Question

A = 1487 -1 2 -2 13 - 4 Find an orthonormal basis of the column space of

Answer 1

In row reduced form, we get the following matrix.

$\left[ \begin{array}{ccc} \phantom{-}1 & \phantom{-}0 & \phantom{-}4 \\ \phantom{-}0 & \phantom{-}1 & \phantom{-}1 \\ \phantom{-}0 & \phantom{-}0 & \phantom{-}0 \\ \end{array} \right]$

For column space we have to notice the pivot columns, i.e. the columns containing leading one (1st and 2nd).

The corresponding columns from the orginal matrix will give us basis for the column space.

So column space contains the vector { v1 , v2 }, where

$\mathbf{v_{1}}=\left[ \begin{array}{c} 1 \\\\ -1 \\\\ 3 \end{array} \right], ~ \mathbf{v_{2}}=\left[ \begin{array}{c} 4 \\\\ 2 \\\\ -8 \end{array} \right].\\ \\ Let~$

$\mathbf{u_1}=\mathbf{v_1}=\left[ \begin{array}{c} 1 \\\\ -1 \\\\ 3 \end{array} \right]$

$Then~~ \mathbf{u_2}=\mathbf{v_2}-\frac{\mathbf{u_1} \cdot \mathbf{v_2}}{\mathbf{u_1} \cdot \mathbf{u_1}}\mathbf{u_1}=\left[ \begin{array}{c} 6 \\\\ 0 \\\\ -2 \end{array} \right]$

And the corresponding orthonomal vectors are

$\mathbf{e_1}=\frac{\mathbf{u_1}}{\sqrt{\mathbf{u_1} \cdot \mathbf{u_1}}}=\left[ \begin{array}{c} \frac{\sqrt{11}}{11} \\\\ - \frac{\sqrt{11}}{11} \\\\ \frac{3 \sqrt{11}}{11} \end{array} \right]\\ \\ and, ~~ \mathbf{e_2}=\frac{\mathbf{u_2}}{\sqrt{\mathbf{u_2} \cdot \mathbf{u_2}}}=\left[ \begin{array}{c} \frac{3 \sqrt{10}}{10} \\\\ 0 \\\\ - \frac{\sqrt{10}}{10} \end{array} \right]\\ \\ \\ Thus~e_1~ and ~e_2~ are~ the ~required~orthonormal~ vectors.$