# If $dim U = k$, why is it that any linearly independent list of vectors in $U$ will have dimension lesser or equat than that?

Suppose I have a a list of linearly independent vectors:

$(v_1,...v_m)$ which are in $U$. Now if $dim U=k$, why is it that $m\leq{k}$?

I am trying to prove it using this theorem:

"In a finite dimensional vector space, the length of every linearly independent list of vectors is less than or equal to the length of every spanning list of vectors."

Why is it that the length of a spanning list of vectors will match up with the dimension of the vector space?

Thanks!

• This comes from the fact that if a homogeneous system of $m$ equations in $n$ unknowns has a unique solution, then $m\geqslant n$. – Pedro Tamaroff Feb 12 '14 at 9:45

The vectors $v_1,\dots, v_m$ are linearly independent, therefore, you can add vectors $w_1,\dots, w_n$ so that $\{v_1, \dots, v_m, w_1, \dots, w_n\}$ is a basis for $U$. Since all bases have $k$ elements, this means that $m+n=k$ for some $n\geq 0$.
This is basically the content of Steinitz's Lemma: if $v_1,...,v_n$ span $V$ and $w_1,...,w_m$ are linearly independent vectors in $V$, then $m\leq n$.