Halmos Chapter one Sections 10, 11, and 12: Subspaces, Calculus of Subspaces, and Dimension of Subspaces
Problem: If and are finite-dimensional subspace with the same dimension , and , then
Proof: Suppose not, then there is some . Then, if is the finite basis for we claim that is l.i. in . To see this suppose that
We may conclude that otherwise
and thus . Thus, may be rewritten as
and thus by the l.i. of we see that this implies that and so is a l.i. set in , but this contradicts that . Thus, there is no and so
Problem: If and are subspaces of a vector space and if every vector in belongs to either or , then either or
Proof: We prove the contrapositive, if are subspaces of such that neither equal then . To prove this, suppose not and . We first see by the condition that that there are and . Now, since we can easily see that and , but where is ? If then by virtue of being a subspace, so is which contradicts our assumption that . Similarly, . Thus, which is a contradiction. It follows that
Problem: If and are vectors such that , then and span the same subspace as and
Proof: We are attempting to show that . To see this, let then there exists such that , but notice that and so this last equation may be restated as . Thus, . Similarly, and so and so a similar reasoning may be applied. Of course, both of these sets equal
Remark: One can clearly see how this extends to general linearly dependent sets.
Problem: Suppose that and are vectors and is a subspace in a vector space ; let be the subspace spanned by and , and let be the subspace spanned by and . Prove that if but not in then
Proof: Let be a basis for , then by we know there exists such that . Now, if we may rewrite this equation as which implies that , which it assumed not to. Thus, and thus
which states precisely that
Remark: We’ve tacitly assumed something here which I’ll prove as a technical lemma, namely:
Lemma: Let be a finite dimensional subspace of a vector space with a basis then
Proof: Recall that
(where means ” is a subspace of”). Thus, noting that
we see from the definition that
conversely, to show the opposite inclusion holds it suffices to prove that
for a fixed but arbitrary subspace containing . But, this is immediate since any subspace containing contains not only but all of the possible linear combinations, and so the above inclusion holds. Thus,
from where the conclusion follows.
Problem: Suppose that , and are subspaces of a vector space .
a) Show that the equation
isn’t necessarily true.
b) Prove that
a) Consider as a vector space and consider the three subspaces of over
But, since and we see that and so is not trivial.
b) Let then and for some and some . But, since and we see that . Thus, and so where and and so
Conversely, let , then where and . But, since is the sum of two elements of , which is a subspace we immediately see that . But, with the same credence we can can see that so that where and , thus . Putting this together we get that .
The conclusion follows.
a) Can it happen that a non-trivial subspace of a vector space has a unique complement?
b) If is an -dimensional subspace of an -dimensional vector space, then every complement of has dimension
a) Yes,there are lots of examples ;). In fact, if
b) By assumption we have that and are subspaces of and so we know there exists bases and for and respectively. We will show that is a basis for . To see this suppose that
but the LHS is an element of and the RHS an element of and since it follows that
and by the l.i. of and this implies that
from where l.i. follows. Now, the fact that
is fairly clear. By assumption for every there exists and such that
It follows that is a basis for and so
from where it follows that
a) Show that if both are three-dimensional subspaces of a five dimensional vecto space, then and are disjoint.
b) If and are finite dimensional subspaces of a vector space, then
Lemma: Let be subspaces of some vector space . Then, if and are respective bases then
Proof: We first prove that
To do this we first remember that
and since evidently a subspace of which contains it is immediate that
Conversely, let . To show that it suffices to show that for a fixed but arbitrary subspace containing . But, clearly implies that , but since is a subspace we see that any linear combination of elements of is contained in , in particular by definition is a linear combination of such vectors and so . The conclusion follows.
We now prove that
To see this, we first begin by remembering a definition, namely:
To see this let , to show that it suffices to show that for a fixed but arbitrary subspace containing , but this is immediate since and so , but since it’s a subspace
Conversely, we merely note that is a subspace which contains and so by definition
and so the conclusion follows.
Lemma: Let be two subspace of (over ) with
then is a subspace and
Proof: The fact that it’s a subspace is elementary, and clear thus it suffices to prove the second claim. To do this let and be bases for and respectively, we claim that is a base for . To do this suppose that
for some , then
but the LHS is an element of and the RHS an element of and so by assumption
and so by the l.i. of and we see that
and so is l.i. Next, we note from the above that
and thus it follows that is a basis for and so in particular
thus the lemma is proven.
Corollary: Given an -dimensional vector space and -dimensional vector spaces and such that then
b) Let be a basis for , then we may extend it to a basis (where ) for and a basis ( same condition) for . We claim that is a basis for . It is evident that
and so it suffices to prove l.i. Suppose that
where not all the coefficients are zero. Now, clearly we must have that and for some since and are l.i. Thus
and so there is some non-zero (considering what we just said) vector
But, this implies that and so and since the two sets are disjoint it follows that which is a contradiction. It follows that is a basis for . Note, that in particular this implies that . So
(the last part using the disjointness of the sets), but this is equal to
But, noticing that , , , and we may put this all together to get