# 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?

• Presumably, you meant to write $\operatorname{pt}(A \otimes B) = \operatorname{Tr}(B)A$ Mar 3, 2021 at 16:26
• I actually meant the opposite way round - sorry. I have edited above! This may simplify your below calculations (which I'm very grateful for), but I discuss further in the comments.. Mar 4, 2021 at 15:34

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 $$M$$.
• You say these $A_{jk}$ are 'block-entries" of A, but I am correct in thinking that the adjacent entries of each $A_{jk}$ are not adjacent elements of $A$ as given by the tensor product. For example $(A_{11})^1_2 = A^1_3$ (where A^a_b is the element in the a-th row and b-th column of $A$ so as not to confuse notations). Mar 4, 2021 at 15:44
• Applying this change, I believe this yields $\sum_{j,k} M_{jk}A_{jk}$ as the result of the partial trace. Just to throw further spanner in the works, I wonder what this reveals if we chose $M$ to have a specific form, perhaps $M = N\otimes 1$ for some $N$ and 1 the identity matrix. This then simplifies the sum to $\sum_{jk} N_{jk}A_{jk}$ where these $N_{jk}$ are now just scalars. This now looks like a matrix valued linear independence equation if we want it to be zero. However, I'm unsure if such a notion exists. Mar 4, 2021 at 15:50
• @CS1994 I am taking the tensor product to be the same as the Kronecker product in this context. With that, for $n=2$, we have $$\sum_{j,k=1}^2 E_{ij} \otimes A_{ij} = \pmatrix{A_{11} & A_{12}\\ A_{21} & A_{22}}.$$ That is what I mean when I say that the $A_{jk}$ are the "block-entries" of $A$. Mar 4, 2021 at 16:44