If \(A\) is Hermitian and satisfies \[ A^5+A=2I, \] prove that \(A=I\).
Solution
1. The real equation\[ t^5+t=2 \]has the unique solution\[ t=1. \]
Set \[ p(t)\coloneqq t^5+t-2. \] For every real \(t\), \[ p'(t)=5t^4+1>0. \] Thus \(p\) is strictly increasing on \(\RR\). Since \[ p(1)=1+1-2=0, \] the value \(1\) is its unique real zero.
2. Every eigenvalue of \(A\) equals \(1\).
Because \(A\) is Hermitian, the spectral theorem gives an orthonormal basis of eigenvectors and all eigenvalues of \(A\) are real. Let \(v\neq0\) be an eigenvector with \[ Av=\lambda v, \qquad \lambda\in\RR. \] Applying the matrix equation to \(v\) gives \[ (A^5+A)v = (\lambda^5+\lambda)v = 2v. \] Hence \[ \lambda^5+\lambda=2. \] Step 1 therefore gives \(\lambda=1\).
3. One has\[ \boxed{A=I}. \]
By the Hermitian spectral theorem, \(A\) is unitarily diagonalizable. Step 2 shows that every diagonal entry in such a diagonalization is \(1\). Thus the diagonal form is \(I\), and conjugating \(I\) by a unitary matrix leaves it unchanged. Hence \(A=I\).
4. Q.E.D.
Step 3 is the required conclusion.