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I'm attempting to prove that

$$ \left[ \begin{array}{c c} A & B \\ C & D \\ \end{array} \right]^\top = \left[ \begin{array}{c c} A^\top & C^\top \\ B^\top & D^\top \\ \end{array} \right]. $$

Intuitively, I can see that it's true. However, when I try to formally prove it, I quickly get lost in the indices. What tricks can I use to keep things straight?

Source: Exercise 2.6.16, P116, Intro to Linear Algebra, 4th Ed by Strang

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up vote 6 down vote accepted

Most people would just claim this is obvious and omit the proof, but if you don't want to do that then perhaps you could first prove that \begin{equation} \begin{bmatrix} M & N \end{bmatrix}^T = \begin{bmatrix} M^T \\ N^T \end{bmatrix} \end{equation} and \begin{equation} \begin{bmatrix} M \\ N \end{bmatrix}^T = \begin{bmatrix} M^T & N^T \end{bmatrix}. \end{equation} Then \begin{align} \begin{bmatrix} A & B \\ C & D \end{bmatrix}^T &= \begin{bmatrix} \begin{bmatrix} A \\ C \end{bmatrix}^T \\ \begin{bmatrix} B \\ D \end{bmatrix}^T \end{bmatrix} \\ &= \begin{bmatrix} A^T & C^T \\ B^T & D^T \end{bmatrix}. \end{align}

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Definition of Transpose is $(A^T)_{ij} = A_{ji}$

Why $\begin{bmatrix} M & N \end{bmatrix}^T = \begin{bmatrix} M^T \\ N^T \end{bmatrix}$? Why NOT $\begin{bmatrix} M & N \end{bmatrix}^T = \begin{bmatrix} M \\ N\end{bmatrix}$?

After transpose, $M$ is in (1, 1) position and $N$ is in (2,1) position. Why still keep the $^T$?

Why $\begin{bmatrix} M \\ N \end{bmatrix}^T = \begin{bmatrix} M^T & N^T \end{bmatrix}$? Why NOT $\begin{bmatrix} M \\ N \end{bmatrix}^T = \begin{bmatrix} M & N \end{bmatrix}$ ?

After transpose - $M$ is in (1, 1) position and $N$ is in (1,2) position. Why still keep the $^T$?

Example $\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}^T = \begin{bmatrix} a_{11}^T & a_{12}^T \\ a_{21}^T & a_{22}^T \end{bmatrix} = \begin{bmatrix} a_{11} & a_{21} \\ a_{12} & a_{22} \end{bmatrix}$. No $^T$ at the end.

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In your example at the end, the $a$s are scalar values, so $a_{11}^T = a_{11}$. But the question is about block matrices and therefore $M = M^T$ doesn’t need to be true, so you cannot omit the $^T$. –  Eike Schulte Oct 1 '13 at 9:14
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