dimensional vector space

**mathbeginner** · November 16th 2010, 11:03 AM

Suppose that K is a finite field with q elements. Let V be an n-dimensional vector space over k. How munch element does v have?

I guess v will have n-q element, but I am not sure.
and if I am right how can I show that v have n-q element???

Pls help!

**TheEmptySet** · November 16th 2010, 11:37 AM

Originally Posted by mathbeginner

I guess v will have n-q element, but I am not sure.
and if I am right how can I show that v have n-q element???

Pls help!

How about this is V is 1 dim vector space how many distinct elements are there?

Okay Now what is V is a 2 dim vector space?

V is 3 dim?

.
.
.
n dim

just follow the pattern

**mathbeginner** · November 16th 2010, 02:38 PM

if only for v is n dim then v will have n element,
but now V is an n-dimensional vector space over k.
will it change the element in V???

**Drexel28** · November 16th 2010, 05:28 PM

Originally Posted by mathbeginner

if only for v is n dim then v will have n element,
but now V is an n-dimensional vector space over k.
will it change the element in V???

You know that since $\dim_{\mathbb{K}} V=n$ that $V\cong \mathbb{K}^n$ . Now, tell me. In general if a set $X$ has $m$ elements, how many elements does $X^m$ have?

**mathbeginner** · November 16th 2010, 05:57 PM

it will also have m element, right?

**Drexel28** · November 16th 2010, 06:07 PM

Originally Posted by mathbeginner

it will also have m element, right?

no..now many ordered pairs oj the set {1,2,3} are there?

**mathbeginner** · November 16th 2010, 06:28 PM

Originally Posted by Drexel28

no..now many ordered pairs oj the set {1,2,3} are there?

3 there are 3 element in the set

**Drexel28** · November 16th 2010, 06:33 PM

Originally Posted by mathbeginner

3 there are 3 element in the set

No. There are $9$ , $(i,j),\text{ }i,j=1,2,3$ . In general if $|X|<\infty$ (there's no need to consider infinite sets for this example) it's true that $\left|X^m\right|=|X|^m$ .

**mathbeginner** · November 16th 2010, 06:44 PM

Originally Posted by Drexel28

No. There are $9$ , $(i,j),\text{ }i,j=1,2,3$ . In general if $|X|<\infty$ (there's no need to consider infinite sets for this example) it's true that $\left|X^m\right|=|X|^m$ .

In general if a set $X$ has $m$ elements, how many elements does $X^m$ have?[/QUOTE]
so you are saying the ans of that is $X^m$ then is $m^m$ since x have m elements?

**Drexel28** · November 16th 2010, 07:27 PM

Originally Posted by mathbeginner

so you are saying the ans of that is $X^m$ then is $m^m$ since x have m elements?

What I'm saying is that if $X$ has $n$ elements then $X^m$ has $n^m$ elements.

**mathbeginner** · November 16th 2010, 07:45 PM

Originally Posted by Drexel28

You know that since $\dim_{\mathbb{K}} V=n$ that $V\cong \mathbb{K}^n$ . Now, tell me. In general if a set $X$ has $m$ elements, how many elements does $X^m$ have?

I also don't really get why $\dim_{\mathbb{K}} V=n$ will become $V\cong \mathbb{K}^n$

**HallsofIvy** · November 17th 2010, 03:07 AM

Then look at simple cases. Way back in the first response TheEmptySet asked you "if V is 1 dim vector space how many distinct elements are there?" You did not answer that- instead you responded about n- dimensions.

If V is a one dimensional vector space, then it has a single basis vector, v, say. All vectors are of the form av where a is from K. If, for example, K is a finite field with 3 elements, 0, 1, 2, say, then the only vectors in the space are 0v, 1v, and 2v- exactly 3 vectors.

If V is a two dimensional vector space, then a basis contains two vectors, u and v, say. All vectors are of the form au+ bv where a and b are from K. Again, if K has only 0, 1, and 2 then the only vectors are 0u+ 0v, 0u+ 1v, 0u+ 2v, 1u+ 0v, 1u+ 1v, 1u+ 2v, 2u+ 0v, 2u+ 1v, and 2u+ 2v- exactly 9 vectors.

Do you get the point? How many vectors would there be if V were three dimensional?

**mathbeginner** · November 17th 2010, 04:55 AM

Originally Posted by HallsofIvy

Then look at simple cases. Way back in the first response TheEmptySet asked you "if V is 1 dim vector space how many distinct elements are there?" You did not answer that- instead you responded about n- dimensions.

If V is a one dimensional vector space, then it has a single basis vector, v, say. All vectors are of the form av where a is from K. If, for example, K is a finite field with 3 elements, 0, 1, 2, say, then the only vectors in the space are 0v, 1v, and 2v- exactly 3 vectors.

If V is a two dimensional vector space, then a basis contains two vectors, u and v, say. All vectors are of the form au+ bv where a and b are from K. Again, if K has only 0, 1, and 2 then the only vectors are 0u+ 0v, 0u+ 1v, 0u+ 2v, 1u+ 0v, 1u+ 1v, 1u+ 2v, 2u+ 0v, 2u+ 1v, and 2u+ 2v- exactly 9 vectors.

Do you get the point? How many vectors would there be if V were three dimensional?

Is it the basis contain 3 vectors and will have 3(vector form K)^3(dim of vector space) am I right?

**Drexel28** · November 17th 2010, 07:45 AM

Originally Posted by mathbeginner

Is it the basis contain 3 vectors and will have 3(vector form K)^3(dim of vector space) am I right?

Is this $3\left(\text{number of things in }\mathbb{K}\right)^3$ or just $\left(\text{number of things in }\mathbb{K}\right)^3$ ?

**HallsofIvy** · November 17th 2010, 07:47 AM

Yes. And if K contains 3 members, then there would be 3^n members of an n dimensional space.

If K contains k members, an n dimensional space over K would contain k^n members.

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