If $\text{tr}(M^3)=\text{tr}(M^2)=\text{tr}(M)=1$ then $M=\mathbf{uu}^\dagger$ for some unit vector $\mathbf{u}$

Question

Consider an $n \times n$ complex Hermitian matrix $M$ which has trace one. I'm trying to show that if $$\text{tr}(M^3)=\text{tr}(M^2)=\text{tr}(M)=1$$ then $M=\mathbf{uu}^\dagger$ for some column vector $\mathbf {u} \in \mathbb C^n$ of unit length. I think it is enough to show that one of the eigenvalues of $M$ is $1$ and the rest are $0$, as $\mathbf {uu}^\dagger$ is a rank one projection matrix. From the trace relations we have $$\sum_i \lambda_i^3=\sum_i \lambda_i^2=\sum_i \lambda_i=1$$ where $\{\lambda_i\}$ are the (real) eigenvalues of $M$, but I'm struggling to show that this implies a single nonzero eigenvalue.

Answer 1

Since $\lambda_j^2 \leq \sum_{i=1}^{n} \lambda_i^2 = 1$, we have $\lambda_j \in [-1, 1]$ for each $j = 1, \cdots, n$. Then $\lambda_i^2(1-\lambda_i) \geq 0$ for all $i$, and

$$ 0 = \sum_{i=1}^{n} \lambda_i^2(1-\lambda_i) $$

tells that $\lambda_i^2(1-\lambda_i) = 0$ for each $i$. So either $\lambda_i = 0$ or $\lambda_i = 1$, and the condition $\sum_{i=1}^{n} \lambda_i = 1$ tells that $\lambda_i = 1$ holds for exactly one of $i = 1,\cdots, n$.