# Question #42ca1

Oct 2, 2016

See below.

#### Explanation:

Calling

$X = \left({x}_{1} , {x}_{2} , {x}_{3} , {x}_{4}\right)$
$S \to \left(\begin{matrix}{s}_{1} \\ {s}_{2}\end{matrix}\right) = \left(\begin{matrix}1 & 2 & 1 & - 1 \\ 0 & - 1 & 3 & 1\end{matrix}\right)$ and
$T \to \left\langle{t}_{0} , X\right\rangle = 0$ with ${t}_{0} = \left(1 , 1 , 0 , 0\right)$

The space $\left(S \bot\right) \cap T$ is

$\left\langle{s}_{1} , {X}_{a}\right\rangle = 0$
$\left\langle{s}_{2} , {X}_{a}\right\rangle = 0$
$\left\langle{t}_{0} , {X}_{a}\right\rangle = 0$

This gives ${X}_{a} = \lambda \left(- 1 , 1 , 0 , 1\right) , \lambda \in \mathbb{R}$ so

$W$ the complement, orthogonal to ${X}_{a}$. This space is given by
${X}_{w} | \left\langle\lambda \left(- 1 , 1 , 0 , 1\right) , {X}_{w}\right\rangle = 0$
This space is then

$W = \left\langle{w}_{0} , X\right\rangle = - {x}_{1} + {x}_{2} + {x}_{4} = 0$

with ${w}_{0} = \left(- 1 , 1 , 0 , 1\right)$

As can be checked

$\det \left(\begin{matrix}{s}_{1} \\ {s}_{2} \\ {t}_{0} \\ {w}_{0}\end{matrix}\right) = 12$

so $\left\{{s}_{1} , {s}_{2} , {t}_{0} , {w}_{0}\right\}$ span ${\mathbb{R}}^{4}$