# Proving Vector space

1. May 17, 2012

### hms.tech

1. The problem statement, all variables and given/known data

V1 is defined as the span of the vectors b1,b2,b3,b4

Prove that V1 is not a vector space.

2. Relevant equations

A set of axioms :
If u and v are tow vectors in the span of b1,b2,b3,b4 then:
1. u + v belongs to V1
2. ku also belongs to V1

3. The attempt at a solution
Don't know how to prove 1. wrong !

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2. May 18, 2012

### tiny-tim

hi hms.tech!
i'm confused

i thought span is defined as the smallest vector subspace that (etc)

what definition of span have you been given?​

3. May 18, 2012

### sharks

To prove that $V_1$ is a vector space, the two closure axioms must be satisfied, as stated in your relevant equations.
$$V_1= \left ( b_1,b_2,b_3,b_4 \right )$$
$$V_1= \left ( \begin{bmatrix}1 \\ 0 \\ 0\\0 \end{bmatrix}, \begin{bmatrix}1 \\ 1 \\ 0\\0 \end{bmatrix},\begin{bmatrix}1 \\ 1 \\1\\0 \end{bmatrix},\begin{bmatrix}1 \\ 1 \\ 1\\1 \end{bmatrix} \right )$$
According to the two properties:
$$1.\;b_1+b_2\not=b_3 \\2.\;k\times b_1\not=b_2$$Also, if you write the 4 vectors as a matrix and reduce to its row echelon form, you will see that it has no free variables, meaning, all the 4 vectors are linearly independent.

4. May 18, 2012

### hms.tech

I think i might have to rephrase the question. Becuase i think i interpreted it wrongly.
so again :

Here is the complete question :

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5. May 18, 2012

### sharks

That's a completely different question from your post #1.$$V_1= \left ( \begin{bmatrix}1 \\ 0 \\ 0\\0 \end{bmatrix}, \begin{bmatrix}1 \\ 1 \\ 0\\0 \end{bmatrix},\begin{bmatrix}1 \\ 1 \\ 1\\0 \end{bmatrix} \right )= \begin{bmatrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix}$$
$$V_2= \left ( \begin{bmatrix}1 \\ 0 \\ 0\\0 \end{bmatrix}, \begin{bmatrix}1 \\ 1 \\ 0\\0 \end{bmatrix},\begin{bmatrix}1 \\ 1 \\ 1\\1 \end{bmatrix} \right )= \begin{bmatrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\\ 0 & 0 & 1 \end{bmatrix}$$Note: vector spaces are sometimes called linear spaces.

Last edited: May 18, 2012
6. May 18, 2012

### hms.tech

yes, i know that vector space is the same thing as a linear space, but again, how to we prove that V1 union V2 is not a vector space ?

7. May 18, 2012

### sharks

$$V_1+V_2\not=kV_1\;or\;V_1+V_2\not=kV_2$$where k is a scalar multiple.

8. May 18, 2012

### hms.tech

what axiom is that, i can't recall any such axiom of a vector space ?

9. May 18, 2012

### sharks

$V_1\cup V_2$ is the sum of all corresponding elements of $V_1$ and $V_2$. Therefore, it is $V_1 + V_2$.

The result of the vector addition, let's call it, $V_3$, is to be tested against the closure axioms to verify if it is a vector space of either $V_1$ or $V_2$.

Now, according to the closure axiom of scalar multiplication, if $V_3$ is a scalar multiple of $V_1$ or $V_2$, then the former is a vector space. You can get the answer easily from here.

10. May 18, 2012

### tiny-tim

the simplest way is to find a in V1 and b in V2 such that a+b is not in V1 and is not in V2

(and the simplest choices for a and b would be one 1 and three 0s)

11. May 19, 2012

### hms.tech

yeah i got it, thanks for the explanation. I took the vectors b3 and b4, added them together and proved that the resultant vector does not lie in v3.