identity matrix

If \[A\] is a matrix with \[n\] rows then \[I_{n}A=A\]. Similarly, if \[A\] is a matrix with \[n\] columns, then \[AI_{n}=A\].

To prove the first statement, let \[A\] be a \[n\times p\] matrix. We need to show that for \[1\le i\le n\] and \[1\le j\le p\], the entry in the \[i\]th row and \[j\]th column of the product \[I_{n}A\] is equal to the entry in the \[i\]th row and \[j\]-th column of \[A\].

\begin{align*} \begin{pmatrix} & & &0& \\ & &\ddots&\vdots& \\ 0&\cdots&0&1&0\\ & & &\vdots& \\ & & &0&\\ \end{pmatrix} \begin{pmatrix} & A_{1j}\\ & \vdots\\ \cdots & A_{ij}\\ &\vdots\\ &A_{nj}\\ \end{pmatrix} &= \begin{pmatrix} & & &\vdots& \\ & & &\vdots& \\ &\cdots&\cdots&\ast&\cdots\\ & & &\vdots& \\ \end{pmatrix} \end{align*} \begin{align*} (I_{n}A)_{ij}&=(0)(A_{1j})+(0)(A_{2j})+\cdots+(1)(A_{ij})+\cdots+(0)(A_{nj})\\ &=(1)(A_{ij})\\ &=A_{ij} \end{align*}

Therefore, \[(I_{n}A)_{ij}=A_{ij}\] for all \[i\] and \[j\], which implies that \[I_{n}A\] and \[A\] have the same entries in each position. Then \[I_{n}A=A\].

The proof for the second statement is similar.