Question

In ${\mathbf{ℝ}}^{\mathbf{m}}$, we define

$\overrightarrow{{\mathbf{e}}_{\mathbf{i}}}\mathbf{=}\left[\begin{array}{c}0\\ 0\\ ⋮\\ 1\\ ⋮\\ 0\end{array}\right]\mathbf{\leftarrow }{\mathit{i}}^{\mathbf{t}\mathbf{h}}\mathbf{}\mathit{c}\mathit{o}\mathit{m}\mathit{p}\mathit{o}\mathit{n}\mathit{e}\mathit{n}\mathit{t}$

If A is an$\mathit{n}\mathbf{\times }\mathit{m}$ matrix, what is$\mathit{A}\overrightarrow{{\mathbf{e}}_{\mathbf{i}}}$ ?

Short answer

The $i^{th}$ column of A is $A\overrightarrow{e_i}$ .

Step-by-step solution

Consider the matrices.

If A is an n x m matrix with row vectors $\overrightarrow{{\mathbf{\omega }}_{\mathbf{1}}\mathbf{,}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\overrightarrow{\mathbf{,}{\mathbf{\omega }}_{\mathbf{n}}}$and $\overrightarrow{\mathbf{x}}$is a vector in ${\mathit{R}}^{\mathbf{m}}$then,

$\mathit{A}\overrightarrow{\mathbf{x}}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{-}{\overrightarrow{\mathbf{\omega }}}_{\mathbf{1}}\mathbf{-}\\ \mathbf{⋮}\\ \mathbf{-}\overrightarrow{{\mathbf{\omega }}_{\mathbf{n}}}\mathbf{-}\end{array}\mathbf{\right]}\overrightarrow{\mathbf{x}}\mathbf{\left[}\begin{array}{c}\mathbf{-}{\overrightarrow{\mathbf{\omega }}}_{\mathbf{1}}\mathbf{.}\overrightarrow{\mathbf{x}}\mathbf{-}\\ \mathbf{⋮}\\ \mathbf{-}\overrightarrow{{\mathbf{\omega }}_{\mathbf{n}}}\mathbf{.}\overrightarrow{\mathbf{x}}\mathbf{-}\end{array}\mathbf{\right]}$

The matrices are,

$$\begin{gathered} A=\left[\begin{array}{ccc}{a}_{11}& \dots & a_{1m}\\ ⋮& \ddots & ⋮\\ a_{n1}& \dots & a_{nm}\end{array}\right] \\ \overrightarrow{e_i}=\left[\begin{array}{c}0\\ 0\\ ⋮\\ 1\\ ⋮\\ 0\end{array}\right]\leftarrow {\text{i}}^{\text{th}}\text{component} \end{gathered}$$

A is the n x m matrix.

Perform the multiplication operation

The $i^{th}$component of $\overrightarrow{e_i}$is 1 and the rest of all the elements are zero. The matrix A has all nonzero elements.

$A\overrightarrow{e}$is basically a multiplication operation of matrices. As all the elements of $\overrightarrow{e}$are zero except the $i^{th}$component, thus, the role="math" localid="1664205251359" $A\overrightarrow{e_1}$will be the $i^{th}$column of A.

Hence, $A\overrightarrow{e_1}$is the $i^{th}$column of A.