Hi

how to show that the dimension of a subvector space ist smaller or at least equal to the dimension of the main vectorspace?

I know that the dimension is the length of the basis but how to use this information here?

thank you

greetings

Printable View

- Nov 21st 2008, 03:44 AMHerbststurmDimension of Vectorspaces
Hi

how to show that the dimension of a subvector space ist smaller or at least equal to the dimension of the main vectorspace?

I know that the dimension is the length of the basis but how to use this information here?

thank you

greetings - Nov 21st 2008, 05:18 AMThePerfectHacker
If is a subspace of then say then it means we can find a basis set for , . But these vectors are linearly independent. The problem is that a set of linearly independent vectors cannot exceede the dimension and we see that exceede the dimension of . This is a contradiction.

- Nov 21st 2008, 09:03 AMHerbststurm
Sorry, I don't understand your post :(

How could you write that the dimension of the subspace is bigger than the main space?

Ad why do you say the base is linear dependent. I thought a base is always a linear indipendent creation system?

greetings - Nov 21st 2008, 09:30 AMThePerfectHacker
Say and as a subspace has . Then this means has a basis of four elements, say, . This is a linearly independent set in . Since because . Therefore, is a linearly independent set in . But it is impossible to have a linearly independent set so that the number of elements in the set exceede the dimension of the space. Here the number of elements are and the dimension of W is and we see that . This is a contradiction. Thus, the basis for must have fewer or equal to than elements.