If a matrix is positive-semidefinite, is Hermitian, and has a trace of 1, is it idempotent?

Question

I have a matrix $A^{n\times n}$ that is positive semi-def. and Hermitian. Also, $Tr(A) = 1$.

I want to show that $A$ is idempotent. Are these properties enough? If so, how would one show this? If not, what else might I need?

Thanks!

Answer 1

WLOG, let $A$ be diagonal (since trace remains invariant under conjugation and all Hermitian matrices are diagonalizable). Then, by positive semi-definiteness, we know that all of the diagonal elements (eigenvalues) are non-negative. The only way a matrix can be idempotent, however, is if all of its eigenvalues are either $0$ or $1$.

So, in order for $A$ to be idempotent, you will also require that $A$ have one eigenvalue equal to $1$ and the rest $0$. In this case, as stated by Tales Rick Perche, $A$ must be a projector.

Answer 2

In physics a positive semi definite operator with trace $1$ is regarded as a density operator. The results we have for that is that it is idempotent if and only if it is a projector.

This can be shown as follows. Given that $A$ is Hermitian, it has a spectral decomposition

$$A = \sum_{k=1}^n \lambda_k \Pi_k,$$

where $\lambda_k$ are real eigenvalues and the $\Pi_k$ are orthogonal projectors into the eigenspaces of $A$ that add up to the identity, that is, $\Pi_k\Pi_j = \delta_{jk}\Pi_{k}$ and $\sum_{k=1}^n\Pi_k = \mathbb{I}$.

Explicitly calculating $A^2$ yields

$$A^2 = \left(\sum_{k=1}^n \lambda_k \Pi_k\right)\left(\sum_{j=1}^n \lambda_j \Pi_j\right) = \sum_{k=1}^n\sum_{j=1}^n \lambda_k \lambda_j \delta_{jk}\Pi_{jk} = \sum_{k=1}^n \lambda_k^2 \Pi_k.$$

It is not hard to see that we can apply this argument inductively to get

$$A^N = \sum_{k=1}^n \lambda_k^N \Pi_k.$$

We then have that $A$ is idempotent if and only if there exists an $N\in\mathbb{N}$ so that $A^N =A$. This is equivalent to

$$0 = A^N -A = \sum_{k=1}^n (\lambda_k^N-\lambda) \Pi_k\Leftrightarrow \lambda_k^N - \lambda_k = 0.$$

Using the fact that the $\lambda_k$s are real and that the trace can be written as the sum of the eigenvalues (that is $\sum_{k=1}^n\lambda_k = 1$), you can see that the equation above is solved if and only if one of them is $1$ and the other ones are zero. This means that $A$ being idempotent happens if and only if there is a $k_0$ such that

$$A = \Pi_{k_0}$$

for the $k_0$ corresponding to the only non vanishing eigenvalue, $\lambda_{k_0}$. Therefore if $A$ is Hermitian semi positive definite and has trace one, we get that $A$ is idempotent if and only if $A$ is a projector.