Matrix Transpose - Notesformsc

Table of Contents

The transpose of a matrix is denoted by $A^T$ is obtained by changing rows into columns or columns to rows of a matrix $A$ . If size of the matrix $A$ is $m \times n$ then the size of the transposed matrix $A^T$ is $n \times m$ .

Transpose Of A Matrix

The element in $i^{th}$ row and $j^{th}$ column of matrix $A$ becomes the $j^{th}$ row and $i^{th}$ column element in matrix $A^T$ .

\begin{aligned} &(A^T)_{ij} = (A)_{ji} \end{aligned}

Let $A$ be a matrix of size $m \times n$ .

\begin{aligned} A = \begin{bmatrix}a & b\\ c & d\\ e & f\end{bmatrix} \end{aligned}

Transpose of matrix $A$ .

\begin{aligned} A^T = \begin{bmatrix}a & c & e\\ b & d & f\end{bmatrix} \end{aligned}

Let us take element ‘c’ which is at 2nd row and 1st column of matrix $A$ ; after transpose operation on matrix $A$ , it is at the position of 1st row and 2nd column of matrix $A^T$ .

Similarly, the element ‘b’ is at the position of first row and second column of matrix $A$ , but after the transpose operation, its position changes to 2nd row and 1st column in matrix $A^T$ .

Example #1

Transpose the following matrix A.

\begin{aligned} A = \begin{bmatrix}3 & 1 & 5\\ 2 & 6 & 9\end{bmatrix} \end{aligned}

The transpose of matrix $A$ is

\begin{aligned} A^T = \begin{bmatrix}3 & 2\\ 1 & 6\\5 & 9\end{bmatrix} \end{aligned}

Example #2

Transpose the following matrix B.

\begin{aligned} &A = \begin{bmatrix}1 & 5\\ 7 & 6\\8 & 4\end{bmatrix} \end{aligned}

The transpose of matrix $A^T$ is

\begin{aligned} A^T = \begin{bmatrix}1 & 7 & 8\\ 5 & 6 & 4\end{bmatrix} \end{aligned}

Symmetric Matrix

When the transpose of the matrix is the original matrix itself, then it is called a Symmetric matrix. Suppose $A$ is a matrix of size $m \times n$ , then the transpose of matrix $A = A^T$ .

All the elements above the diagonal is a mirror image of elements below the diagonal elements. That is, $A = [a_{ij}]_{m \times n}$ is symmetric matrix if $(A)_{ij} = (A)_{ji}$ for all i and j.

\begin{aligned} A = \begin{bmatrix}a_{11} & p & q\\p & a_{22} & r\\q & r & a_{33}\end{bmatrix} \end{aligned}

The elements of $(A)<em>{ij} = (A)</em>{ji}$ . The transpose of such a matrix is,

\begin{aligned} A^T = \begin{bmatrix}a_{11} & p & q\\p & a_{22} & r\\q & r & a_{33}\end{bmatrix} \end{aligned}

Therefore,

\begin{aligned} A = A^T \end{aligned}

What Are The Properties Of A Transpose Of A Matrix ?

In this section, we shall discuss about the properties of a transpose of a matrix. There are 4 interesting properties of a transpose as listed below.

$(A^T)^T = A$ , where $A$ is a matrix of size $m \times n$ or $n \times n$ .
$(A + B)^T = A^T + B^T$ , where $A$ and $B$ are of same size, that is, $m \times n$ or $n \times n$ .
$(rA)^T = rA^T$ , where $A$ is matrix of size $m \times n$ or $n \times n$ and $r$ is a real number.
$(AB)^T = B^TA^T$ , where $A$ and $B$ are matrices of size $A_{m \times n}$ and $B_{n \times p}$ .

Let us verify each of the statement.

#1 : $(A^T)^T = A$

The transpose of a transpose of matrix $A$ is the original matrix $A$ .

\begin{aligned} Let \hspace{5px}A = \begin{bmatrix}2 & 3\\-1 & 5\end{bmatrix} \end{aligned}

Transpose of matrix $A$ .

\begin{aligned} A^T = \begin{bmatrix}2 & -1\\3 & 5\end{bmatrix} \end{aligned}

Transpose of $A^T$ .

\begin{aligned} (A^T)^T = \begin{bmatrix}2 & 3\\-1 & 5\end{bmatrix} \end{aligned}

From the results above, it is clear that $(A^T)^T = A$ where $A$ is a matrix of size $m \times n$ or $n \times n$ .

#2 : $(A + B)^T = A^T + B^T$

The transpose of sum of two matrices $A$ and $B$ of same size $m \times n$ or $n \times n$ is equal to sum of transpose of matrices $A$ and $B$ .

Let $A$ and $B$ be two matrices of same size. Then

\begin{aligned} &A = \begin{bmatrix}1 & 5\\-2 & 3\end{bmatrix} B = \begin{bmatrix}2 & 1\\5 & -1\end{bmatrix}\\\\ &(A + B) = \begin{bmatrix}3 & 6\\3 & 2\end{bmatrix} \end{aligned}

Transpose of matrix $(A + B)^T$ .

\begin{aligned} (A + B)^T = \begin{bmatrix}3 & 3\\6 & 2\end{bmatrix} \end{aligned}

Now, we shall take transpose of matrix $A$ and matrix $B$ and add them together to obtain $A^T + B^T$ .

\begin{aligned} A = \begin{bmatrix}1 & 5\\-2 & 3\end{bmatrix} B = \begin{bmatrix}2 & 1\\5 & -1\end{bmatrix} \end{aligned}

Transpose of A.

\begin{aligned} A^T = \begin{bmatrix}1 & -2\\5 & 3\end{bmatrix} \end{aligned}

Transpose of B.

\begin{aligned} B^T = \begin{bmatrix}2 & 5\\1 & -1\end{bmatrix} \end{aligned}

Sum of $A^T$ and $B^T$ .

\begin{aligned} A^T + B^T = \begin{bmatrix}3 & 3\\6 & 2\end{bmatrix} \end{aligned}

#3 : $(rA)^T = rA^T$

A transpose of the product of matrix $A$ with scalar $r$ is equal to the product of scalar $r$ and transpose of matrix $A$ where size of the matrix $A$ is $m \times n$ or $n \times n$ and $r$ is a real number.

\begin{aligned} &Let \hspace{5px} A = \begin{bmatrix}2 & 3\\1 & 7\end{bmatrix} \hspace{4px} and \hspace{5px} r = 2\\\\ &(rA) = \begin{bmatrix}4 & 6\\2 & 14\end{bmatrix} \end{aligned}

Transpose of $rA$ ,

\begin{aligned} (rA)^T = \begin{bmatrix}4 & 2\\6 & 14\end{bmatrix} \end{aligned}

Similarly, let us take transpose of $A$ .

\begin{aligned} A^T = \begin{bmatrix}2 & 1\\3 & 7\end{bmatrix} \end{aligned}

The product $rA^T$ is,

\begin{aligned} rA^T = \begin{bmatrix}4 & 2\\6 & 14\end{bmatrix} \end{aligned}

Therefore, $(rA)^T = rA^T$ .

The output of both the products are equal and the property $(rA)^T = rA^T$ is true for all matrices.

#4 : $(AB)^T = B^TA^T$

The transpose of product of two defined ( $A_{m \times n}$ and $B_{n \times p}$ ) matrices $A$ and $B$ is equal to the product of transpose of matrix $B$ and transpose of matrix $A$ . Let us verify this claim with the help of an example.

\begin{aligned} &Let \hspace{5px}A = \begin{bmatrix}1 & 5\\2 & 1\end{bmatrix} \hspace{5px} and \hspace{5px} B = \begin{bmatrix}3 & -1\\2 & 3\end{bmatrix}\\\\ &AB = \begin{bmatrix}3 + 10 & -1 + 15\\6 + 2 & -2 + 3\end{bmatrix}= \begin{bmatrix}13 & 14\\8 & 1\end{bmatrix} \end{aligned}

Transpose of $AB$ .

\begin{aligned} (AB)^T = \begin{bmatrix}13 & 8\\14 & 1\end{bmatrix} \end{aligned}

Similarly, the transpose of matrix $B$ and matrix $A$ is,

\begin{aligned} &B^T = \begin{bmatrix}3 & 2\\-1 & 3\end{bmatrix} \hspace{5px} and \hspace{5px} A^T = \begin{bmatrix}1 & 2\\5 & 1\end{bmatrix}\\\\ &B^TA^T = \begin{bmatrix}3 + 10 & 6 + 2\\-1 + 15 & -2 + 3\end{bmatrix}\\\\ &B^TA^T = \begin{bmatrix}13 & 8\\14 & 1\end{bmatrix}\\\\ &Therefore, \hspace{5px}(AB)^T = B^TA^T \end{aligned}

Once again, the product of both sides of the equation of the property $(AB)^T = B^TA^T$ holds true. The property is valid.

In the next, post we will discuss more about symmetric and skew-symmetric matrices.