# Do matrices $A$ and $B$ exist such that $AXB$ turns only $X_{ij}$ to $0$, with other entries unchanged?

Let $$X$$ be an $$n \times n$$ matrix. Given $$i$$ and $$j$$, do $$n \times n$$ matrices $$A$$ and $$B$$ exist such that $$AXB$$ turns the $$i$$-th row, $$j$$-th column entry of $$X$$ to $$0$$, with other entries unchanged?

I tend to believe there aren't such matrices, but how to prove it?

Update: Sorry for my implicit description. I mean only one entry $$X_{ij}$$ turns to $$0$$, with other entries not changed. I have tried the case for $$n=2$$, which shows $$A$$ and $$B$$ don't exist.

• As in, the same $A$ and $B$ for all $X$ (for given $n,i,j$)? May 4, 2022 at 14:54
• yes, the same A and B for all X (for given n,i,j) May 4, 2022 at 15:43
• What is the motivation for this question? May 4, 2022 at 15:56
• Well briefly,I get one matrix $\left( \begin{matrix}A & B\\C&0 \end{matrix}\right)$,I hope to turn A = 0 and finally get $\left( \begin{matrix}B+C & 0\\0&0 \end{matrix}\right)$ by using the transformation $X$ ->$MXN$(which ensures one algorithm can be applied. May 4, 2022 at 16:11
• In last comment $A$,$B$ and $C$ are all functions of $X$ ,M and N are constant matrices,which means we can't do that by getting inverse of $A$ May 4, 2022 at 16:24

Let $$E_i$$ be a matrix of zeros except the entry $$(i,i)$$ which is equal to one.

Then, the map $$X\mapsto X-E_iXE_j$$ does exactly what you want. We can prove that there are no such matrices $$A,B$$ such that $$AXB=X-E_iXE_j$$ holds. We have that

$$AXB=Y$$ if and only if $$(B^T\otimes A)\mathrm{vec}(X)=\mathrm{vec}(Y)$$ where $$\mathrm{vec}(\cdot)$$ is the vectorization operator and $$\otimes$$ denotes the Kronecker product.

We would like $$\mathrm{vec}(Y)=S\mathrm{vec}(X)$$ where $$S$$ is a diagonal matrix of ones on the diagonal except for one entry, which is equal to 0. This entry corresponds to that entry in $$X$$ that we would like to set to zero while the others remain unchanged.

Therefore, we need to have that

$$((B^T\otimes A)-S)\mathrm{vec}(X)=0.$$

Since this expression must be true for all $$X$$, then the above equality holds if and only if

$$(B^T\otimes A)-S=0.$$

Expanding the above expression yields

$$\begin{bmatrix}b_{11}A & b_{21}A & \ldots & b_{n1}A\\ \vdots\\ b_{1n}A & b_{2n}A & \ldots & b_{nn}A \end{bmatrix}-S=0$$

Since $$S$$ is a diagonal matrix we must have $$b_{ij}=0$$ for all $$i\ne j$$. Moreover, all but one blocks are identity blocks in $$S$$ by construction. Assume that the last block has one zero diagonal entry while the others entries are equal to one, and that the other blocks are identity blocks.

Then, we must have that $$b_{ii}A=I$$ for all $$i=1,\ldots,n-1$$. This implies that $$A$$ must be diagonal with nonzero entries. However, this contradicts the fact that $$b_{nn}A$$ must have one zero entry on the diagonal.

Therefore, there is no such matrices $$A$$ and $$B$$.

Note, however, that some restrictions of the problem may lead to the existence of such matrices. One restriction is to consider a subclass of matrices and an obvious one is the class of diagonal matrices $$X$$. Another restriction would be to consider one specific entry to be set to zero and not look for the existence of $$A$$ and $$B$$ so that any entry can be set to zero.

We can restrict the set of matrices as follows by assuming that $$\mathrm{vec}(X)=Mz$$ for some full column rank $$M$$ and vector $$z$$.

In that case, the problem becomes

$$((B^T\otimes A)-S)M=0.$$

In other words, $$M$$ should span a subspace of the null-space of $$(B^T\otimes A)-S$$.

• Thanks for your proof very much! May 4, 2022 at 18:03

When $$n>1$$, no. Assume the contrary. Then $$Ae_ie_j^TB=0$$, where $$\{e_1,\ldots,e_n\}$$ is the standard basis of $$\mathbb F^n$$ (and $$\mathbb F$$ denotes the underlying field). Thus either $$Ae_i=0$$ or $$e_j^TB=0$$. Without loss of generality, assume that $$Ae_i=0$$. Then $$Ae_ie_k^TB$$ is also zero for every other $$k$$, meaning that $$X\mapsto AXB$$ will also turn the $$(i,k)$$-th element of $$e_ie_k^T$$ to zero. Hence we arrive at a contradiction.

Yes, but of course the size of X has to be fixed if you intend to find specific A and B matricies, for an m×m matrix X, letting A be the m×m identity matrix with its ith 1 omitted, and let B be the the m×m identity matrix with its jth 1 ommitted, these matrices satisfy your conditions, and they are in fact the only matrices which have this property for a general matrix X

To help you understand why this works, it might help to think about different views of matrix multiplication, one is for example that the ij entry of a matrix X which is the product of A and B, is the dot product of the ith row of A with the jth column of B, then multiplying a matrix A with the identity matrix gives as its ij entry the projection of the ith row of A onto the jth standard unit vector, which is just the jth component of that row

• Thanks a lot,but sorry that my description is implicit.I mean only one entry $X_{ij}$ turns to 0,with other entries not changed. May 4, 2022 at 15:49