Prove that if W is a subspace of a finite dimensional vector
space V, then Wis finite dimensional.
Proof. Assume that V is a vector space of finite dimension
n and that W is a subspace of V.
Case 1. If W has the zero vector as its only vector then W
is finite-dimensional by definition.
Case 2. So assume that W is not just {0}. Since
every vector in W also belongs to V, V is not just {0}
either. Hence V is n-dimensional with n > 0. This means that
V has a basis with exactly n vectors in it . Moreover, it tells
us that independent sets in V cannot have more than n vectors,
so independent sets in W cannot have more than n vectors, a key
fact that we will use a little later. The other key fact that
we will use is the Plus portion of the Plus/Minus Theorem. Now
we proceed as follows.
Choose some nonzero vector w1 in W. If
every vector in W is a scalar multiple of
w1, then {w1} is a basis
for W (independence follows from the general theorem
ku=0 and k nonzero ==> u=0) and W is
finite-dimensional.
If some vector in W fails to be a scalar multiple of
w1 then choose a vector
w2that belongs to W but is not a scalar
multiple of w1. By the Plus part of the
Plus/Minus Theorem, the set {w1,
w2} is linearly independent. If every
vector in W is a linear combination of w1
and w2 then {w1,
w2} is a basis for W and W is
finite-dimensional.
If some vector in W fails to be a linear combination of
w1 and w2 , then choose
a vector w3 that belongs to W but is not a
linear combination of w1 and
w2 . By Plus/Minus, the set
{w1, w2 ,
w3 } is independent. If it also spans W
then it is a basis for W and W is finite-dimensional. If
not, we choose yet another vector that is not a linear
combination of those already chosen.
This process cannot continue indefinitely, since
independent sets in W cannot have more than n vectors. Thus
at some point we arrive at a basis for W (with no more than
n vectors in it) and conclude that W is
finite-dimensional.
Note that this proof also shows that the dimension of W
cannot exceed the dimension of V.
True or False? Either prove true or give a counterexample to
show not.
If W is a subspace of a finite-dimensional vector space V, and B
is a basis for V then some subset of B is a basis for
W.
This is false. I'd hoped that recognizing its
falsity would help you reject one potential "false proof" for the
preceding problem. Alas, that particular false proof is just too
tempting. A suggestion for you: test things out with simple
examples whenever you can. You'll be less likely to believe your
own "false proofs". Just as you may use examples from your
personal life to argue against pronouncements that others make in
a discussion, test linear algebra "pronouncements" against your
own increasing store of linear algebra examples.
An example to show it is false. Let V be
R2 with its usual operations. Let B be the
standard basis {(1,0), (0,1)}. Let W be the subspace {(x,x):
x is a real number}. Neither of the basis vectors in B belongs to
W, so we can't choose vectors in B to make up a basis for W.
So what would be true concerning bases for W and bases for
V? If W is a subspace of a finite-dimensional vector space V, then
each basis for W can be expanded, if necessary, to a basis for V.
(For instance, given the basis {(1,1)} for the W in the example we
could insert the vector (1,0), say, and obtain a basis for
V=R2. Some bases for V contain bases for B but
some do not.