# Transpose of block matrix

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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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 $$\begin{bmatrix} M & N \end{bmatrix}^T = \begin{bmatrix} M^T \\ N^T \end{bmatrix}$$ and $$\begin{bmatrix} M \\ N \end{bmatrix}^T = \begin{bmatrix} M^T & N^T \end{bmatrix}.$$ 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