Question

The n × n identity matrix is the matrix with diagonal entries are all 1’s and the rest are all 0’s. Show that, for any n×n matrix A, we have AI=IA=A.

Answer 1

Express the matrix $I$ like $(\delta_{ij})_{n\times n}$, where $\delta_{ij}=1$ if $i=j$ and $\delta_{ij}=0$ if $i\neq j$. ($\delta_{ij}$ is known as kronecker's delta)

In the same form we express the matrix $A=(a_{ij})_{n\times n}$.

The firs index indicate the row and the second the column.

By the multiplication of matrices we have $AI=(c_{ij})_{n\times n}$, where

$c_{ij}=\sum_{k=1}^n a_{ik}\delta _{kj} = a_{ij}$

because only $\delta_{jj}$ is non-zero in the last sum.

therfore we have $AI=(c_{ij})_{n\times n}=(a_{ij})_{n\times n}=A$.

In the same manner we have

$IA=(d_{ij})_{n\times n}$, where

$d_{ij}=\sum_{k=1}^n \delta _{ik}a_{kj} = a_{ij}$

And so, $IA=A$