Sub-space?.. :D

**brd_7** · November 15th 2008, 08:01 AM

Is this a subspace, if so, why and how?

1) set of all vectors from R^n which have a first coordinate of 0.

2) set of all vectors from R^2 of form

$\begin{bmatrix}x\\y \end{bmatrix}$ with x - y =2

3) set of all vectors from R^3 of form

$\begin{bmatrix}x\\2y-x\\x-y \end{bmatrix}$

4) set of all vectors $\begin{bmatrix}x1\\x2\\x3\\x4\\x5 \end{bmatrix}$ from R^5 satisfying x1 + 2x2 = x3 & x4=2x5

**Scopur** · November 15th 2008, 06:18 PM

Just check if it satisfies scalar multiplication and addition properties. Also check if the Zero vector is contained in your set(all though this is redundant.) Can you do this?

**brd_7** · November 16th 2008, 01:29 AM

Not really.. I havent been given an example of how to go about tackeling a question like this.. So a starting help would be appreciated

**clic-clac** · November 16th 2008, 03:11 AM

For instance, the first example is very simple:
Let $n\geq2$ (for easier notations) be an integer and $A$ the set of vectors in $\mathbb{R}^{n}$ which have a first coordinate of 0.

Clearly, $0_{\mathbb{R}^{n}}=(0,...,0)$ belongs to $A$ . $(\star)$

Let $x=(x_{1},...,x_{n})$ and $y=(y_{1},...,y_{n})$ be two elements in A. That means $x_{1}=y_{1}=0$ .
Let $\lambda$ be a real number, then
$\lambda x + y=\lambda(0,x_{2},...,x_{n})+(0,y_{2},...,y_{n})=( \lambda .0,\lambda x_{2},...,\lambda x_{n})+(0,y_{2},...,y_{n})=$ $(0,\lambda x_{2}+y_{2},...,\lambda x_{n}+y_{n}) \in A$

Thus every linear combination of vectors in $A$ is in $A$ . $(\star \star)$

$(A \subset \mathbb{R}^{n}), (\star)$ and $(\star \star) \Rightarrow A$ is a subspace of $\mathbb{R}^{n}$

**brd_7** · November 16th 2008, 12:16 PM

This does make sense.. I think ive successfully completed questions 2 and 3.. Ive attempted question 4, but became stuck on a number of occassions, so any help there would be much appreciated!

Many thanks

**clic-clac** · November 16th 2008, 12:18 PM

Maybe show your attempts for 4)

**Skerven** · November 18th 2008, 12:11 AM

Originally Posted by brd_7

Is this a subspace, if so, why and how?
...
4) set of all vectors $\begin{bmatrix}x1\\x2\\x3\\x4\\x5 \end{bmatrix}$ from R^5 satisfying x1 + 2x2 = x3 & x4=2x5

$\forall\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{ bmatrix}=\begin{bmatrix}x_1\\x_2\\x_1 + 2x_2\\2x_5\\x_5 \end{bmatrix}\in\mathbb{R}^{5}$
$c_1*\begin{bmatrix}x_1\\x_2\\x_1+2x_2\\2x_5\\x_5\e nd{bmatrix}=A\subset\mathbb{R}^5$
$\forall c_1\in\mathbb{R}$
$\Rightarrow$ A is a subspace of $\mathbb{R}^{5}$

I'm doing it wrong.

**HallsofIvy** · November 18th 2008, 03:19 AM

Originally Posted by Skerven

$\forall\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{ bmatrix}=\begin{bmatrix}x_1\\x_2\\x_1 + 2x_2\\2x_5\\x_5 \end{bmatrix}\in\mathbb{R}^{5}$
$c_1*\begin{bmatrix}x_1\\x_2\\x_1+2x_2\\2x_5\\x_5\e nd{bmatrix}=A\subset\mathbb{R}^5$
$\forall c_1\in\mathbb{R}$
$\Rightarrow$ A is a subspace of $\mathbb{R}^{5}$

I'm doing it wrong.

Only because just showing "if v in S, then av in S" is not enough to show that S is a subspace. You also have to show "if u and v in S then u+ v is in S".

Let u= $\begin{bmatrix}x_1\\x_2\\x_1 + 2x_2\\2x_5\\x_5 \end{bmatrix}$ and let v= [tex]\begin{bmatrix}y_1\\y_2\\y_1 + 2y_2\\2y_5\\y_5 \end{bmatrix}[/quote]

Then u+ v= $\begin{bmatrix}x_1+ y_1\\x_2+ y_2\\(x_1 + 2x_2)+ (y_1+ 2y_2\\2x_5+ 2y_5\\x_5+ y_5 \end{bmatrix}$
$= \begin{bmatrix}x_1+ y_1\\x_2+ y_2\\(x_1 + y_1)+ 2(x_2+ y_2)\\2(x_5+ y_5)\\x_5+ y_5 \end{bmatrix}$
which is again of the same form.

By the way, Scopur, showing that the 0 vector is in the set is not entirely redundant. If S= {}, the empty set, then it is trivial that "if v is in S, then cv is in S" and "if u and v are in S then u+ v is in S" but the standard definition of subspace requires that it be non-empty. You must show that there is some vector in the set and it is usually easiest to show that the 0 vector is in the set.

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