# Thread: Field and vector space

1. ## Field and vector space

Let F be a field with $\displaystyle q$ elements and let V be a vector space above F with dimension of $\displaystyle n$.

1. How many different basis do V have?
2. How many subspaces of dimension of $\displaystyle k$ do V have?

2. Originally Posted by Also sprach Zarathustra
Let F be a field with $\displaystyle q$ elements and let V be a vector space above F with dimension of $\displaystyle n$.

1. Who many different basis do V have?
2. Who many subspaces of dimension of $\displaystyle k$ do V have?

It is not "who" but "how"...!נו באמת

Anyway: how many elements in V can be the first element in the basis? (all but the zero vector): how many elements can be the second element in a basis? (all but the scalar multiples of the first one, and there are q multiples like these), etc.

Try now to do the second part by yourself.

Tonio

3. Shalom! I am mitbaesh beatzmi...
Can you please write the whole answer for the question.
1. q^n-1 ?
2.?

4. p.s:
What is MHF expert mean?

5. Originally Posted by Also sprach Zarathustra
p.s:
What is MHF expert mean?
Math Help Forum Expert.

Tonio

6. Originally Posted by Also sprach Zarathustra
Shalom! I am mitbaesh beatzmi...
Can you please write the whole answer for the question.
1. q^n-1 ?
2.?

No, not the whole answer but some highlights: there are $\displaystyle q^n -1$ choices for the first vector (since zero cannot be part of a basis), then there are $\displaystyle q^n-q$ choices for the second vector (all the available vectors minus those that are a scalar multiple of the first one), then there are $\displaystyle q^n-q^2$ choices for the third one (all the possible vectors minus all the scalar multiples of the first two), etc.
All in, there are $\displaystyle (q^n-1)(q^n-q)(q^n-q^2)$...try to end the argument by yourself now.

Tonio

Ps ...הרבה יותר כיף לגלות את הפתרון בעצמך, אפילו אם זה קשה, מאשר לקבל את הכל מוכן

7. And what about 2?

8. Originally Posted by tonio
no, not the whole answer but some highlights: There are $\displaystyle q^n -1$ choices for the first vector (since zero cannot be part of a basis), then there are $\displaystyle q^n-q$ choices for the second vector (all the available vectors minus those that are a scalar multiple of the first one), then there are $\displaystyle q^n-q^2$ choices for the third one (all the possible vectors minus all the scalar multiples of the first two), etc.
All in, there are $\displaystyle (q^n-1)(q^n-q)(q^n-q^2)$...try to end the argument by yourself now.

Tonio

ps ...הרבה יותר כיף לגלות את הפתרון בעצמך, אפילו אם זה קשה, מאשר לקבל את הכל מוכן
גם אתה מישראל?

9. Originally Posted by Defunkt
גם אתה מישראל?

אלא מה? מי עוד כותב עם הסימנים המעצבנים הללו

Tonio

10. יפה יפה... אני מניח שכבר סיימת תואר ראשון?