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Is this identity true for all possible values of $A,B$ and $C$? $$A^TBC = C^TBA$$ where either

  1. $A,B,C$ are square matrices of same size
  2. $A$ and $C$ are vectors of size $n$ and $B$ is square matrix of size $n\times n$
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Yes I guess. Can we use that for above proof? –  gibraltar Nov 21 '12 at 16:34
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Have you tried any examples? –  Jonas Meyer Nov 21 '12 at 16:40
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The identity would imply that a symmetric $n\times n$ matrix commutes with every $n\times n$ matrix (take $C$ the identity and $A$ symmetric). Any chance? –  Andrea Mori Nov 21 '12 at 18:24
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1 Answer 1

up vote 7 down vote accepted

Recall, to prove that a statement is not an identity (an equality that holds for all $A, B, C$), you need only find one counterexample that satisfies the premise, but fails to satisfy the equality/identity.


$(1)$ Does the identity hold for all $A, B, C$ where $A, B, C$ are $n\times n$ square matrices for some $n$? For example, consider the following matrices:

Let $$A = \begin{bmatrix}\ 1 &2\\ 1 & 0\\ \end{bmatrix} ,\quad B= \begin{bmatrix}\ \,\,1 & 1 \\ -1 & 2\\ \end{bmatrix},\quad C= \begin{bmatrix}\ 2 & -1\\ \,\,0 & 1\\ \end{bmatrix}.\ $$

  • You'll first need to find $A^T$ and $C^T$.

  • Then compute $A^TBC$ and $C^TBA$.

  • Does $A^TBC = C^TBA$?


$(2)$ Does the identity hold for all $A, B, C$ when $A$ and $C$ are each an $n\times 1$ column vector and $B$ is an $n\times n$ square matrix? ($A$ and $C$ must be $n\times 1$ column vectors if $A^TBC$ and $C^TBA$ are to be defined.)

  • Note that $A^TBC$ and $C^TBA$ each evaluate to a scalar.
  • Try experimenting with different possibilities for $A, B, C$. Start simple; for example, try working with $n = 2$.
  • Again, if you can find $A, B, C$ such that $A^TBC \neq C^TBA$, then $(2)$ cannot hold for all such $A, B, C$.

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