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.

    \( \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix} \ne \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} \)

Difficulty level: Easy task (using definitions and simple reasoning)
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