# Field and vector space

• Dec 2nd 2009, 02:44 PM
Also sprach Zarathustra
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?
• Dec 2nd 2009, 03:13 PM
tonio
Quote:

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
• Dec 2nd 2009, 03:28 PM
Also sprach Zarathustra
Shalom! I am mitbaesh beatzmi...
1. q^n-1 ?
2.?
• Dec 2nd 2009, 03:32 PM
Also sprach Zarathustra
p.s:
What is MHF expert mean?
• Dec 2nd 2009, 07:15 PM
tonio
Quote:

Originally Posted by Also sprach Zarathustra
p.s:
What is MHF expert mean?

Math Help Forum Expert.

Tonio
• Dec 2nd 2009, 07:21 PM
tonio
Quote:

Originally Posted by Also sprach Zarathustra
Shalom! I am mitbaesh beatzmi...
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
• Dec 3rd 2009, 12:47 AM
Also sprach Zarathustra
• Dec 3rd 2009, 03:04 AM
Defunkt
Quote:

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? :)
• Dec 3rd 2009, 03:59 PM
tonio
Quote:

Originally Posted by Defunkt
גם אתה מישראל? :)

אלא מה? מי עוד כותב עם הסימנים המעצבנים הללו(Wink)

Tonio
• Dec 3rd 2009, 04:09 PM
Defunkt
יפה יפה... אני מניח שכבר סיימת תואר ראשון?