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 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.
 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.

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

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

TuringTest
 2 years ago
Best 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?

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1the vectors in H are 4tuples. Any basis vector will also be a 4tuple

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1so\[\vec v_1=\left[\begin{matrix}1\\0\\3\\5\end{matrix}\right]\]\[\vec v_2=\left[\begin{matrix}0\\1\\5\\4\end{matrix}\right]\]??? (linear algebra review needed obviously)

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

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