# Linear algebra - orthogonal basis

Question- Find an orthogonal basis for the set of solutions of the equations:

$x_1-2x_2+3x_3-x_4=0$

$2x_1+x_2+x_3+2x_4=0$

(sorry for all the underscores, still haven't worked out how to insert equations)

So, I put this into a matrix, row reduced and got $x_1$ and $x_2$ as functions of $x_3$ and $x_4$.

$x_1 = -\frac{1}{5} x_3 - \frac{3}{5} x_4$

$x_2 = \frac{7}{5} x_3 -\frac{4}{5} x_4$

Then let $s=x_3$ and $t=x_4$ to get a basis,

Basis.image

However I believe this is not orthogonal as the dot product should equal zero.

Any help is appreciated, I'm not sure if I'm doing any of this correctly, thanks.

• You need to add a \$before and after the mathematical expression. For example, adding a \$ before and after x_2 results in $x_2$. Aug 24 '17 at 5:00

The equations you provided ($x_1 - 2 x_2 + 3 x_3 - x_4=0$ and $2 x_1 + x_2 + x_3 + 2 x_4=0$) should result in $$x_1 = -x_3 -\frac{3}{5}x_4,$$ and $$x_2 = x_3 -\frac{4}{5}x_4.$$
Suppose you have computed a basis (the vectors you have in the linked image) for the solutions set to be $\{u,v\}$. Then you want a basis $\{u,w\}$ such that $u \perp w$. Since $w \in \text{Span}(\{u,v\})$, so you can simple choose $$w=v-\text{proj}_uv=v-\left(\frac{u \cdot v}{\|u\|^2}\right)u.$$ It is easy to see that $w \perp u$.