Let $A=({a}_{ij})$ be a $m\times n$ matrix. The transpose of $A$, written ${A}^{T}$, is the $n\times m$ matrix whose $i,j$ entry is ${a}_{ji}$.
You can think of the transpose as being obtained by reflecting $A$ in the south east diagonal starting in the top left hand corner, or as the matrix whose columns are the transposes of the rows of $A$, or the matrix whose rows are the transposes of the columns of $A$.
If $A=\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right)$ then ${A}^{T}=\left(\begin{array}{cc}1& 4\\ 2& 5\\ 3& 6\end{array}\right)$.
If $A=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)$ then ${A}^{T}=\left(\begin{array}{cc}1& 3\\ 2& 4\end{array}\right)$.
If $A=\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)$ then ${A}^{T}=\left(\begin{array}{ccc}1& 2& 3\end{array}\right)$.
It’s common to use transposes when we want to think geometrically, because if $\mathbf{x}\in {\mathbb{R}}^{n}$ then ${\mathbf{x}}^{T}\mathbf{x}$ is equal to
$${x}_{1}^{2}+{x}_{2}^{2}+\mathrm{\cdots}+{x}_{n}^{2}$$ |
which is the square of the length of $\mathbf{x}$. (As usual, we have identified the $1\times 1$ matrix ${\mathbf{x}}^{T}\mathbf{x}$ with a number here).
When $\mathbf{z}$ is a complex column vector, that is, an element of ${\u2102}^{n}$ for some $n$, this doesn’t quite work. If $\mathbf{z}=\left(\begin{array}{c}1\\ i\end{array}\right)$ for example, then ${\mathbf{z}}^{T}\mathbf{z}=0$, which is not a good measure of the length of $\mathbf{z}$. For this reason, when people work with complex vectors they often use the conjugate transpose ${A}^{H}$ defined to be the matrix whose entries are the complex conjugates of the entries of ${A}^{T}$. With this definition, for a complex vector $\mathbf{z}=\left(\begin{array}{c}{z}_{1}\\ \mathrm{\vdots}\\ {z}_{n}\end{array}\right)$ we get
$${\mathbf{z}}^{H}\mathbf{z}={|{z}_{1}|}^{2}+\mathrm{\cdots}+{|{z}_{n}|}^{2}.$$ |