Recall that a square matrix with 1's on the diagonal (starting from the top-left down to the bottom-right) and 0's everywhere else is called an identity matrix. In this segment, we will look at some properties of this matrix. But first, let us define the identity matrix more formally.
Let \(\mathbb{F}\) denote a field. Let \(n\) be a positive integer. The \(n\times n\) identity matrix over \(\mathbb{F}\), denoted \(I_n\), is the matrix such that \[ (I_n)_{i,j} = \left \{ \begin{array}{ll} 1 & \text{if } i = j \\ 0 & \text{otherwise}. \end{array}\right.\] For example, \(I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}\) and \(I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\).
Often times, the subscript \(n\) is omitted (i.e. \(I\) is written in place of \(I_n\)) if it is clear from the context what the dimension of the identity matrix is.
One important property of the identity matrix is the following:
If \(A \in \mathbb{F}^{n\times m}\) and \(B \in \mathbb{F}^{p\times n}\), then \[I_n A = A~~~~\text{and }~~~~B~I_n= B.\] In other words, the identity matrix acts like the multiplicative identity in a field.
\[\begin{bmatrix} 1& 0 \\ 0 & 1\end{bmatrix} \begin{bmatrix} a & b \\ c & d\end{bmatrix} = \begin{bmatrix} 1\cdot a+ 0\cdot c & 1\cdot b + 0 \cdot d \\ 0\cdot a+ 1\cdot c & 0 \cdot b + 1 \cdot d \end{bmatrix} = \begin{bmatrix} a & b \\ c & d\end{bmatrix} \]
We now prove that \(I_n A = A\). The proof that \(B~I_n = B\) is left as an exercise.
Let \(C = I_n A\). Recall that \(C_{i,j}\) is given by the \(i\)th row of \(I_n\) multiplied by the \(j\)th column of \(A\). That is \[C_{i,j} = \sum_{p = 1}^n (I_n)_{i,p} A_{p,j}.\] Since \((I_n)_{i,p} = 0\) whenever \(p \neq i\), we get \[C_{i,j} = (I_n)_{i,i} A_{i,j} = A_{i,j}.\] In other words, the \((i,j)\)-entry of \(C\) is precisely the \((i,j)\)-entry of \(A\) for all \(1\leq i \leq n\) and \(1 \leq j \leq m\). Thus, \(C = A\) as desired.
Let \(A = \begin{bmatrix} 2 & 0 \\ 0 & -3 \end{bmatrix}\). Find a matrix \(B\) such that \(A B = B A = I_2\).
Let \(\mathbb{F}\) denote a field. Let \(p\) and \(n\) be positive integers. Prove that if \(B \in \mathbb{F}^{p\times n}\), then \(B~I_n = B\).