Suppose $|\psi\rangle$ is a pure state of a composite system with components $A$ and $B.$
Prove that the Schmidt number of $|\psi\rangle$ is equal to the rank of the reduced density matrix $\rho_A \equiv \operatorname{tr}_B(|\psi\rangle\langle\psi|).$
(Note that the rank of a Hermitian operator is equal to the dimension of its support.)
I understand that the Schmidt number is the number of non-zero Schmidt coefficients, but I don't know how to generally find that and then proceed with the proof from there.
How do I find the Schmidt number generally? What is a support?
If the Schmidt number is $r$, then $\lvert\psi\rangle$ can, in some basis, be written as $$\lvert\psi\rangle=\sum_{k=1}^r \lambda_k \lvert k\rangle\otimes\lvert k\rangle, \,\,\lambda_k\in\mathbb R.$$
It follows that $$\lvert\psi\rangle\!\langle\psi\rvert=\sum_{jk}^r\lambda_j\lambda_k \lvert j\rangle\!\langle k\rvert\otimes \lvert j\rangle\!\langle k\rvert,$$ and finally $$\rho_A\equiv\operatorname{tr}_B(\lvert\psi\rangle\!\langle\psi\rvert) = \sum_{i=1}^r (1\otimes\langle i\rvert)\,\,\lvert\psi\rangle\!\langle\psi\rvert\,\,(1\otimes\lvert i\rangle) = \sum_{j=1}^r \lambda_j^2 \lvert j\rangle\!\langle j\rvert.$$ This means that $\rho_A$ has exactly $r$ nonvanishing eigenvalues, and thus its rank is $r$.