Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

nadian63

  • 2 years ago

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.

  • This Question is Closed
  1. nadian63
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I don't know anything

  2. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  3. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  4. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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?

  5. Zarkon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  6. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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)

  7. Zarkon
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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 \[\left[\begin{matrix}r\\s\\t\\u\end{matrix}\right]=\left[\begin{matrix}-3x+5y\\-x+4y\\x\\y\end{matrix}\right]\] \[=\left[\begin{matrix}-3\\-1\\1\\0\end{matrix}\right]x+\left[\begin{matrix}5\\4\\0\\1\end{matrix}\right]y\]

  8. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  9. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    meant a -1*

  10. nadian63
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you so much!

  11. Not the answer you are looking for?
    Search for more explanations.

    Search OpenStudy
    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.