1. ## Sub-space?.. :D

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

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

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

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

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

2. 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?

3. 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

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

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

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

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

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

5. 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

6. Maybe show your attempts for 4)

7. Originally Posted by brd_7
Is this a subspace, if so, why and how?
...
4) set of all vectors $\displaystyle \begin{bmatrix}x1\\x2\\x3\\x4\\x5 \end{bmatrix}$ from R^5 satisfying x1 + 2x2 = x3 & x4=2x5
$\displaystyle \forall\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{ bmatrix}=\begin{bmatrix}x_1\\x_2\\x_1 + 2x_2\\2x_5\\x_5 \end{bmatrix}\in\mathbb{R}^{5}$
$\displaystyle c_1*\begin{bmatrix}x_1\\x_2\\x_1+2x_2\\2x_5\\x_5\e nd{bmatrix}=A\subset\mathbb{R}^5$
$\displaystyle \forall c_1\in\mathbb{R}$
$\displaystyle \Rightarrow$ A is a subspace of $\displaystyle \mathbb{R}^{5}$

I'm doing it wrong.

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

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

Then u+ v= $\displaystyle \begin{bmatrix}x_1+ y_1\\x_2+ y_2\\(x_1 + 2x_2)+ (y_1+ 2y_2\\2x_5+ 2y_5\\x_5+ y_5 \end{bmatrix}$
$\displaystyle = \begin{bmatrix}x_1+ y_1\\x_2+ y_2\\(x_1 + y_1)+ 2(x_2+ y_2)\\2(x_5+ y_5)\\x_5+ y_5 \end{bmatrix}$
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.