Vanishing partial trace Given a square matrix $A$ (could be chosen to be unitary, and possibly also involutive if it helps), then how do we find a matrix $M$ such that
$$\text{pt}(MA) = 0?$$
On a pure tensor, the partial trace above acts as $\text{pt}(A\otimes B) = \text{Tr}(A) B$ and can then be extended to general $n^2 \times n^2$ matrices by linear combinations.
In the case of a full trace, we of course can say for example that if $MA = BC - CB$ for some $C$ and $B$, then the trace vanishes, and there are also other possiblities. But what about the case of a partial trace?
 A: Express $A$ in the form
$$
A = \sum_{j,k = 1}^n E_{jk} \otimes A_{jk}
$$
where $E_{jk}$ denotes the matrix with $1$ as its $j,k$ entry and zeros elsewhere. Equivalently, each $A_{jk}$ is an $n \times n$ "block-entry" of $A$. Partition $M$ in the same fashion.
We can write the $p,q$ entry of $\operatorname{pt}(MA)$ as
$$
\operatorname{tr}\left(\sum_{j=1}^n M_{pj} A_{jq}\right) = \operatorname{tr}(M C_{pq}^T),
$$
where $C_{pq}$ denotes the matrix whose block-entries are zero except for the $p$-th row, and the block-entries of the $p$th row are $A_{1q}^T,\dots,A_{nq}^T$. Now, in terms of the vectorization operator, we have
$$
\operatorname{tr}(MC_{pq}^T) = \operatorname{vec}(C_{pq})^T \operatorname{vec}(M).
$$
With that, we see that $\operatorname{vec}(M)$ must be a solution to the equation $C x = 0$, where $C$ is the matrix whose rows are $\operatorname{vec}(C_{pq})^T$ for all $1 \leq p,q \leq n$.

To simplify the above a bit: we can write
$$
C_{pq} = \sum_{j=1}^n E_{pj} \otimes A_{jq}^T,
$$
so that
$$
\operatorname{vec}(C_{pq}) = \sum_{j=1}^n \operatorname{vec}(E_{pj} \otimes A_{jq}^T) = \sum_{j=1}^n \sum_{k=1}^n e_j \otimes e_k \otimes e_p \otimes A^T_{jp}[k],
$$
where $M[k]$ denotes the $k$th column of the matrix $M$.
