Given the function $$ E(M) = \sum_{i=1}^N \sum_{a=1}^K \left( M_{ia} \cdot \left\lVert\sum_{i=1}^N M_{ia}\cdot x_i\right\rVert_2^2 \right) $$ $x$ is a given constant matrix, $x_i$ is a the $n_\text{th}$ column of that. $M_{ia} \in \{0,1\}$ and $\sum_{i=1}^N M_{ia}=\frac{N}{K} $, and we also have $\sum_{a=1}^N M_{ia}=1 \\\\$.
The question is, can $E$ of $M$ be further simplified to a form using only quadratic and linear terms of $M$?
Again, thank you so much.
Yes, $$E(M) = \frac{N}{K} \lVert X M \rVert_F^2$$ writing $X$ for the matrix $\begin{bmatrix} x_1 & \dots & x_N \end{bmatrix}$. Note $\sum_{i=1}^N M_{ia}x_i = X M e_a$ where $e_a$ is the $a$th standard basis vector. The constraints imply $M$ is a binary $N \times K$ matrix with a single nonzero per row and each column having the same number $m = N/K$ of nonzeros. Let the single nonzero in row $i$ be in column $j_i$. Only the term with $a=j_i$ is nonzero in the sum over $a$ so that $E(M) = \sum_{i=1}^N \lVert X M e_{j_i} \rVert_2^2$. Finally note that each column index $j$ will be visited exactly $m$ times in the sum over $i$, so that $E(M) = m \sum_{j=1}^K \lVert X M e_j \rVert_2^2 = m \lVert X M \rVert_F^2$.
