# Thread: Dimension of Vector Space

1. ## Dimension of Vector Space

Let (v1,v2,v3.....,v14) be a basis for a vector space V over the field R.

(a) Is V finite-dimensional? if so, what is its dimension?

V is finete-dimension and its dimension is 14.

Why is it 14? because there are 14 vectors in the basis of V? how many elements/components does v1,...,v14 have?

thanks for any help.

2. A basis for a vector space has three properties:
1) It spans the vector space.
2) The vectors in it are independent.
3) The number of vectors in any basis for the same vector space have the same number of vectors.

Further, you can show that if any two of those is true, the third is also.

A vector space is defined to be finite dimensional if it has a finite basis and, in that case, the dimension of the vector space is defined to be the number of vectors in a basis.

I have no idea what you mean by "elements/components" if not the vectors themselves.

3. Originally Posted by charikaar
Let (v1,v2,v3.....,v14) be a basis for a vector space V over the field R.

(a) Is V finite-dimensional? if so, what is its dimension?

V is finete-dimension and its dimension is 14.

Why is it 14? because there are 14 vectors in the basis of V? how many elements/components does v1,...,v14 have?

thanks for any help.
$\displaystyle v_1,v_2,...,v_{14}$ have one component.

let v1=$\displaystyle (1,0,0,0,0,0,0,0,0,0,0,1,055,14,5)^T$

what do we call each entry of v1?

thanks

5. Originally Posted by charikaar

let v1=$\displaystyle (1,0,0,0,0,0,0,0,0,0,0,1,055,14,5)^T$

what do we call each entry of v1?

thanks
coordinates...or components as you said...

6. Originally Posted by Raoh
$\displaystyle v_1,v_2,...,v_{14}$ have one component.

do you mean v1 is not equal to $\displaystyle (1,0,0,0,0,0,0,0,0,0,0,1,055,14,5)^T$ in the above example.

sorry I can't get my head around this stuff!

7. Originally Posted by charikaar
do you mean v1 is not equal to $\displaystyle (1,0,0,0,0,0,0,0,0,0,0,1,055,14,5)^T$ in the above example.

sorry I can't get my head around this stuff!
hmm,i don't think so,from where you got those coordinates..??

8. I chose them randomly....

thanks for your help.

9. i correct, it depends on your vector space $\displaystyle V (V=\vec{\mathbb{R}}^n)$ which is over $\displaystyle \mathbb{R}$.
that means when for example $\displaystyle n = 1$,your vectors will have one component,$\displaystyle n=2$ two components...

10. Vectors in a general vector space do not have "components".

After you have defined a basis for the vector space, then you can talk about "components". If, for example, your basis is $\displaystyle \{v_1, v_2, \cdot\cdot\cdot, v_{14}\}$, then any vector, v, in that space can be written $\displaystyle v= a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_{14}v_{14}$. In that case, the "components of v" in this basis, are the scalars $\displaystyle a_1, a_2, \cdot\cdot\cdot, a_{14}$.

Any n dimensional vector space is isomophic to $\displaystyle R^n$ but it is wrong to say it "is" $\displaystyle R^n$. Exactly what that isomorphism is and how each vector would written in "components" depends, as I said before, on the choice of basis.

11. Originally Posted by HallsofIvy
Vectors in a general vector space do not have "components".

After you have defined a basis for the vector space, then you can talk about "components". If, for example, your basis is $\displaystyle \{v_1, v_2, \cdot\cdot\cdot, v_{14}\}$, then any vector, v, in that space can be written $\displaystyle v= a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_{14}v_{14}$. In that case, the "components of v" in this basis, are the scalars $\displaystyle a_1, a_2, \cdot\cdot\cdot, a_{14}$.

Any n dimensional vector space is isomophic to $\displaystyle R^n$ but it is wrong to say it "is" $\displaystyle R^n$. Exactly what that isomorphism is and how each vector would written in "components" depends, as I said before, on the choice of basis.
...but they have coordinates,and that what "charikaar" meant.