Here's the question you clicked on:

## nadian63 Group Title Let H be the subspace of R^4 given by H = {(r, s, t, u)^T | r, s, t, u ∈ R, r − 2s + t + 3u = 0 and s + t − 4u = 0 } Find a basis for H and determine dim H. one year ago one year ago

• This Question is Closed
1. nadian63 Group Title

I don't know anything

2. TuringTest Group Title

I'm afraid my linear algebra is a bit shoddy, but @Zarkon surely know what do to, though he remains silent.

3. TuringTest Group Title

don't we just set up the augmented matrix and collect the s,t, and u parts as separate vectors?

4. TuringTest Group Title

$\left[\begin{matrix}1&-2&1&3\\0&1&1&-4\end{matrix}\right]=\binom 00$$\left[\begin{matrix}1&0&3&-5\\0&1&1&-4\end{matrix}\right]=\binom 00$$x_1=-3t+5u$$x_2=-t+4ut$$x_3=t$$x_4=u$$\vec x=t\binom{-3}{-1}+u\binom54$am I on the right track?

5. Zarkon Group Title

the vectors in H are 4-tuples. Any basis vector will also be a 4-tuple

6. TuringTest Group Title

so$\vec v_1=\left[\begin{matrix}1\\0\\-3\\-5\end{matrix}\right]$$\vec v_2=\left[\begin{matrix}0\\1\\5\\4\end{matrix}\right]$??? (linear algebra review needed obviously)

7. Zarkon Group Title

$\left[\begin{matrix}1&0&3&-5\\0&1&1&-4\end{matrix}\right]$ which uses the variables r,s,t,u so t and u are free variables...call them x and y respectively then r=-3x+5y and s= -x+4y then $\left[\begin{matrix}r\\s\\t\\u\end{matrix}\right]=\left[\begin{matrix}-3x+5y\\-x+4y\\x\\y\end{matrix}\right]$ $=\left[\begin{matrix}-3\\-1\\1\\0\end{matrix}\right]x+\left[\begin{matrix}5\\4\\0\\1\end{matrix}\right]y$

8. TuringTest Group Title

oh, I meant a -3 in the first vector, but put a 5 instead well, I was almost close... thanks Zarkon

9. TuringTest Group Title

meant a -1*

10. nadian63 Group Title

Thank you so much!