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Is this statement true :$A=\left[ {\begin{array}{cc}P&Q\\R&S\\\end{array}}\right]^T=\left[ {\begin{array}{cc}P^T&Q^T\\R^T&S^T\\\end{array}}\right]$Justify the answer

I know this statement is wrong. I have tested it on $4\times 4 $ matrices and it is not true.

Also if $P=P^T,Q=Q^T,R=R^T,S=S^T,$ then also suppose we have a $4\times 4 $ matrix

$B=\left[ {\begin{array}{cc}1&0&0&1\\0&1&1&0\\1&0&0&1\\0&1&1&0\end{array}}\right]$

Now if we consider $\{B_{11},B_{12},B_{21},B_{22}\}$ be $P$ ,$\{B_{13},B_{14},B_{23},B_{24}\}$ as $Q$ and $\{B_{31},B_{32},B_{41},B_{42}\}$ as $R$ and $\{B_{33},B_{34},B_{43},B_{44}\}$ as $S$, we will get $\left[ {\begin{array}{cc}P&Q\\R&S\\\end{array}}\right]^T\ne\left[ {\begin{array}{cc}P^T&Q^T\\R^T&S^T\\\end{array}}\right]$ although $P=P^T,Q=Q^T,R=R^T,S=S^T$

Please give some empirical evidence for why it is not true.

Saradamani
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  • What else are you looking for, really? If you've tested it on $4 \times 4$ matrices and found counterexamples, then you already have all the empirical evidence you need... – Hans Lundmark May 14 '18 at 11:34
  • You can also test on a $2\times2$ matrix, and $1\times1$ blocks. For the formula to be true, you would have to swap $Q^T$ and $P^T$ – Jean-Claude Arbaut May 14 '18 at 11:35

3 Answers3

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Some reasons that $$ \begin{bmatrix}P^T&Q^T\\R^T&S^T\end{bmatrix} $$ can't be the transpose:

  1. The dimensions of the blocks don't line up. In other words, the number of rows of $P^T$ and the number of rows of $Q^T$ might not be the same (similarly for the number of columns of $P^T$ and $R^T$ and many other locations). For example $$ \begin{bmatrix} \begin{bmatrix} 1&2 \end{bmatrix}&\begin{bmatrix}3\end{bmatrix}\\ \begin{bmatrix}4&5\\6&7\end{bmatrix}&\begin{bmatrix}8\\9\end{bmatrix} \end{bmatrix} $$ would become, under your formula, $$ \begin{bmatrix} \begin{bmatrix} 1\\2 \end{bmatrix}&\begin{bmatrix}3\end{bmatrix}\\ \begin{bmatrix}4&6\\5&7\end{bmatrix}&\begin{bmatrix}8&9\end{bmatrix} \end{bmatrix}. $$ This has all of the dimensions messed up (there aren't clear rows or columns in the matrix).

  2. The entry in the upper right corner of the matrix must end up in the lower left corner under the transpose, but this doesn't happen as the upper right corner is in $Q$, so it ends up in the lower left corner of $Q^T$, which is towards the middle of the matrix. For example, the matrix $$ \begin{bmatrix} \begin{bmatrix} 1 \end{bmatrix} & \begin{bmatrix} 2 \end{bmatrix}\\ \begin{bmatrix} 3 \end{bmatrix}& \begin{bmatrix} 4 \end{bmatrix} \end{bmatrix} $$ would become, under your formula, $$ \begin{bmatrix} \begin{bmatrix} 1 \end{bmatrix} & \begin{bmatrix} 2 \end{bmatrix}\\ \begin{bmatrix} 3 \end{bmatrix}& \begin{bmatrix} 4 \end{bmatrix} \end{bmatrix}. $$ However, this is the same matrix as you started with, the upper right corner's entry hasn't moved.

To get the transpose, must also interchange the blocks.

$$ \begin{bmatrix}P&Q\\R&S\end{bmatrix}^T=\begin{bmatrix}P^T&R^T\\Q^T&S^T\end{bmatrix}. $$

Michael Burr
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$$A=\left[ {\begin{array}{cc}P&Q\\R&S\\\end{array}}\right]^T=\left[ {\begin{array}{cc}P^T&Q^T\\R^T&S^T\\\end{array}}\right]$$is not true.

The first row of the matrix on the left and the first column of the matix on the right should match and they do not.

The correct form is $$A=\left[ {\begin{array}{cc}P&Q\\R&S\\\end{array}}\right]^T=\left[ {\begin{array}{cc}P^T&R^T\\Q^T&S^T\\\end{array}}\right]$$

Mohammad Riazi-Kermani
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The top right element should end-up bottom-left, so no scheme that leaves the blocks at the same location can work.