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Math Help - Subspaces

  1. #1
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    Subspaces

    Hello MHF, would appreciate some help with the following equestions;
    Determine which of the following sets are subspaces. Give reasons for your answers.

    (a) H=\left\{\left[\begin{array}{cc}s\\2s\\0\end{array}\right]:s\in\mathbb{R},s\geq0 \right\}

    (b) L=\left\{\left[\begin{array}{cc}s+2t+1\\s+3t\end{array}\right]:s,t\in\mathbb{R} \right\}
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  2. #2
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    Hello,
    Quote Originally Posted by Robb View Post
    Hello MHF, would appreciate some help with the following equestions;
    Determine which of the following sets are subspaces. Give reasons for your answers.
    You have to check that for any u,v belonging to the set, u+v belongs to the set, and for any real number \lambda , \lambda u belongs to the set.

    (a) H=\left\{\left[\begin{array}{cc}s\\2s\\0\end{array}\right]:s\in\mathbb{R},s\geq0 \right\}
    So let u=\begin{bmatrix} s\\2s\\0 \end{bmatrix} and v=\begin{bmatrix} s'\\2s'\\0\end{bmatrix}, where s,s' \in\mathbb{R} and \geq 0.

    u+v=\begin{bmatrix} s+s' \\ 2(s+s') \\ 0 \end{bmatrix}
    Does this belong to H ?

    \lambda\in\mathbb{R}
    Consider the case when \lambda< 0

    (b) L=\left\{\left[\begin{array}{cc}s+2t+1\\s+3t\end{array}\right]:s,t\in\mathbb{R} \right\}
    Let u=\begin{bmatrix}s+2t+1\\s+3t\end{bmatrix} and v=\begin{bmatrix}s'+2t'+1\\s'+3t'\end{bmatrix}

    u+v=\begin{bmatrix} (s+s')+2(t+t')+2\\(s+s')+3(t+t') \end{bmatrix}

    Can the first coordinate be in the form m+2n+{\color{red}1} ?

    The problem here is that s and t are the same in the first and the second coordinates.
    If there weren't, you could have just written (s+s'+1)+2(t+t')+1
    But you would have to get (s+s'+1)+3(t+t') as your second coordinate...
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  3. #3
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    Quote Originally Posted by Robb View Post
    Hello MHF, would appreciate some help with the following equestions;
    Determine which of the following sets are subspaces. Give reasons for your answers.

    (a) H=\left\{\left[\begin{array}{cc}s\\2s\\0\end{array}\right]:s\in\mathbb{R},s\geq0 \right\}

    (b) L=\left\{\left[\begin{array}{cc}s+2t+1\\s+3t\end{array}\right]:s,t\in\mathbb{R} \right\}
    A subset of a vector space is a subspace if and only if it is "closed" under vector addition and scalar multiplication.

    If u and v are two members of H, they are of the form u= \left[\begin{array}{c}s \\ 2s \\ 0\end{array}\right] and v= \left[\begin{array}{c}t\\ 2t\\ 0\end{array}\right] for some positive numbers s and t. Their sum is u+ v=\left[\begin{array}{c}s+t \\ 2s+ 2t \\ 0\end{array}\right]=\left[\begin{array}{c}s \\ 2(s+t) \\ 0\end{array}\right] while, for any number k, k v= \left[\begin{array}{c}ks \\ 2(ks) \\ 0\end{array}\right]. What happens if k is negative?

    If u and v are two members of L, they are of the form u= \left[\begin{array}{c}s+ 2t+ 1 \\ s+ 3t\end{array}\right] and v= \left[\begin{array}{c}x+ 2y+ 1 \\ x+ 3y\end{array}\right] where s, t, x, y can be any real numbers. Their sum is u+ v= \left[\begin{array}{c}s+ 2t+ 1+ x+ 2y+ 1 \\ s+ 3t+ x+ 3y\end{array}\right]= \left[\begin{array}{c}(s+x)+ 2(t+y)+ 2 \\ (s+x)+ 3(t+y)\end{array}\right] and, for a real number k, ku= \left[\begin{array}{c}k(s+ 2t+ 1) \\ k(s+ 3t)\end{array}\right]= \left[\begin{array}{c}ks+ 2kt+ k) \\ (ks+ 3lt)\end{array}\right]. Can you see that those are NOT of the correct form for L?
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  4. #4
    Moo
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    To Robb and HallsOfIvy : using \begin{bmatrix} ... \end{bmatrix} is shorter than going with arrays
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