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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
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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nadian63Best ResponseYou've already chosen the best response.0
I don't know anything
 one year ago

TuringTestBest 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

TuringTestBest 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

TuringTestBest 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

ZarkonBest 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

TuringTestBest 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

ZarkonBest 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

TuringTestBest 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
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