Page 40 - Linear Algebra Done Right

Chapter 2. Finite-Dimensional Vector Spaces
26
Step 1
The list (w 1 ,...,w n ) spans V, and thus adjoining any vector to it
produces a linearly dependent list. In particular, the list
(u 1 ,w 1 ,...,w n )
is linearly dependent. Thus by the linear dependence lemma (2.4),
we can remove one of the w’s so that the list B (of length n)
consisting of u 1 and the remaining w’s spans V.
Step j
The list B (of length n) from step j−1 spans V, and thus adjoining
any vector to it produces a linearly dependent list. In particular,
the list of length (n + 1) obtained by adjoining u j to B, placing it
just after u 1 ,...,u j−1 , is linearly dependent. By the linear depen-
dence lemma (2.4), one of the vectors in this list is in the span of
the previous ones, and because (u 1 ,...,u j ) is linearly indepen-
dent, this vector must be one of the w’s, not one of the u’s. We
can remove that w from B so that the new list B (of length n)
consisting of u 1 ,...,u j and the remaining w’s spans V.
After step m, we have added all the u’s and the process stops. If at
any step we added a u and had no more w’s to remove, then we would
have a contradiction. Thus there must be at least as many w’s as u’s.

Our intuition tells us that any vector space contained in a finite-
dimensional vector space should also be finite dimensional. We now
prove that this intuition is correct.

2.7 Proposition: Every subspace of a finite-dimensional vector
space is finite dimensional.

Proof: Suppose V is finite dimensional and U is a subspace of V.
We need to prove that U is finite dimensional. We do this through the
following multistep construction.

Step 1
If U ={0}, then U is finite dimensional and we are done. If U =
{0}, then choose a nonzero vector v 1 ∈ U.

Step j
If U = span(v 1 ,...,v j−1 ), then U is finite dimensional and we are