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nadian63

  • 3 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.

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  1. nadian63
    • 3 years ago
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    I don't know anything

  2. TuringTest
    • 3 years ago
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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.

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

  4. TuringTest
    • 3 years ago
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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?

  5. Zarkon
    • 3 years ago
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    the vectors in H are 4-tuples. Any basis vector will also be a 4-tuple

  6. TuringTest
    • 3 years ago
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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)

  7. Zarkon
    • 3 years ago
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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 \[\left[\begin{matrix}r\\s\\t\\u\end{matrix}\right]=\left[\begin{matrix}-3x+5y\\-x+4y\\x\\y\end{matrix}\right]\] \[=\left[\begin{matrix}-3\\-1\\1\\0\end{matrix}\right]x+\left[\begin{matrix}5\\4\\0\\1\end{matrix}\right]y\]

  8. TuringTest
    • 3 years ago
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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

  9. TuringTest
    • 3 years ago
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    meant a -1*

  10. nadian63
    • 3 years ago
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    Thank you so much!

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