## Halmos Chapter one Sections 10, 11, and 12: Subspaces, Calculus of Subspaces, and Dimension of Subspaces

1.

**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

2.

**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

3.

**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.

4.

**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

for any

**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.

5.

**Problem: **Suppose that , and are subspaces of a vector space .

**a) **Show that the equation

isn’t necessarily true.

**b) **Prove that

**Proof:**

**a) **Consider as a vector space and consider the three subspaces of over

and

Notice that

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.

6.

**Problem:**

**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

**Proof:**

**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

then

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

since

It follows that is a basis for and so

from where it follows that

7.

**Problem: **

**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

**Proof:**

**a)**

**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 if

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

as desired.

Nice. What do you do when you can’t figure out how to prove something? Or does that never happen? : )

Comment by Ragnarok | September 28, 2010 |

Ha! It happens a lot. When, it does I like to think about it. Do something else, you know. If I can’t I sometimes get impatient and look at the first line of a proof in a book or something haha.

Comment by drexel28 | September 29, 2010 |