Question

A-o 2 13 -2 Use Gram-Schmidt process to find a matrix Q with the same column space 2019 Pablo Soberón Use the columns of Q to find the projection of2onto C(A)

Answer 1

First part: We first check that the columns of the matrix are linearly independent.

To see this, suppose

2 3003

Then,

a + 3b a +3b - 2c

Since $\begin{align*}a+4c=0=a+2c\end{align*}$ , we get $\begin{align*}2c=a+4c-a-2c=0-0\end{align*}$ , so that $\begin{align*}c=0\end{align*}$ . Then, $\begin{align*}a=a+4c=0\end{align*}$ and $\begin{align*}3b=a+3b=0\end{align*}$ shows $\begin{align*}a=b=c=0\end{align*}$ . hence, the columns are linearly independent.

We now perform Graham-Schmidt. Let

3003-1111 551 .5 3003 2

$\begin{align*}v_3&=\begin{pmatrix}0\\4\\2\\-2\end{pmatrix}+{\frac 43}\begin{pmatrix}1.5\\-1.5\\-1.5\\1.5\end{pmatrix}-\begin{pmatrix}1\\1\\1\\1\end{pmatrix}\\ &=\begin{pmatrix}1\\1\\-1\\-1\end{pmatrix}\end{align*}$

Finally, we let

$\begin{align*}u_1&={\frac {v_1}{||v_1||}}\\ &=\begin{pmatrix}0.5\\0.5\\0.5\\0.5\end{pmatrix}\\ u_2&={\frac {v_2}{||v_2||}}\\ &=\begin{pmatrix}0.5\\-0.5\\-0.5\\0.5\end{pmatrix}\\ u_3&={\frac {v_3}{||v_3||}}\\ &=\begin{pmatrix}0.5\\0.5\\-0.5\\-0.5\end{pmatrix}\end{align*}$

Because of Graham-Schmidt, these three vectors
11 1 1 u21 из
form an orthonormal basis for the column space of . We let be the matrix whose columns are
11 1 1 u21 из
. Thus,

$\begin{align*}Q&=\begin{pmatrix}0.5&0.5&0.5\\0.5&-0.5&0.5\\0.5&-0.5&-0.5\\0.5&0.5&-0.5\end{pmatrix}\end{align*}$

Second part: By orthonormality, the projection is

0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5

where

$\begin{align*}a&=\langle\begin{pmatrix}4\\-1\\2\\-5\end{pmatrix},\begin{pmatrix}0.5\\0.5\\0.5\\0.5\end{pmatrix}\rangle\\ &=0\\ b&=\langle\begin{pmatrix}4\\-1\\2\\-5\end{pmatrix},\begin{pmatrix}0.5\\-0.5\\-0.5\\0.5\end{pmatrix}\rangle\\&=-1\\ c&=\langle\begin{pmatrix}4\\-1\\2\\-5\end{pmatrix},\begin{pmatrix}0.5\\0.5\\-0.5\\-0.5\end{pmatrix}\rangle\\ &=3\end{align*}$

Hence, the projection is

$\begin{align*}-\begin{pmatrix}0.5\\-0.5\\-0.5\\0.5\end{pmatrix}+3\begin{pmatrix}0.5\\0.5\\-0.5\\-0.5\end{pmatrix}\end{align*}$