Notation - Transpose of Block Matrices [Lay P121 Q2.4.12]

Question

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

$1.$ Why $\begin{bmatrix} M & N \end{bmatrix}^T = \begin{bmatrix} M^T \\ N^T \end{bmatrix}$, and NOT $\begin{bmatrix} M \\ N\end{bmatrix}$?

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

For example, pursuant to Git Gud's comment, I tried $\begin{bmatrix} \begin{bmatrix} 1 & 5 \\ 3 & 7 \\ \end{bmatrix} & \begin{bmatrix} 2 & 6 \\ 4 & 8 \\ \end{bmatrix} \end{bmatrix}^T $.
This means transposing the entries (the two 2 by 2 matrices) once, so: $ \begin{bmatrix} \begin{bmatrix} 1 & 5 \\ 3 & 7 \\ \end{bmatrix} \\ \begin{bmatrix} 2 & 6 \\ 4 & 8 \\ \end{bmatrix} \end{bmatrix}$ ?

$2.$ Why $\begin{bmatrix} M \\ N \end{bmatrix}^T = \begin{bmatrix} M^T & N^T \end{bmatrix}$, and NOT $ \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 $^T$?

Per contra, this example contains not one $^T$ at the end. I recast it as a question.
I see user Eike Schulte's comment and brook that $c^T = c$ for all complex numbers $c$. $\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}^T = \begin{bmatrix} a_{11}^T & a_{12}^T \\ a_{21}^T & a_{22}^T \end{bmatrix} = \begin{bmatrix} a_{11} & a_{21} \\ a_{12} & a_{22} \end{bmatrix}$

Answer 1

Given two matrices $M_{m\times n},N_{m\times p}$, there are two ways to interpret the entity $\begin{bmatrix} M & N \end{bmatrix}$.

One is the $m\times (n+p)$ matrix whose $(i,j)$ entry is $\begin{cases} (M)_{(i,j)}, &\text{if }j\leq n\\ (N)_{(i, j-n)}, &\text{if }j\ge n+1\end{cases}$.

In this case I'd rather denote the matrix described above as $\begin{bmatrix} M \mid N \end{bmatrix}$, (the augmented matrix). This is standard notation.

The other is a $1\times 2$ matrix whose first entry is the matrix $M$ and whose second entry is the matrix $N$.

Under the first interpretation one has $$\begin{bmatrix} M & N \end{bmatrix}^T=\begin{bmatrix} M \mid N \end{bmatrix}^T=\begin{bmatrix} M^T \\ \overline{N^T} \end{bmatrix}.$$

Under the second interpretation one has $$\begin{bmatrix} M & N \end{bmatrix}^T=\begin{bmatrix} M \\ N \end{bmatrix}.$$

The second interpretation is very uncommon. Most of the time it's safe to assume one is under the first interpretation.

An example: Let $M=\begin{bmatrix} 1 & 0 & 1\\ 2 & 3 & 5\end{bmatrix}_{2\times 3}$ and $N=\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix}_{2\times 2}$.

The first interpretation yields the matrix $A$ where $$(A)_{ij}=\begin{cases} (M)_{(i,j)}, &\text{if }j\leq 3\\ (N)_{(i, j-n)}, &\text{if }j\ge 4\end{cases}, \text{ for all }(i,j)\in \{1,2\}\times\{1,2,3,4,5\}.$$

That is $$ A=\left[\begin{array}{ccc|cc} (M)_{11} & (M)_{12} & (M)_{13} & (N)_{14} & (N)_{15}\\ (M)_{21} & (M)_{22} & (M)_{23} & (N)_{24} & (N)_{25} \end{array}\right]=\left[\begin{array}{ccc|cc} 1 & 0 & 1 & 1 & 1\\ 2 & 3 & 5 & 0 & 1 \end{array}\right].$$

Transposing yields $A^T=\left[\begin{array}{cc}1 & 2\\ 0 & 3\\ 1 & 5\\ \hline 1 & 0\\ 1 & 1 \end{array}\right]=\left[\begin{array}{c}M^T\\ \hline N^T \end{array}\right]$.

The second interpretation gives $A_{1\times 2}=\left[\begin{matrix} (A)_{11} & (A)_{12}\end{matrix}\right]_{1\times 2}$ where $(A)_{11}=M$ and $(A)_{12}=N$, transposing: $$\left(A^T\right)_{2\times 1}=\begin{bmatrix} (A)_{11}\\ (A)_{21}\end{bmatrix}_{2\times 1}=\begin{bmatrix} M\\ N\end{bmatrix}_{2\times 1}=\begin{bmatrix}\begin{bmatrix} 1 & 0 & 1\\ 2 & 3 & 5\end{bmatrix}_{2\times 3}\\ \begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix}_{2\times 2} \end{bmatrix}_{2\times 1}.$$

Here the entries in the matrix just happen to be matrices themselves, you can create matrices in which their entries are whatever you want. For instance, $\begin{bmatrix} 1 & \begin{bmatrix} 1 & 2\\ 3 & 4\end{bmatrix} & \spadesuit\\ \implies & \huge{〠} & +\end{bmatrix}$ is a matrix.

Answer 2

If you look at row of people lined up in front of a mirror, not only will their mirror images appear in the opposite order (the image of the person closest to the mirror comes first) but also the image of each individual person will be a mirror image (if facing towards the mirror, the mirror image will be facing out of the mirror) just as if the person were alone. The same applies to transposition of a collection of blocks.