I'm trying to work out a proof for the following proposition:

Let $A \in \mathbb{R}^{n,n}$ a real, symmetric matrix with eigenvalues $\lambda_1 \ge \lambda_2 \ge \cdots \ge \lambda_n$, then

$$\max \lbrace {\rm tr} ( V^t A V ) \mid V \in \mathbb{R}^{n,k}, \, V^t V={\rm Id}_k \rbrace = \sum\limits_{i=1}^k \lambda_k $$

I've already found a proof myself which I've posted at the end of this question. However, I feel I've still missed a very important point: This theorem should be a direct consequence of the **min/max**- or **Courant-Fischer-characterization of eigenvalues** (https://en.wikipedia.org/wiki/Min-max_theorem). Although, after realizing that $\text{tr}(V^TAV) = \sum_{i=1}^k <Av_i,v_i>$, this approach looks very promising, I was not able to continue the proof.

For $k=1$ the statement is directly proven by Courant-Fischer. For other values of $k$ inductive reasing comes to mind. However, it is not clear to me, why this 'greedy approach' (maximize $k-1$ summands first and then choose the $k$-th accordingly) reaches the true maximum and how the minimum of the Courant-Fischer characterization isn't an issue.

I'd be very thankful for hints and solutions!

**different proof without min/max-theorem:**
Choose an orthonormal basis consisting of eigenvectors, writeall $v_i$'s in this representation and calculate

$\text{tr}(V^TAV) = \sum_{i=1}^n \lambda_i (\sum_{j=1}^k \alpha_{i,j}^2) = \sum_{i=1}^n h_i \lambda_i$

with $h_i := \sum_{j=1}^k \alpha_{i,j}^2$. Since $V^TV=\text{Id}_k$ we have $0 \le h_i \le 1$ and $\sum_{i=1}^n=k$. This shows that the maximum is attained for $h_1 = ... = h_k =1$ and $h_{k+1} = ... = h_n = 0$, which proofs the claim.