Linear Algebra: Foundations to Frontiers

Subsection 4.2.2 Transposing a partitioned matrix

We know that

The above example illustrates a general rule: When transposing a partitioned matrix (matrix partitioned into submatrices), you transpose the matrix of blocks, and then you transpose each block. In other words, let \(A \in \Rmxn \) be partitioned as follows:

where \(A_{i,j} \in \R^{m_i \times n_j} \text{.}\) Then

We now discuss a number of special cases that you may encounter.

• Each submatrix is a scalar.

  If

  \begin{equation*} A = \left( \begin{array}{c | c | c | c} \alpha_{0,0} \amp \alpha_{0,1} \amp \cdots \amp \alpha_{0,N-1} \\ \hline \alpha_{1,0} \amp \alpha_{1,1} \amp \cdots \amp \alpha_{1,N-1} \\ \hline \vdots \amp \vdots \amp \amp \vdots \\ \hline \alpha_{M-1,0} \amp \alpha_{M-1,1} \amp \cdots \amp \alpha_{M-1,N-1} \end{array} \right) \end{equation*}

  then

  \begin{equation*} \begin{array}{rcl} A^T \amp = \amp \left( \begin{array}{c | c | c | c} \alpha_{0,0}^T \amp \alpha_{1,0}^T \amp \cdots \amp \alpha_{M-1,0}^T \\ \hline \alpha_{0,1}^T \amp \alpha_{1,1}^T \amp \cdots \amp \alpha_{M-1,1}^T \\ \hline \vdots \amp \vdots \amp \amp \vdots \\ \hline \alpha_{0,N-1}^T \amp \alpha_{1,N-1}^T \amp \cdots \amp \alpha_{M-1N-1}^T \end{array} \right) \\ \amp = \amp \left( \begin{array}{c c c c} \alpha_{0,0} \amp \alpha_{1,0} \amp \cdots \amp \alpha_{M-1,0} \\ \alpha_{0,1} \amp \alpha_{1,1} \amp \cdots \amp \alpha_{M-1,1} \\ \vdots \amp \vdots \amp \amp \vdots \\ \alpha_{0,N-1} \amp \alpha_{1,N-1} \amp \cdots \amp \alpha_{M-1,N-1} \end{array} \right). \end{array} \end{equation*}

  This is because the transpose of a scalar is just that scalar.

• The matrix is partitioned by rows.

  If

  \begin{equation*} A = \left( \begin{array}{c c c c} \widetilde a_{0}^T \\ \hline \widetilde a_{1}^T \\ \hline \vdots \\ \hline \widetilde a_{m-1}^T \end{array} \right), \end{equation*}

  where each \(\widetilde a_i^T \) is a row of \(A \text{,}\) then

  \begin{equation*} \begin{array}{rcl} A^T \amp = \amp \left( \begin{array}{c c c c} \widetilde a_{0}^T \\ \hline \widetilde a_{1}^T \\ \hline \vdots \\ \widetilde a_{m-1}^T \end{array} \right)^T \\ \amp = \amp \left( \begin{array}{c | c | c | c} \left( \widetilde a_{0}^T\right)^T \amp \left( \widetilde a_{1}^T\right)^T \amp \cdots \amp \left( \widetilde a_{m-1}^T\right)^T \end{array} \right) \\ \amp = \amp \left( \begin{array}{c | c | c | c} \widetilde a_{0} \amp \widetilde a_{1} \amp \cdots \amp \widetilde a_{m-1} \end{array} \right). \end{array} \end{equation*}

  This shows that rows of \(A \text{,}\) \(\widetilde a_i^T \text{,}\) become columns of \(A^T \text{:}\) \(\widetilde a_{i} \text{.}\)

• The matrix is partitioned by columns.

  If

  \begin{equation*} A = \left( \begin{array}{c |c |c |c} a_{0} \amp a_{1} \amp \cdots \amp a_{n-1} \end{array} \right), \end{equation*}

  where each \(a_j\) is a column of \(A \text{,}\) then

  \begin{equation*} \begin{array}{rcl} A^T \amp = \amp \left( \begin{array}{c |c |c |c} a_{0} \amp a_{1} \amp \cdots \amp a_{n-1} \end{array} \right)^T \\ \amp = \amp \left( \begin{array}{c c c c} a_{0}^T \\ \hline a_{1}^T \\ \hline \vdots \\ a_{n-1}^T \end{array} \right) . \end{array} \end{equation*}

  This shows that columns of \(A \text{,}\) \(a_j \text{,}\) become rows of \(A^T \text{:}\) \(a_j^T \text{.}\)

• \(2 \times 2 \) blocked partitioning.

  If

  \begin{equation*} A = \left( \begin{array}{c |c} A_{TL} \amp A_{TR} \\ \hline A_{BL} \amp A_{BR} \end{array} \right), \end{equation*}

  then

  \begin{equation*} A^T = \left( \begin{array}{c |c} A_{TL}^T \amp A_{BL}^T \\ \hline A_{TR}^T \amp A_{BR}^T \end{array} \right) . \end{equation*}

• \(3 \times 3 \) blocked partitioning.

  If

  \begin{equation*} A = \left( \begin{array}{c |c |c} A_{00} \amp a_{01} \amp A_{02} \\ \hline a_{10}^T \amp \alpha_{11} \amp a_{12}^T \\ \hline A_{20} \amp a_{21} \amp A_{22} \end{array} \right), \end{equation*}

  then

  \begin{equation*} \begin{array}{rcl} A^T \amp = \amp \left( \begin{array}{c |c |c} A_{00} \amp a_{01} \amp A_{02} \\ \hline a_{10}^T \amp \alpha_{11} \amp a_{12}^T \\ \hline A_{20} \amp a_{21} \amp A_{22} \end{array} \right)^T \\ \amp = \amp \left( \begin{array}{c |c |c} A_{00}^T \amp \left( a_{10}^T\right)^T \amp A_{20}^T \\ \hline a_{01}^T \amp \alpha_{11}^T \amp a_{21}^T \\ \hline A_{02}^T \amp \left( a_{12}^T \right) ^T \amp A_{22}^T \end{array} \right) \\ \amp = \amp \left( \begin{array}{c |c |c} A_{00}^T \amp a_{10} \amp A_{20}^T \\ \hline a_{01}^T \amp \alpha_{11} \amp a_{21}^T \\ \hline A_{02}^T \amp a_{12} \amp A_{22}^T \end{array} \right) . \end{array} \end{equation*}

Anyway, you get the idea!!!