# Transpose of a Matrix

  

## Definition of the Transpose of a Matrix

If $M$ is an $m \times n$ matrix, then the transpose of $M$, denoted by $M^T$, is the $n \times m$ matrix obtained by interchanging the rows and columns of matrix $M$.
These are examples of the transpose of matrices.
a)
$A = \begin{bmatrix} 1 & -2 & -3\\ \end{bmatrix}$
Matrix $A$ has one row and a size (or order) $1 \times 3$. The transpose of matrix $A$ is obtained by interchanging the row of the matrix into a column. Hence the transpose of matrix $A$ has a size $3 \times 1$ and denoted by $A^T$ is given by
$A^T = \begin{bmatrix} 1\\ -2\\ -3 \end{bmatrix}$

b)
$B = \begin{bmatrix} 9\\ -2\\ 6 \\ 0 \end{bmatrix}$
The transpose of matrix $B$, which has one column and a size $4 \times 1$, is obtained by interchanging the column of the matrix into a row. Hence the transpose of matrix $B$ has an order $1 \times 4$ and denoted by $B^T$ is given by
$B^T = \begin{bmatrix} 9& -2& 6& 0 \end{bmatrix}$

c)
$C = \begin{bmatrix} -3 & 0 & - 2\\ 6 & 4 & 0 \end{bmatrix}$
Matrix $C$ has a size $2 \times 3$. The transpose of this matrix is obtained by interchanging the rows of the matrix into columns (or columns into rows). Hence the transpose $C^T$ of matrix $C$ has an order $3 \times 2$ and is given by
$C^T = \begin{bmatrix} -3 & 6 \\ 0 & 4 \\ - 2 & 0 \end{bmatrix}$

d)
$D = \begin{bmatrix} 2 & - 5 & 9 \\ -7 & 0 & 9 \\ 1 & -2 & 11 \end{bmatrix}$
The transpose of matrix $D$ with order $3 \times 3$ is obtained by interchanging the rows of the matrix into columns (or columns into rows). Hence the transpose $D^T$ of matrix $D$ has an order $3 \times 3$ and is given by
$D^T = \begin{bmatrix} 2 & -7 & 1 \\ - 5 & 0 & -2 \\ 9 & 9 & 11 \end{bmatrix}$
Note that the rows the transpose of a given matrix are the columns of the matrix and the columns of the transpose are the rows of the matrix.

## Properties of the Transpose of Matrices

Some of the most important properties of the transpose of matrices are given below.

1.   $(A^T)^T = A$.
2.   $(AB)^T = B^T A^T$
3.   $(A+B)^T = A^T + B^T$
4.   $(k A)^T = k A^T$ , k is a real number.
5.   $(A^T)^{-1} = (A^{-1})^T$
6.   $Det(A^T) = Det(A)$.
7.   $A^T = A$ if and only if $A$ is a symmetric matrix.
8.   $A^{-1} = A^T$ if and only if $A$ is an orthogonal (square) matrix.

## Examples with Solutions

Example 1
Find the transpose of the matrices:
a) $A = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$      b) $B = \begin{bmatrix} -2 & 0 & 1 \\ \end{bmatrix}$      c) $C = \begin{bmatrix} 5 & -7 \\ 1 & -4 \\ 0 & -1 \\ 7 & -4 \\ \end{bmatrix}$

Solution
We find the transpose of a matrix by interchanging the rows and columns as follows:
a) $A^T = \begin{bmatrix} 0 & 1\\ 0 & 0 \\ 1 & 0 \end{bmatrix}$      b) $B^T = \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}$      c) $C^T = \begin{bmatrix} 5 & 1 & 0 & 7 \\ -7 & -4 & -1 & -4 \end{bmatrix}$

Example 2
Matrices $A$ and $B$ are given by $A = \begin{bmatrix} -1 & 0 & 1 \\ 1 & 2 & 0 \end{bmatrix}$ , $B = \begin{bmatrix} -1 \\ 1 \\ 2 \end{bmatrix}$.
Show that $(AB)^T = B^T A^T$ (verify property 2 above).

Solution
Calculate $AB$
$AB = \begin{bmatrix} -1 & 0 & 1 \\ 1 & 2 & 0 \end{bmatrix} \begin{bmatrix} -1 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \end{bmatrix}$
Determine $(AB)^T$
$(AB)^T = \begin{bmatrix} 3 & 1 \end{bmatrix}$

Determine $A^T$ and $B^T$
$A^T = \begin{bmatrix} -1 & 1 \\ 0 & 2 \\ 1 & 0 \end{bmatrix}$ , $B^T = \begin{bmatrix} -1 & 1 & 2 \end{bmatrix}$
Calculate $B^T A^T$
$B^T A^T = \begin{bmatrix} -1 & 1 & 2 \end{bmatrix} \begin{bmatrix} -1 & 1 \\ 0 & 2 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 3 & 1 \end{bmatrix}$
Hence $(AB)^T = B^T A^T$

Example 3
Let matrix $A = \begin{bmatrix} -1 & 0\\ 1 & 2 \end{bmatrix}$
Show that $(A^T)^{-1} = (A^{-1})^T$ (verify property 5 above).

Solution
Determine $A^T$
$A^T = \begin{bmatrix} -1 & 1\\ 0 & 2 \end{bmatrix}$
Use the formula of the inverse of a $2 \times 2$ matrix $\begin{bmatrix} x & y \\ z & w \\ \end{bmatrix}^{-1} = \dfrac{1}{xw - yz} \begin{bmatrix} w & - y \\ - z & x \\ \end{bmatrix}$ to find $(A^T)^{-1}$
$(A^T)^{-1} = -\dfrac{1}{2} \begin{bmatrix} 2 & 0\\ 1 & -1 \end{bmatrix}$

$\quad \quad = \begin{bmatrix} -1 & 0\\ -\dfrac{1}{2} & \dfrac{1}{2} \end{bmatrix}$

Use the same formula of the inverse of a $2 \times 2$ matrix given above to find $A^{-1}$
$A^{-1} = -\dfrac{1}{2} \begin{bmatrix} 2 & 1\\ 0 & -1 \end{bmatrix}$

$\quad \quad = \begin{bmatrix} -1 & -\dfrac{1}{2}\\ 0 & \dfrac{1}{2} \end{bmatrix}$

We now determine $(A^{-1}) ^T$
$(A^{-1}) ^T = \begin{bmatrix} -1 & 0 \\ -\dfrac{1}{2} & \dfrac{1}{2} \end{bmatrix}$

Hence we conclude that $(A^T)^{-1} = (A^{-1})^T$.

Example 4
Let matrix $A = \begin{bmatrix} -1 & 0 & 1\\ 1 & 2 & 2 \\ -2 & 0 & 1 \end{bmatrix}$
Show that $Det(A^T) = Det(A)$ (verify property 6 above)

Solution
Calculate $Det(A)$ using the upper row
Calculate $Det(A) = -1 ( 2 \times 1 - 0) + 1 (0 - 2(-2) ) = 2$

Determine $A^T$
$A^T \begin{bmatrix} -1 & 1 & -2 \\ 0 & 2 & 0 \\ 1 & 2 & 1 \end{bmatrix}$
Calculate $Det(A^T)$ using the leftmost column
$Det(A^T) = -1 ( 2 \times 1 - 0) + 1 ( 0 - 2(-2)) = 2$
Hence we conclude that $Det(A^T) = Det(A)$

Example 5
Use property 7 above to show that matrix $A = \begin{bmatrix} 0 & -1 & 3\\ -1 & 2 & 5 \\ 3 & -5 & 1 \end{bmatrix}$ is not a symmetric matrix.

Solution
Determine $A^T$
$A^T = \begin{bmatrix} 0 & -1 & 3 \\ -1 & 2 & -5 \\ 3 & 5 & 1 \end{bmatrix}$
We can verify that the entry at $A_{2,3} = 5$ and the entry at $A^T_{2,3} = - 5$, therefore matrices $A^T$ and $A$ are not equal which means that matrix $A$ is not symmetric.

## Questions (with solutions given below)

• Part 1
Given matrix $A = \begin{bmatrix} 1 & 1 & -3 \\ -1 & 0 & 0 \end{bmatrix}$, calculate and show that the matrices $A^T A$ and $A A^T$ are both symmetric.
• Part 2
Given the matrices $A = \begin{bmatrix} -5 & -2 \\ - 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 3 \\ 2 & -3 \end{bmatrix}$ , find the matrix $(A+B)^T$
• Part 3
Let matrix $A = \begin{bmatrix} 0 & \dfrac{5}{3 \sqrt 5} & \dfrac{2}{3} \\ \dfrac{1}{\sqrt 5} & - \dfrac{4}{3 \sqrt 5} & \dfrac{2}{3} \\ \dfrac{2}{\sqrt 5} & \dfrac{2}{3 \sqrt 5} & - \dfrac{1}{3} \end{bmatrix}$. Show that $A A^T = A^T A = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and therefore $A^{-1} = A^T$.