$A$ be $n \times n$ Hermitian matrix with $A^5+A^3+A=3I_n$ $implies$ $A=I_n$ ?

Question

Let $A$ be a Hermitian matrix of size $n$ such that $A^5+A^3+A=3I_n$ , then is it true that $A=I_n$ ? What I got is if $a$ is an eigenvalue then $a^5+a^3+a-3=0=(a-1)(a^4+a^3+2a^2+2a+3)$ this doesn't seem to get anywhere , Please help .

Answer 1

1. Any Hermitian matrix is diagonalizable and has real eigenvalues.

2. The only real solution of $x^5+x^3+x-3=0$ is $x=1$.

3. Conclude.

Answer 2

Cayley-Hamilton theorem gives you, for $p\left( x \right) = {x^5} + {x^3} + x - 3$, and taking into consideration that $p\left( 1 \right) = 0$, conclude that the characteristic polynomial of $A$ must be divisible by $p$. You can pick it up from there.

Answer 3

As $A$ is a Hermitian matrix, so there exists some unitary matrix $P$, such that $P^{-1}AP=diag\{\lambda_1,\lambda_2,\cdots,\lambda_n\}$, where $\lambda_1,\lambda_2,\cdots,\lambda_n$ are real numbers, so $3I=P^{-1}(A^5+A^3+A)P=diag\{(\lambda_1)^5+(\lambda_1)^3+\lambda_1,\cdots,(\lambda_n)^5+(\lambda_n)^3+\lambda_n\}$, so $\lambda_1=\lambda_2=\cdots=\lambda_n=1\Rightarrow P^{-1}AP=I\Rightarrow A=I$.