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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

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  1. nadian63 Group Title
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    I don't know anything

    • one year ago
  2. TuringTest Group Title
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    I'm afraid my linear algebra is a bit shoddy, but @Zarkon surely know what do to, though he remains silent.

    • one year ago
  3. TuringTest Group Title
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    don't we just set up the augmented matrix and collect the s,t, and u parts as separate vectors?

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

    • one year ago
  5. Zarkon Group Title
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    the vectors in H are 4-tuples. Any basis vector will also be a 4-tuple

    • one year ago
  6. TuringTest Group Title
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    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
  7. Zarkon Group Title
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    \[\left[\begin{matrix}1&0&3&-5\\0&1&1&-4\end{matrix}\right]\] 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
  8. TuringTest Group Title
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    oh, I meant a -3 in the first vector, but put a 5 instead well, I was almost close... thanks Zarkon

    • one year ago
  9. TuringTest Group Title
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    meant a -1*

    • one year ago
  10. nadian63 Group Title
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    Thank you so much!

    • one year ago
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