# Homework Help: Basis for subspace

1. Jun 29, 2009

### JG89

1. The problem statement, all variables and given/known data

Suppose T is a linear operator on R^4 such that T(a,b,c,d) = (a + b, b - c, a + c, a + d). Find a basis for the T-cyclic subspace of R^4 generated by z = (1, 0, 0, 0)

2. Relevant equations

3. The attempt at a solution

I found a basis, but I don't think the method I used was the most efficient way. To save you the tedious calculations, I'll post an outline of what I did.

First denote the T-cyclic subspace in question by W. Then since W is a subspace of R^4,
1<= dim(W) <= 4. Now dim(W) cannot be equal to 1, for then any single vector in W would be a basis for W (since a set consisting of a single vector is linearly independent) and (1, 0, 0, 0) does NOT span W (I've computed enough vectors in W to know this). So we have
2<= dim(W) <= 4. The set { (1, 0, 0, 0), (1, 0, 1, 1) } is linearly independent, both those vectors are in W, and so if dim(W) = 2, then that set should span W, but it doesn't.
So we have 3 <= dim(W) <=4. What I did now was take 4 vectors that span W and showed that they aren't linearly independent, and so this must mean that dim(W) = 3, and so I took those 4 vectors that I found that span W, removed one, and showed that the other 3 are linearly independent, which is my basis.

Last edited: Jun 29, 2009
2. Jun 29, 2009

### Staff: Mentor

I don't know the answer to your question, but some of the things you have written are incorrect.
1. The set {0} is a subspace of R4 (and lots of other vector spaces), but the dimension of this subspace is 0.
2. The set {0} is linearly dependent, not linearly independent. The equation a0 = 0 has an infinite number of solutions.

3. Jun 29, 2009

### JG89

1. The dimension of W cannot be 0, because then W = {0} which I know isn't true.
2. I should have said that a set consisting of a single NON-ZERO vector is linearly independent. Either way, W still cannot be one dimensional since it isn't spanned by (1, 0, 0, 0)

Surely there has to be a way to find a basis without going through all the stuff I did?

4. Jun 29, 2009

### Dick

Your description of what you are doing is a little unclear, but if you mean that {z,T(z)} is linearly independent that's ok. Then you looked at {z,T(z),T(T(z))}, right? Just keep adding T^n(z) for increasing values of n until you get a linearly dependent set. Then you know you've added one too many.

5. Jun 29, 2009

### JG89

Beautiful - that's much simpler than what I was trying to do. Thanks.

6. Jun 29, 2009

### JG89

Just one question...

You said to keep increasing the value of n for T^n(z) until I get a linearly dependent set. Say the set {z, T(z), T^2(z)} is linearly independent and also suppose that when I choose n = 3, then the set {z, T(z), T^2(z), T^3(z)} is linearly dependent. That doesn't necessarily mean that any set consisting of 4 vectors from the T-cyclic subspace is going to be linearly dependent, does it? So then is it possible that there is another value of n, say n = 56, so that the set {z, T(z), T^2(z), T^56(z)} is linearly independent?

7. Jun 29, 2009

### Dick

Good question. If {z,T(z),T^2(z),T^3(z)} is linearly dependent then you can write T^3(z)=a*z+b*T(z)+c*T^2(z). What does that tell you about T^4(z) etc?

8. Jun 29, 2009

### JG89

If T^3(z) = a*z+b*T(z)+c*T^2(z) then T^4(z) = a*T(z) + b*T^2(z) + c*T^3(z)
= a*T(z) + b*T^2(z) + c*(a*z+b*T(z)+c*T^2(z))

= (c*a)*z + a*T(z) + b*T^2(z) + c*a*T(z) + c^2*T^2(z)

with at least one of a, b, or c being non-zero (For if a = b = c =0 then T^4(z) would be the zero vector and the set {z, T^(z), T^2(z), T^4(z)} would be linearly dependent).

If a is non-zero, then T^4(z) = a*T(z) + ... the rest with a being non-zero

If b is non-zero, then T^4(z) = b*T^2(z) + ... the rest with b being non-zero

If c is non-zero, then T^4(z) = c^2*T^2(z) + ... the rest with c^2 being non-zero

And so T^4(z) can be written as a linear combination of the vectors z, T(z), T^2(z) and so the set {z, T(z), T^2(z), T^4(z)} is linearly dependent, and by induction the set {z, T(z), T^2(z), T^n(z)} is linearly dependent for all positive integers n >= 3

Last edited: Jun 29, 2009
9. Jun 29, 2009

### Dick

That's a bit overcomplicated. But sure, if T^3(z) can be expressed in terms of T^2(z), T(z) and z, then so can T^4(z) and so on. Just apply T and replace powers greater than 2 with linear combinations of T^2(z), T(z) and z.