Suppose that \(n\times m\) matrix A is given. Then the transposed matrix of A is \(m\times n\) matrix whose column are formed from the corresponding rows of A, and we denote the transposed matrix by \({}^{t}A\).
$$A=\begin{bmatrix}a_{11} &a_{12}&\cdots&a_{1m}\\a_{21} &a_{22}&\cdots&a_{2m}\\\vdots &\vdots & \ddots & \vdots\\a_{n1} &a_{n2}&\cdots&a_{nm}\\\end{bmatrix}\ \ \ \ \ \Longrightarrow\ \ \ \ \ A^{t}=\begin{bmatrix}a_{11} &a_{21}&\cdots&a_{n1}\\a_{12} &a_{22}&\cdots&a_{n2}\\\vdots &\vdots & \ddots & \vdots\\a_{1m} &a_{2m}&\cdots&a_{nm}\\\end{bmatrix}$$
Let us denote the \((i,j)\)-entry in A by \([A]_{ij}\). Then we have
$$[A]_{ij}=a_{ij}\ \ \text{and }\ \ [A^{t}]_{ij}=[A]_{ji}=a_{ji}$$
Example
Let $$A=\begin{bmatrix}a&b\\c&d\\\end{bmatrix}, \ \ \ B=\begin{bmatrix}1&2&3\\4&5&6\\\end{bmatrix}, \ \ \ C=\begin{bmatrix}1&2\\3&4\\5&6\\\end{bmatrix}$$
Then$$A^{t}=\begin{bmatrix}a&c\\b&d\end{bmatrix},\ \ \ B^{t}=\begin{bmatrix}1&4\\2&5\\3&6\\\end{bmatrix},\ \ \ C^{t}=\begin{bmatrix}1&3&5\\2&4&6\\\end{bmatrix}$$
Properties
| \((A^{t})^{t}=A\) | Proof |
| \((A+B)^{t}=A^{t}+B^{t}\) | Proof |
| \((kA)^{t}=kA^{t}\) | Proof |
| \((AB)^{t}=B^{t}A^{t}\) | Proof |
| \((A^{t})^{-1}=(A^{-1})^{t}\) | Proof |