The transpose of a matrix is denoted by
Transpose Of A Matrix
The element in
\begin{aligned} &(A^T)_{ij} = (A)_{ji} \end{aligned}
Let
\begin{aligned} A = \begin{bmatrix}a & b\\ c & d\\ e & f\end{bmatrix} \end{aligned}
Transpose of matrix
\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
Similarly, the element ‘b’ is at the position of first row and second column of matrix
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
\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
\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
All the elements above the diagonal is a mirror image of elements below the diagonal elements. That is,
\begin{aligned} A = \begin{bmatrix}a_{11} & p & q\\p & a_{22} & r\\q & r & a_{33}\end{bmatrix} \end{aligned}
The elements of
\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.
, where is a matrix of size or . , where and are of same size, that is, or . , where is matrix of size or and is a real number. , where and are matrices of size and .
Let us verify each of the statement.
#1 :
The transpose of a transpose of matrix
\begin{aligned} Let \hspace{5px}A = \begin{bmatrix}2 & 3\\-1 & 5\end{bmatrix} \end{aligned}
Transpose of matrix
\begin{aligned} A^T = \begin{bmatrix}2 & -1\\3 & 5\end{bmatrix} \end{aligned}
Transpose of
\begin{aligned} (A^T)^T = \begin{bmatrix}2 & 3\\-1 & 5\end{bmatrix} \end{aligned}
From the results above, it is clear that
#2 :
The transpose of sum of two matrices
Let
\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
\begin{aligned} (A + B)^T = \begin{bmatrix}3 & 3\\6 & 2\end{bmatrix} \end{aligned}
Now, we shall take transpose of matrix
\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
\begin{aligned} A^T + B^T = \begin{bmatrix}3 & 3\\6 & 2\end{bmatrix} \end{aligned}
#3 :
A transpose of the product of matrix
\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
\begin{aligned} (rA)^T = \begin{bmatrix}4 & 2\\6 & 14\end{bmatrix} \end{aligned}
Similarly, let us take transpose of
\begin{aligned} A^T = \begin{bmatrix}2 & 1\\3 & 7\end{bmatrix} \end{aligned}
The product
\begin{aligned} rA^T = \begin{bmatrix}4 & 2\\6 & 14\end{bmatrix} \end{aligned}
Therefore,
The output of both the products are equal and the property
#4 :
The transpose of product of two defined (
\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
\begin{aligned} (AB)^T = \begin{bmatrix}13 & 8\\14 & 1\end{bmatrix} \end{aligned}
Similarly, the transpose of matrix
\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
In the next, post we will discuss more about symmetric and skew-symmetric matrices.