Matrix Transpose

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.

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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.

  1. (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, where A and B are of same size, that is, m \times n or n \times n.
  3. (rA)^T = rA^T, where A is matrix of size m \times n or n \times n and r is a real number.
  4. (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.

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\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.

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