Show that \({I_n}A = A\) when \(A\) is \(m \times n\) matrix. (Hint: Use the (column) definition of \({I_n}A\).)

The matrix of A is\(A = \left( {\begin{aligned}{*{20}{c}}{{a_1}}&{{a_2}}&{...}&{{a_n}}\end{aligned}} \right)\).

Here, \({a_i} \in {\mathbb{R}^m}\) for all \(i = 1,2,...,n\).

The matrix of \({I_n}\) is \({I_n} = \left( {\begin{aligned}{*{20}{c}}{{e_1}}&{{e_2}}&{...}&{{e_n}}\end{aligned}} \right)\).

Here, \({e_i}\) denotes the \(n \times 1\) matrix in which only the i^th place has 1, and all other places have 0.

\(\begin{aligned}{c}A{I_n} = A\left( {\begin{aligned}{*{20}{c}}{{e_1}}&{{e_2}}&{...}&{{e_n}}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{A{e_1}}&{A{e_2}}&{...}&{A{e_n}}\end{aligned}} \right)\end{aligned}\)

Note that \(A{e_i} = {a_i}\) for all \(i = 1,2,...,n\).

Matrix \(A{I_n}\) reduces to \(A{I_n} = \left( {\begin{aligned}{*{20}{c}}{{a_1}}&{{a_2}}&{...}&{{a_n}}\end{aligned}} \right) = A\).

Hence, \(A{I_n} = A\) is shown.

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