Does $ABC = ADC \implies B = D$?

Question

Let's assume that we have 4 matrices: $A, B, C, D$. None of them is equal to the zero matrix. Is the statement $ABC = ADC \implies B = D$ always true?

I don't know how to prove it, and I haven't found any counterexample either.

Answer 1

As said in the comments, if $A,C$ are invertible, then $B=D$.

If, for example, $A$ is not invertible, let $Av=0$. Then take $D = vv^T$ and $B=0$. You get $$0 = ABC = Avv^TC = ADC.$$

In the same way, if $C$ is not invertible, take $w^TC = 0$, $D=ww^T$, $B=0$. You get again $$0 = ABC = Aww^TC = ADC.$$

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If you want any other examples, just take, respectively, $D = B + vv^T$ for the first case, and $D = B+ww^T$ for the second case, and you will get again $$ ABC = ADC $$ so you just have to find $B$ so that $ABC\ne 0$

Answer 2

Hint: Consider $A = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}$, $B = \begin{bmatrix}0 \\ 1 \end{bmatrix}$, $D = \begin{bmatrix}0 \\ 2 \end{bmatrix}$, and $C=[1]$.

Answer 3

Here is a concrete counterexample: $$\begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&0\\0&1\end{pmatrix}\begin{pmatrix}1&0\\0&1\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}=\begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&0\\0&2\end{pmatrix}\begin{pmatrix}1&0\\0&1\end{pmatrix}$$

Answer 4

As a simple counterexample take: $$ A=\begin{bmatrix} 1&0\\0&0 \end{bmatrix} \quad B=\begin{bmatrix} b&0\\0&d \end{bmatrix}\quad C=\begin{bmatrix} 1&0\\0&1 \end{bmatrix}\quad D=\begin{bmatrix} b&0\\d&0 \end{bmatrix} $$

Answer 5

Let $A\in k^{n\times m}$ and $C\in k^{d\times s}$ and consider the linear map $$\phi^C_A: k^{m\times d}\to k^{n\times s}\\ \phi^C_A(X)=AXC$$

Then, $\dim\ker \phi^C_A=\operatorname{rk}C\cdot\dim\ker A + (d-\operatorname{rk}C)n$.

This is best seen if we look at the matrices as linear functions $f_A\in\operatorname{Lin}(k^m,k^n)$, $f_C\in\operatorname{Lin}(k^s,k^d)$, and likewise $\phi^C_A(X)=f_A\circ X\circ f_C$ becomes a function from $\operatorname{Lin}(k^d,k^m)$ to $\operatorname{Lin}(k^s,k^n)$. Consider a subspace $W\subseteq k^s$ such that $W\oplus\operatorname{im} f_C=k^s$. Then, the map $$\Psi:\ker\phi^C_A\to \operatorname{Lin}(\operatorname{im} f_C,\ker f_A)\oplus \operatorname{Lin}(W,k^n)\\\Psi(X)=(\left.X\right\rvert_{\operatorname{im} f_C}, \left.X\right\rvert_{W})$$ is linear and bijective (use the fact that $X$ satisfies $g\circ X\circ f=0$ if and only if $\operatorname{im}\left.X\right\rvert_{\operatorname{im} f}\subseteq\ker g$).

So, for any matrix $B$, the matrices $X$ such that $AXC=ABC$ form an affine subspace in $k^{m\times d}$ of positive dimension as soon as either $A$ is singular or $C$ isn't full-row-rank.