Let \(A\) be an \(n\times n\) Hermitian matrix satisfying \[ A^5+A^3+A=3I. \] Prove that \(A=I\).
Solution
Let \[ p(t)=t^5+t^3+t-3. \]
1. The polynomial \(p\) has exactly one real zero, namely \(1\).
For every \(t\in\RR\), \[ p'(t)=5t^4+3t^2+1>0. \] Thus \(p\) is strictly increasing on \(\RR\). Since \[ p(1)=1+1+1-3=0, \] the unique real zero of \(p\) is \(1\).
2. Every eigenvalue \(\lambda\) of \(A\) is real and satisfies\[ p(\lambda)=0. \]
Because \(A\) is Hermitian, the spectral theorem implies that all of its eigenvalues are real. If \(Av=\lambda v\) with \(v\neq0\), then the matrix equation gives \[ 0 = (A^5+A^3+A-3I)v = (\lambda^5+\lambda^3+\lambda-3)v = p(\lambda)v. \] Since \(v\neq0\), one has \(p(\lambda)=0\).
3. Every eigenvalue of \(A\) is equal to \(1\).
By step 2, every eigenvalue is a real zero of \(p\). Step 1 shows that the only such zero is \(1\).
4. One has\[ \boxed{A=I}. \]
By the spectral theorem there is a unitary matrix \(U\) and a real diagonal matrix \(D\) such that \[ A=UDU^*. \] Step 3 shows that every diagonal entry of \(D\) is \(1\), so \(D=I\). Hence \[ A=UIU^*=I. \]
5. Q.E.D.
Step 4 is the required conclusion.