Transpose of a Matrix

Activity:

Let us consider a matrix:

$$A = \begin{bmatrix} 2 & 3 & 4 \\ 5 & 6 & 7 \end{bmatrix}$$

Observation:

Matrix $A$ has 2 rows and 3 columns
The order of matrix $A$ is $2 \times 3$
It is a rectangular matrix (as rows ≠ columns)

Thinking:

What will happen if we change (write) element of rows as columns and vice-versa?

$$\begin{bmatrix} 2 & 5 \\ 3 & 6 \\ 4 & 7 \end{bmatrix} $$

Key Observation:

A new matrix is formed with reverse order $3 \times 2$

Conclusion:

We call this new matrix a transpose of the given matrix $A$.

Definition:

A new matrix formed by interchanging rows and columns of a given matrix is called transpose of a matrix.

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

Its transpose is written as $A'$ or $A^t$ with order $n \times m$.

Properties of Transpose of Matrices

Property 1:

$(A + B)^t = A^t + B^t$

$$\text{Let } A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 6 \\ 7 & 9 \end{bmatrix}$$

$$A + B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} + \begin{bmatrix} 1 & 6 \\ 7 & 9 \end{bmatrix} = \begin{bmatrix} 2+1 & 3+6 \\ 4+7 & 5+9 \end{bmatrix} = \begin{bmatrix} 3 & 9 \\ 11 & 14 \end{bmatrix}$$

$$(A + B)^t = \begin{bmatrix} 3 & 11 \\ 9 & 14 \end{bmatrix}$$

$$A^t = \begin{bmatrix} 2 & 4 \\ 3 & 5 \end{bmatrix}, \quad B^t = \begin{bmatrix} 1 & 7 \\ 6 & 9 \end{bmatrix}$$

$$A^t + B^t = \begin{bmatrix} 2 & 4 \\ 3 & 5 \end{bmatrix} + \begin{bmatrix} 1 & 7 \\ 6 & 9 \end{bmatrix} = \begin{bmatrix} 3 & 11 \\ 9 & 14 \end{bmatrix}$$

$$\therefore (A + B)^t = A^t + B^t \,\checkmark$$

Property 2:

$(A - B)^t = A^t - B^t$

$$\text{Let } A = \begin{bmatrix} 6 & 2 \\ 3 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 4 & 5 \\ 0 & 7 \end{bmatrix}$$

$$A - B = \begin{bmatrix} 6 & 2 \\ 3 & 1 \end{bmatrix} - \begin{bmatrix} 4 & 5 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 6-4 & 2-5 \\ 3-0 & 1-7 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ 3 & -6 \end{bmatrix}$$

$$(A - B)^t = \begin{bmatrix} 2 & 3 \\ -3 & -6 \end{bmatrix}$$

$$A^t = \begin{bmatrix} 6 & 3 \\ 2 & 1 \end{bmatrix}, \quad B^t = \begin{bmatrix} 4 & 0 \\ 5 & 7 \end{bmatrix}$$

$$A^t - B^t = \begin{bmatrix} 6 & 3 \\ 2 & 1 \end{bmatrix} - \begin{bmatrix} 4 & 0 \\ 5 & 7 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ -3 & -6 \end{bmatrix}$$

$$\therefore (A - B)^t = A^t - B^t \,\checkmark$$

Property 3:

$(KA)^t = KA^t$ (where $K$ is a scalar)

$$\text{Let } K = 2, \quad A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$

$$KA = 2 \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix}$$

$$(KA)^t = \begin{bmatrix} 2 & 6 \\ 4 & 8 \end{bmatrix}$$

$$A^t = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$$

$$KA^t = 2 \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 6 \\ 4 & 8 \end{bmatrix}$$

$$\therefore (KA)^t = KA^t \,\checkmark$$

Property 4:

$(A^t)^t = A$

$$\text{Let } A = \begin{bmatrix} 3 & 4 \\ 5 & 6 \end{bmatrix}$$

$$A^t = \begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}$$

$$(A^t)^t = \begin{bmatrix} 3 & 4 \\ 5 & 6 \end{bmatrix} = A$$

$$\therefore (A^t)^t = A \,\checkmark$$

Property 5:

If $A = A^t$, then the matrix is called a symmetric matrix.

$$\text{Let } A = \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix}$$

$$A^t = \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix} = A$$

$$\therefore A = A^t \,\checkmark$$

Note: If a matrix is equal to its transpose then such a matrix are called symmetric matrix.

Property 6:

Transpose of Identity Matrix and Null Matrix:

$$\text{Let } I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \text{ (Identity matrix)}$$

$$I^t = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$

Transpose of an identity matrix is identity matrix itself.

$$\text{Let } N = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \text{ (Null matrix)}$$

$$N^t = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = N$$

Transpose of null matrix is null matrix itself.