Transpose of a Matrix

Definition of the Transpose of a Matrix

If M is an m × n matrix, then the transpose of M , denoted by M^T, is the n × m matrix obtained by interchanging the rows and columns of matrix M .
These are examples of the transpose of matrices.
a)

Matrix A has one row and a size (or order) 1 × 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 × 1 and denoted by A^T is given by


b)

The transpose of matrix B , which has one column and a size 4 × 1 , is obtained by interchanging the column of the matrix into a row. Hence the transpose of matrix B has an order 1 × 4 and denoted by B^T is given by


c)

Matrix C has a size 2 × 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 × 2 and is given by


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 \( A^{-1} = A^T \).
  

Solutions to the Above Questions

• Part 1
  Given matrix \( A = \begin{bmatrix} 1 & 1 & -3 \\ -1 & 0 & 0 \end{bmatrix} \).
  Determine \( A^T \)
  \( A^T = \begin{bmatrix} 1 & -1 \\ 1 & 0 \\ -3 & 0 \end{bmatrix} \)
  
  Calculate \( A^T A \)
  \( A^T A = \begin{bmatrix} 1 & -1 \\ 1 & 0 \\ -3 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 & -3 \\ -1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 1 & -3\\ 1 & 1 & -3\\ -3 & -3 & 9 \end{bmatrix} \)
  
  Calculate \( A A^T \)
  \( A A^T = \begin{bmatrix} 1 & 1 & -3 \\ -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 1 & 0 \\ -3 & 0 \end{bmatrix} = \begin{bmatrix} 11 & -1\\ -1 & 1 \end{bmatrix} \)
  Both matrices \( A^T A \) and \( A A^T \) are symmetric.
  
  
• Part 2
  Calculate \( A+B \)
  \( A+B = \begin{bmatrix} -5 & -2 \\ - 1 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 3 \\ 2 & -3 \end{bmatrix} = \begin{bmatrix} -5 & 1\\ 1 & -1 \end{bmatrix} \)
  Note that the matrix \( A+B \) is symmetric and therefore
  \( (A + B)^T = A + B = \begin{bmatrix} -5 & 1\\ 1 & -1 \end{bmatrix}\)
  
  
• Part 3
  Given matrix \( A = \begin{bmatrix} 0 & \dfrac{5}{3 \sqrt 5} & \dfrac{2}{3} \\ \dfrac{1}{\sqrt 5} & - \dfrac{4}{3 \sqrt 5} & \dfrac{2}{3} \\ \dfrac{2}{\sqrt 5} & \dfrac{2}{3 \sqrt 5} & - \dfrac{1}{3} \end{bmatrix} \)
  Determine \( A^T \)
  \( A^T = \begin{bmatrix} 0 & \dfrac{1}{\sqrt 5} & \dfrac{2}{\sqrt 5}\\ \dfrac{5}{3 \sqrt 5} & - \dfrac{4}{3 \sqrt 5} & \dfrac{2}{3 \sqrt 5} \\ \dfrac{2}{3} & \dfrac{2}{3} &- \dfrac{1}{3} \end{bmatrix} \)
  
  Calculate \( A A^T \)
  \( A A^T = \begin{bmatrix} 0 & \dfrac{5}{3 \sqrt 5} & \dfrac{2}{3} \\ \dfrac{1}{\sqrt 5} & - \dfrac{4}{3 \sqrt 5} & \dfrac{2}{3} \\ \dfrac{2}{\sqrt 5} & \dfrac{2}{3 \sqrt 5} & - \dfrac{1}{3} \end{bmatrix} \begin{bmatrix} 0 & \dfrac{1}{\sqrt 5} & \dfrac{2}{\sqrt 5}\\ \dfrac{5}{3 \sqrt 5} & - \dfrac{4}{3 \sqrt 5} & \dfrac{2}{3 \sqrt 5} \\ \dfrac{2}{3} & \dfrac{2}{3} &- \dfrac{1}{3} \end{bmatrix} \)
  
  \( \quad \quad = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \)
  
  Calculate \( A^T A\)
  \( A^T A = \begin{bmatrix} 0 & \dfrac{1}{\sqrt 5} & \dfrac{2}{\sqrt 5}\\ \dfrac{5}{3 \sqrt 5} & - \dfrac{4}{3 \sqrt 5} & \dfrac{2}{3 \sqrt 5} \\ \dfrac{2}{3} & \dfrac{2}{3} &- \dfrac{1}{3} \end{bmatrix} \begin{bmatrix} 0 & \dfrac{5}{3 \sqrt 5} & \dfrac{2}{3} \\ \dfrac{1}{\sqrt 5} & - \dfrac{4}{3 \sqrt 5} & \dfrac{2}{3} \\ \dfrac{2}{\sqrt 5} & \dfrac{2}{3 \sqrt 5} & - \dfrac{1}{3} \end{bmatrix} \)
  
  \( \quad \quad = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \)
  
  It is shown that \( A A^T = A^T A = I_3 \) and by definition of the inverse matrix, \( A^{-1} = A^T \).
  Note that these type of matrices are called orthogonal matrices.

More References and links

• Matrices with Examples and Questions with Solutions.
  
• Symmetric Matrix.
  
• Determinant of a Square Matrix.
  
• Inverse Matrix Questions with Solutions.
  
• Elementary Linear Algebra - 7 th Edition - Howard Anton and Chris Rorres
• Introduction to Linear Algebra - Fifth Edition (2016) - Gilbert Strang
• Linear Algebra Done Right - third edition, 2015 - Sheldon Axler
• Linear Algebra with Applications - 2012 - Gareth Williams