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
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 $\displaystyle V$ is a subspace of $\displaystyle W$ then say $\displaystyle \text{dim}(V) > \text{dim}(W)$ then it means we can find a basis set for $\displaystyle V$, $\displaystyle \{v_1,...,v_n\}$. But these vectors are linearly independent. The problem is that a set of linearly independent vectors cannot exceede the dimension and we see that $\displaystyle v_1,...,v_n$ exceede the dimension of $\displaystyle W$. This is a contradiction.
Say $\displaystyle \text{dim}(W) = 3$ and $\displaystyle V\subseteq W$ as a subspace has $\displaystyle \text{dim}(V) = 4$. Then this means $\displaystyle V$ has a basis of four elements, say, $\displaystyle \{ v_1,v_2,v_3,v_4\}$. This is a linearly independent set in $\displaystyle V$. Since $\displaystyle v_i \in V \implies v_i \in W$ because $\displaystyle V\subseteq W$. Therefore, $\displaystyle \{ v_1,v_2,v_3,v_4\}$ is a linearly independent set in $\displaystyle W$. 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 $\displaystyle 4$ and the dimension of W is $\displaystyle 3$ and we see that $\displaystyle 4>3$. This is a contradiction. Thus, the basis for $\displaystyle V$ must have fewer or equal to than $\displaystyle 3$ elements.