For $n\times n$ matrices, is it true that $AB=CD\implies AEB=CED$?

Question

If $A,B,C,D,E$ are $n\times n$ matrices, does $AB=CD$ imply $AEB=CED$?

I only know that $AB=CD \implies ABE=CDE$, but I don't see how you can sandwhich $E$ within it.

Also, if $AB=CD=0$, does $\det(AB)=\det(CD)=0$?

I think this should be true because $AB$ and $CD$ are the same matrices and $\det(0)=0$

Answer 1

The first question is not true as the following counterexample shows:

$$ A = B = \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix} \\ C = D = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\\ E = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$ We get $$ AB = 0 = CD$$ but $$AEB = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\\ CED = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$$

Answer 2

$A= \begin{bmatrix} 1 &0 \\ 0& 0 \end{bmatrix}$ ,$B= \begin{bmatrix} 0 &0 \\ 0& 1 \end{bmatrix}$ ,$C=D=0$ ,$E= \begin{bmatrix} 1 &2 \\ 4& 3 \end{bmatrix}$. then that following equality is not true.

Answer 3

The "sandwich" statement doesn't generally hold. For example, take $$ A = B = I = \pmatrix{1&0\\0&1}, \quad C = D = \pmatrix{0&1\\1&0} $$ Verify that $AB = CD$. However, if we take $$ E = \pmatrix{1&0\\0&0} $$ we find that $AEB \neq CED$

Answer 4

I don't think this works by sandwiching..... $$A=\begin{pmatrix} 1 & 3 \\ 0 & 1 \\ \end{pmatrix}, B=\begin{pmatrix} 1 & 2 \\ 0 & 1 \\ \end{pmatrix}, C=\begin{pmatrix} 1 & 1 \\ 0 & 1 \\ \end{pmatrix}, D=\begin{pmatrix} 1 & 4 \\ 0 & 1 \\ \end{pmatrix},$$ Then we have that $AB=CD$ by matrix multiplication, however if $$E=\begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix}$$ Then $AEB=\begin{pmatrix} 4 & 12\\ 1 & 3 \\ \end{pmatrix}\neq \begin{pmatrix} 2 & 10\\ 1 & 5 \\ \end{pmatrix}=CED$.

Answer 5

Let $A$ and $D$ be the identity matrix. Does $B=C$ imply $EB=CE$?

Answer 6

The first statement is only true if $E$ commutes with $B$ and with $D$ or if $E$ commutes with $A$ and with $C$