# Subspace of a finite dimensional space is finite dimensional

I was reading a proof about subspaces of finite dimensional spaces are finite dimensional. The key was to add and remove vectors and then make a counting conclusion.

An excerpt (adapted from Axler) is given below

This is going to sound very stupid, but it would seem the whole proof is based on the idea that $span(v_1, \dots, v_{j-1})$ exists. How do we know $U$ is not infinite dimensional? Isn't there something wrong by assuming $U = span(v_1, \dots, v_{j-1})$

• That's the induction step. You've already picked $j-1$ vectors in the previous steps. That's why it's called "Step $j$". – wj32 Jun 29 '13 at 5:33
• @wj32, oh...okay. – Hawk Jun 29 '13 at 5:41

The span of $\{v_1,\dots,v_{j-1}\}$ is just the set of all linear combinations of the vectors $v_1,\dots,v_{j-1}$; this certainly exists, whether or not it is equal to the given subspace $U$. Note that Step $j$ does not say that $U=\operatorname{span}\{v_1,\dots,v_{j-1}\}$; it says that if $U=\operatorname{span}\{v_1,\dots,v_{j-1}\}$, then we’re done, because (as you’ll see if you read further) $\{v_1,\dots,v_{j-1}\}$ is then a finite linearly independent set of vectors spanning $U$, i.e., a basis for $U$. In fact, in that case we can say that $\dim U=j-1$. If, however, the vectors $v_1,\dots,v_{j-1}$ do not span $U$, then we can find a vector $v_j\in U\setminus\operatorname{span}\{v_1,\dots,v_{j-1}\}$. Because $v_j\notin\operatorname{span}\{v_1,\dots,v_{j-1}\}$, the new, bigger set $\{v_1,\dots,v_{j-1},v_j\}$ is linearly independent, and we go on to Step $j+1$:
• if $U=\operatorname{span}\{v_1,\dots,v_j\}$, then $\{v_1,\dots,v_j\}$ is a basis for $U$, $\dim U=j$, and $U$ is therefore finite-dimensional;
• if $U\ne\operatorname{span}\{v_1,\dots,v_j\}$, then we can find a $v_{j+1}\in U\setminus\operatorname{span}\{v_1,\dots,v_j\}$, and we construct the new, still bigger linearly independent set $\{v_1,\dots,v_j,v_{j+1}\}$ and go on to Step $j+2$.
Now recall that $V$ is finite-dimensional. Let $n=\dim V$. Then we know that if $A$ is a linearly independent set of vectors in $V$, then $|A|\le n$: you cannot find a set of more than $n$ linearly independent vectors in $V$. Thus, the process described in the proof must stop by Step $n$: any further steps would give us at least $n+1$ linearly independent vectors $\{v_1,\dots,v_{n+1}\}$, and that’s impossible.
• Basically, there is no worry about infinite dimensionality since we are actually just picking vectors in $U$. – Hawk Jun 29 '13 at 5:42
• @sidht: We don’t know at the start that $U$ is finite-dimensional. The thing is that we’re picking linearly independent vectors in V, which we do know is finite-dimensional. – Brian M. Scott Jun 29 '13 at 5:44
• Well even if $U$ isn't finite dimensional, what stops me from taking vectors in $U$? Because in the proof, it actually says choose $v_1 \in U$ no? – Hawk Jun 29 '13 at 5:47
• Actually you wrote the same thing! *a vector v_j \in U$– Hawk Jun 29 '13 at 5:54 • @sidht: Yes, we’re choosing vectors in$U$. But$U\subseteq V$, so that means that these vectors are also in$V$. And it’s the fact that they are in V (and linearly independent) that tells us that there can be only finitely many of them, not the fact that they are in$U$. It’s only after we know that there are only finitely many of them that we can use the fact that they span$U$to conclude that$U\$ is also finite-dimensional. – Brian M. Scott Jun 29 '13 at 6:00