Identity Matrix

Scalars have something called a multiplicative identity. Multiplying any scalar by the multiplicative identity doesn’t change it. In the case of scalars, it’s the number $1$ .

That is because for any scalar $a$ :

1a = a.

With matrices, there is a similar concept: the identity matrix.

The identity matrix is always a square matrix and is usually denoted by the symbol $I_n$ , where $n$ is the number of rows (and at the same time the number of columns, since it is a square matrix).

$I_n$ is also referred to the identity matrix of order $n$ .

The identity matrix always has $1$ s along its main diagonal (the diagonal going from top-left to bottom-right) and $0$ s in all the other spots. That is

I_n = \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{bmatrix}.

Just like the scalar identity, multiplying a matrix by the identity matrix does not change it. That is, for some matrix $A$ of shape $m \times n$ ,

A I_n = A

and

I_m A = A.

The reason for having to use two different sizes for the identity matrix based on whether it is the first or the second operand in the multiplication is because of matrix conformability for multiplication, which I go over in my lesson about matrix multiplication.

Sometimes the subscript with the dimension of the identity matrix is omitted whenever it is possible to deduce the size from context.

Example

Here is a concrete numerical example to illustrate:

Take the matrix

A = \begin{bmatrix} 2 & -1 \\ 0 & 4 \\ 7 & 2 \end{bmatrix}.

Let’s multiply it by

I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}.

That is

\begin{align*} AI_2 &= \begin{bmatrix} 2 \cdot 1 + (-1) \cdot 0 & 2 \cdot 0 + (-1) \cdot 1 \\ 0 \cdot 1 + 4 \cdot 0 & 0 \cdot 0 + 4 \cdot 1 \\ 7 \cdot 1 + 2 \cdot 0 & 7 \cdot 0 + 2 \cdot 1 \end{bmatrix} \\ &= \begin{bmatrix} 2 & -1 \\ 0 & 4 \\ 7 & 2 \end{bmatrix} \end{align*}

Notice how

AI_2 = A.