## Halmos Chapter one Section five, six, and seven: Linear Dependence, Linear Combinations, and Bases

1.

**Problem:**

**a) **Prove that the four vectors , and are linearly dependent, but any three are linearly independent

**b) **If the vectors and in are given by and , prove that and are dependent, but any three of them are linearly independent

**Proof:**

**a) **Clearly all four are l.d. (linearly dependent) since . Now, clearly are l.i. (linearly independent) and so it remains to show that are l.i. We do this only for the first case, since the others are done similarly. So, suppose that

comparison of the third coordinates tells us that , and thus from where l.i. follows.

**b) **Clearly they are l.d. since . We can prove that any three are l.i. in much the same way as we can the tuples (clearly this is not coincidence, the variables when treated as formal objects are glorified placeholders for the coefficients) and so we once again prove one. Namely, if

then in particular as well as that

Now, noting that for all we in particular may note that it’s true for all . And, for such we see that

and in particular, . Finally, repeating this process again shows that from where it all cascades back to show that

2.

**Problem: **Prove that if is considered as a vector space over , then a necessary and sufficient condition that the vectors and in be l.i. is that the real number is irrational.

**Proof:** This is evident. Suppose that but was not l.i., then there exists nonzero for which (neither can be zero since one being zero implies the other must be zero, considering ) but, this in particular means that , which is a contradiction. Conversely, suppose that is l.i. but . Then, there exists such that . Clearly, and so this violates the l.i. of

3.

**Problem: **Is it true that if , and are l.i. vectors, then so are ?

**Proof: **Yes. Note that if

then the l.i. of tells us that the system of equations

But,

upon which cancellation gives from where the rest follows.

*Remark:* This clearly generalized, but it’s too late. I’ll come back later and think about it.

4.

**Problem:**

**a) **Under what conditions on the scalar are the vectors and in l.d.?

**b) **Under what conditions on the scalar are the vectors and in l.d.?

**c) **What is the answer to b) for

**Solution:**

**a) **We first note that

So, we are looking to see when this expression equals zero. Clearly, setting this equal to zero gives us

Now, since a quick check would show that they would have to then be zero. So, we may compare these two equations and arrive at . Thus, they are l.d. precisely when they coincide.

**b) **We note first that

thus, if we set this equal to zero we get the following three equations

We then note that

So, if we assume that we arrive at . Thus and so says that and so it follows that . Thus, the only possibility is that . Checking this we find that and are linearly dependent.

**c) **The above implies that the same conclusion must be drawn.

5.

**Problem: **Prove the following

**a) **The vectors and in are l.d. implies

**b) **Find a similar necessary condition for the l.d. of three ectors in .

**c) **Is there a set of three l.i. vectors in ?

**Proof:**

**a) **We first note that

and thus we are afforded the equations

or in matrix form

Now, if were invertible then

and thus the vectors are l.i. It follows that is not invertible, or equivalently

**b) **For three vectors we follow the same line of logic and note that the determinant formed by having the the three vectors as rows must be zero.

**c) **Yes, what about and

6.

**Problem:** Prove the following

**a) **Under what conditions on the scalars are the vectors l.d.

**b) **Under what conditions on the scalars and are the vectors and l.d. in ?

**c) **Generalize to

**Proof:**

**a) **By the last problem we see that they are l.d. iff

**b) **By the last problem we see they are l.d. iff

**c) **It is clear following the same logic that are l.d. iff

in other words, iff

7.

**Problem: **Prove the following

**a) **Find two bases in such that the only vectors common to both are and

**b) **Find two bases in that nave no vectors in common, so that one of them contains the vectors and and the other one contains teh vectors and

**Proof:**

**a) **Consider . To see that this set is l.i. we note that

clearly implies that and the fact that quickly follows. Also, if then taking and we can readily see that . Thus, is, in fact, a basis for .

Using the same process we can see that forms a basis, and

**b) **One can check that from last time and work.

8.

**Problem:**

**a) **Under what conditions on the scalar do the vectors and form a basis of ?

**b) **Under what conditions on the scalar do the vectors and form a basis of

**Proof:**

**a) **Note that if this set of vectors is surely not l.i. Thus, we may assume that . We note then that if

then the three equations

hold. Namely, we see that

and thus by we see that . Thus, if that these vectors are l.i. But, suppose that were such that

We see that and so insertion of this into shows that , and insertion of this into gives that . But, inserting these into gives

which is a contradiction. Thus, these vectors can never be a basis.

9.

**Problem: **If is the set consisting of the six vectors find two different maximal independt subsets of .

**Proof: **It is tedious, but one can prove that and are two such setes.

10.

**Problem:** Let be a vector space. Prove that has a basis.

**Proof: **We can prove something even stronger. We can prove that given a set of l.i. vectors that there is a basis of such that . To do this, let

To prove this we first note that is a partially ordered set. Also, given some chain we can easily see that is an upper bound. To see that we let (remember we’re dealing with the arbitrary notion of l.i., not necessarily the finite one). Then, by definition there exists such that . Now, it clearly follows (from being a chain) that

but, that means that is contained within an element of , namely it is the subset of a set of l.i. vectors, and thus l.i. Thus, . So, evoking Zorn’s lemma we see that admits a maximal element . We claim that . To see this, suppose not. Then, there exists some . Now, let be a finite subset of . Clearly if then it is a l.i. set, and if for some , then we see that

since otherwise

which contradicts that . But, clearly implies (by the l.i. of ) that . Thus, is l.i. and so is contained in . But, this contradicts the maximality of . It follows that no such exists, in other words

(since . So, taking we see that must admit a basis.

Old post, I know, but I’m working through some of the early exercises in Halmos (slowly) and just came across your site. I’m happy to see it, and it’s in my bookmarks.

I think you’re mistaken on one of these, however. For #3 — “If it’s true that {x,y,z} are l.i., then so are {x+y,y+z,z+x}.” (From your note at the end, it seems like you were a bit unsure of the proof.) I am almost certain this is false. As a counterexample, think of the vector space consisting of 3-tuples over the most trivial field (containing 0 and 1 only, with 1+1=0, equivalent to the Z_2 field Halmos defines in the first batch of exercises). Let x=(1,0,0), y=(0,1,0), z=(0,0,1) — clearly l.i. We then have:

x+y=(1,1,0)

y+z=(0,1,1)

z+x=(1,0,1)

The sum (x+y)+(y+z)+(z+x) then equals (0,0,0). The mistake in the proof comes in juggling the equalities — cancellation doesn’t give alpha_3=0, it gives alpha_3=-alpha_2=-alpha_3.

Hope this helps. I spent most of a page trying to prove the statement true before thinking about this example. I look forward to checking some of your later proofs as I work forward, so the postings are very much appreciated.

Comment by jh | March 10, 2011 |

Dear Jackie,

Thank you for pointing that out. I must admit that I was anxious to get to some of the later stuff and to be quite frank Halmos wasn’t my main goal and so I kind of rushed, but I have a compulsion to do problems.

Now that I have padded my ego after making a stupid mistake I agree with what you said 100%…good job. 🙂

Best,

Alex

Comment by drexel28 | March 10, 2011 |

Hello, Another comment.

I think your solution to 4(b) is incorrect. (1) – (3) = (a_1 – a_3)x. The answer should be (I believe) x = 0, +sqrt(2), -sqrt(2). That would mean part (c) needs fixed too.

Comment by tyler | March 27, 2012 |