An identity for the sum of $k$ smallest eigenvalues Let $V$ be a real vector space of $n(<\infty)$ dimensional equipped with an inner product $\langle,\rangle$. Let $B:V\to V$ be a symmetric linear transform on $V$. Let $\lambda_1\leq\cdots\leq\lambda_n$ be the eigenvalues of $B$. Then how to prove that
\begin{eqnarray}
\lambda_1+\cdots+\lambda_k=\min\left\{\sum_{j=1}^k \langle e_j, Be_j\rangle\ 
|\ (e_j)_{j=1}^k\text{ is an orthonormal system of }V\right\}
\end{eqnarray}? Thank you. 

Usually when you have a symmetric map, respectively a symmentric matrix, you first diagonalize it and then you read the rest of the question. In this case: pick an orthonormal basis of $V$, consisting of eigenvectors of $B$. In this basis the matrix of $B$ is $D = diag(\lambda_1, \cdots \lambda_n)$ and hence in your expression above 
$$RHS = \min \{ \sum_{j=1}^k \langle f_j, Df_j\rangle | (f_j)_{j=1}^k \textrm{is an orthonormal system}\}$$
and I renamed $e_j$ to $f_j$ for clarity. Now, if we pick $f_j$ to be the stanard basis vectors $e_j$ we get $$ \sum_{j=1}^k \langle e_j, De_j\rangle = \lambda_1 + \ldots + \lambda_k$$ Thus, $LHS \geq RHS$  Now we need to show equality.
Proceed by induction on $k \leq n$ The case $k = n$ is a restatement of the claim $tr(D) = tr (P^tDP)$ where the columns of $P$ are $(f_j)_{j=1}^n$
Assume we know the statement for $k+1 \leq n$ 
$$\lambda_1 + \cdots + \lambda_{k+1} = \min \{  \sum_{j=1}^{k+1} \langle f_j, Df_j\rangle \} $$ 
If we fix $f_{k+1}$we are going to increase the RHS, as the minimum will be on a smaller set. So, choose $f_{k+1} = e_{k+1}$ Then $\langle e_{k+1}, De_{k+1}\rangle = \lambda_{k+1}$ so we can subtract $\lambda_{k+1}$ from both sides f the equation and get: 
$$\lambda_1 + \cdots + \lambda_{k} \leq \min \{  \sum_{j=1}^{k} \langle f_j, Df_j\rangle \} $$ 
But we already showed the reverse inequality, so actually 
$$\lambda_1 + \cdots + \lambda_{k} = \min \{  \sum_{j=1}^{k} \langle f_j, Df_j\rangle \} $$ 
and we have the claim for $k\geq 1$, so the induction is complete. 
Note: try induction from $k$ to $k+1$ to see why induction from $k+1$ to $k$ is easier!
