# Dimension of Vectorspaces

• November 21st 2008, 03:44 AM
Herbststurm
Dimension 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
• November 21st 2008, 05:18 AM
ThePerfectHacker
Quote:

Originally Posted by Herbststurm
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

If $V$ is a subspace of $W$ then say $\text{dim}(V) > \text{dim}(W)$ then it means we can find a basis set for $V$, $\{v_1,...,v_n\}$. But these vectors are linearly independent. The problem is that a set of linearly independent vectors cannot exceede the dimension and we see that $v_1,...,v_n$ exceede the dimension of $W$. This is a contradiction.
• November 21st 2008, 09:03 AM
Herbststurm
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
• November 21st 2008, 09:30 AM
ThePerfectHacker
Quote:

Originally Posted by Herbststurm
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

Say $\text{dim}(W) = 3$ and $V\subseteq W$ as a subspace has $\text{dim}(V) = 4$. Then this means $V$ has a basis of four elements, say, $\{ v_1,v_2,v_3,v_4\}$. This is a linearly independent set in $V$. Since $v_i \in V \implies v_i \in W$ because $V\subseteq W$. Therefore, $\{ v_1,v_2,v_3,v_4\}$ is a linearly independent set in $W$. 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 $4$ and the dimension of W is $3$ and we see that $4>3$. This is a contradiction. Thus, the basis for $V$ must have fewer or equal to than $3$ elements.