Linear Algebra proof involving traces and rank one projections

Question

How does one prove that if $T$ is a non-negative linear operator over a finitely generated Hilbert space that satisfies $tr(T^2)=tr(T)=1$, then $T$ must be a rank one projection? I am sorry to give no more details or any work at all. I don't even know where to start! My issue comes from a claim done in the following lecture http://pirsa.org/displayFlash.php?id=12090000

Answer 1

We're considering a non-negative operator $T$ in a finite-dimensional Hilbert space (let's call the dimension $n$).

The operator $T$ is non-negative, so it is hermitian and all of its eigenvalues $\lambda_1, \dots \lambda_n$ are $\geq 0$ (eigenvalues counted with multiplicities, i.e. they need not to be distinct).

Since the trace of $T$ equal to the sum of the eigenvalues, we have ${\rm tr} T=\sum_{i=1}^n \lambda_i = 1.$ Since all of the summands are nonnegative, we have $\lambda_i \in [0,1]$ for all $i=1,\dots ,n$

The eigenvalues of $T^2$ are $\lambda_1^2 , \dots \lambda_n^2$, so we have ${\rm tr T^2}=\sum_{i=1}^n \lambda_i^2 = 1.$

For $a\in (0,1),$ we have $a^2 < a,$ but the sum above isn't getting any smaller ($\sum_{i=1}^n \lambda_i = \sum_{i=1}^n \lambda_i^2$) . So we have that every eigenvalue needs to be either equal to $0$ or $1$, but since ${\rm tr} T = 1,$ only one eigenvalue can be equal to $1$ and the other need to be equal to $0$.

That is, $T$ is the identity on the span of the one-dimensional eigenspace to the eigenvalue $1$ and zero on its orthogonal complement, i.e. a projection with rank $1$.