Non-commutative product
Task number: 2422
For an arbitrary asymmetric square matrix \(\mathbf A\) construct a matrix \(\mathbf B\) s.t. their products do not commute, i.e. \(\mathbf A\mathbf B\ne\mathbf B\mathbf A\).
Do products commute when both matrices are symmetric?
Hint
Use facts from the previous exercise.
Resolution
There are several solutions to this problem.
If \(a_{i,j}\ne a_{j,i}\), we can choose \(\mathbf B\) s.t. \(b_{i,j}=b_{j,i}=1\), and \(b_{k,l}=0\) otherwise. Then we get \((\mathbf A\mathbf B)_{i,i}=a_{i,j}\ne a_{j,i}=(\mathbf B\mathbf A)_{i,i}\).
Another option is to choose \(\mathbf B\) s.t. the only nonzero element is \(b_{i,i}=1\). Then the two products cannot agree on elements with indices \(i\) and \(j\). We get \((\mathbf A\mathbf B)_{i,j}=0=(\mathbf B\mathbf A)_{i,j}=a_{i,j}\), while \((\mathbf B\mathbf A)_{j,i}=0=(\mathbf A\mathbf B)_{j,i}=a_{j,i}\), but at least one of \(a_{i,j}\) and \(a_{j,i}\) is distinct from zero.
Even on symmetric matrices the product does not commute (even the result need not to be symmetric), e.g.
\( \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix} \ne \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} \)
