Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
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
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

nadian63 Group TitleBest ResponseYou've already chosen the best response.0
I don't know anything
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
I'm afraid my linear algebra is a bit shoddy, but @Zarkon surely know what do to, though he remains silent.
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
don't we just set up the augmented matrix and collect the s,t, and u parts as separate vectors?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[\left[\begin{matrix}1&2&1&3\\0&1&1&4\end{matrix}\right]=\binom 00\]\[\left[\begin{matrix}1&0&3&5\\0&1&1&4\end{matrix}\right]=\binom 00\]\[x_1=3t+5u\]\[x_2=t+4ut\]\[x_3=t\]\[x_4=u\]\[\vec x=t\binom{3}{1}+u\binom54\]am I on the right track?
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
the vectors in H are 4tuples. Any basis vector will also be a 4tuple
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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)
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
\[\left[\begin{matrix}1&0&3&5\\0&1&1&4\end{matrix}\right]\] 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 \[\left[\begin{matrix}r\\s\\t\\u\end{matrix}\right]=\left[\begin{matrix}3x+5y\\x+4y\\x\\y\end{matrix}\right]\] \[=\left[\begin{matrix}3\\1\\1\\0\end{matrix}\right]x+\left[\begin{matrix}5\\4\\0\\1\end{matrix}\right]y\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
oh, I meant a 3 in the first vector, but put a 5 instead well, I was almost close... thanks Zarkon
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
meant a 1*
 one year ago

nadian63 Group TitleBest ResponseYou've already chosen the best response.0
Thank you so much!
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.