Is there a matrix $C$ that can make $AB$ return the same result as $BA$ where $C$ is based on $B$?

Question

I am given an arbitrary matrix $A$ that I will be multiplying by a rotational matrix $B$ ( both $4\times4$ )

Is there any matrix $C$, based only on manipulation of matrix $B$, that when doing $A(BC)$ will produce the same result as $BA$?

$BA = A(BC)$

I am trying to find some sort of abstract solution for making matrix multiplication commutative

If it can not only be based on matrix $B$, is there any way to find that matrix $C$ if using matrix $A$ as well?

To clarify, let's define a matrix $D = BC$ . Is there a method to obtain a matrix $C$ such that $BA = AD$

I am not sure the statement: (1)For all $A,B$, there is a $C=C(A,B)$ such that $BA=ABC$. (2) For all $A,B$, there is a $C=C(B)$ such that $BA=ABC$. I consider the statement is the former. — Camford Oxbridge, Jul 06 '20 at 02:14
The ideal solution would be for statement 2, as I believe it could then be applied to statement 1 — David W, Jul 06 '20 at 02:25
If $C$ denpends only on $B$, then the answer depends on $A$. In my view, $C$ should be a function of both $A$ and $B$. — Camford Oxbridge, Jul 06 '20 at 02:38

score 2 · Accepted Answer · answered Jul 06 '20 at 01:31

2

Let's start by considering $BA = ABC$ now by multiplying by $B^{-1}A^{-1}$ on the left we get $B^{-1}A^{-1}BA = C$. Now this is only satisfactory when we are working with invertible matrices... If you want something stronger than that I don't think it's possible for a general case.

answered Jul 06 '20 at 01:31

TdotA

400

They are invertible yes. My concern with this answer is the order of matrix multiplication. In ABC, BC must be multiplied first. So A(BC) is the correct form. With that in mind, wouldn't the removal factor be the inverse of (BC)? Does that affect the outcome? – David W Jul 06 '20 at 02:38
I'm not sure if I understand your question, but note that A(BC) = (AB)C and thus you can just proceed as I wrote. https://people.math.carleton.ca/~kcheung/math/notes/MATH1107/wk05/05_associativity_of_matrix_multiplication.html#:~:text=Matrix%20multiplication%20is%20associative&text=If%20A%20is%20an%20m,%3D(AB)C. – TdotA Jul 06 '20 at 03:20

score 2 · Answer 2 · answered Jul 06 '20 at 01:59

2

To signify that $C$ only depends on $B,$ let $C=f(B).$ This gives us $BA=ABf(B)$ for all $A.$ Plugging in $A=I,$ we get $f(B)=I.$ Thus, $BA=AB,$ which means $B$ commutes with every matrix. This is only possible if $B$ is a multiple of the identity, and in this case $C=I$ indeed works.

answered Jul 06 '20 at 01:59

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This seems to imply in practice that the only way it's possible is if you are dealing with identity matrixes. Is that the case? – David W Jul 06 '20 at 02:33
You statement is that for all $B$ there is $C=C(B)$ such that for all $A$, $BA=ABC$. – Camford Oxbridge Jul 06 '20 at 02:34

Is there a matrix $C$ that can make $AB$ return the same result as $BA$ where $C$ is based on $B$?

2 Answers2