Subspaces

**Robb** · May 31st, 2009, 01:34 AM

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) $L=\left\{\left[\begin{array}{cc}s+2t+1\\s+3t\end{array}\right]:s,t\in\mathbb{R} \right\}$

**Moo** · May 31st, 2009, 02:20 AM

Hello,

Originally Posted by Robb

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...

**HallsofIvy** · May 31st, 2009, 02:26 AM

Originally Posted by Robb

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?

**Moo** · May 31st, 2009, 02:28 AM

To Robb and HallsOfIvy : using \begin{bmatrix} ... \end{bmatrix} is shorter than going with arrays

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