipply them by orthogonal matrices $Q$ as $Q^TDQ$.