Linearly independent sets within repeated powers of a linear operator

[Fractal20]

Homework Statement

Suppose that T:W -> W is a linear transformation such that T^m+1 = 0 but T^m ≠ 0. Suppose that {w₁, ... , w_p} is basis for T^m(W) and T^m(u_k) = w_k, for 1 ≤ k ≤ p. Prove that {Tⁱ(u_k) : 0 ≤ i ≤ m, 1 ≤ j ≤ p} is a linearly independent set.

Homework Equations

The Attempt at a Solution

By definition of a basis w₁, ..., w_p are linearly independent. Now suppose T^{m - 1}(u_k) is not linearly independent from the w's. Then it can be written as some sum of linear combinations of the w's which is equivalent to saying T^m-1(u_k) = c₁T^mu₁ + ... + c_pT^mu_p. If both sides are left multiplied by T then we have T^m = the linear combination of T^{m + 1} of the u_i's which by definition are all 0. But then we have T^m(u_k) = 0 = w_k by definition but this is a contradiction since w_k cannot be 0 if it is linearly independent of the other w_i's.

From here, it seems like this same process can be applied backwards but I am not sure how it can be rigorously done in an elegant manner. I think I can use induction and say given that T^q(u_k) is linearly independent from {T^r(u_k): q + 1 ≤ r ≤ m, 1 ≤ k ≤ p} then T^{q - 1}(u_k) is linearly independent for all k for if it wasn't then ... the same argument as the base case but applying repeated left multiples of T to keep creating 0 terms on the right eventually yielding the same contradiction of having w_k = 0. This seems bulky and I am not confident it works. Also, I'm not use to using induction in a downwards trend and don't know if I would need to do an additional bottom base case for T¹ specifically.

Is this weird attempt at induction valid? Do you have any more elegant approaches? Thanks!

For background, I am starting grad school in the fall and have to take a placement exam in linear algebra and vector calculus. This is a question off of one of the earlier placement exams.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[voko]

Your idea is right. But it is really... well.. backwards :)

Say some Tⁱu_k are linearly dependent. At least one of them has the minimal i. Does it give you any ideas?

 

[Fractal20]

Thanks! I think that helps nicely. I was a little unsure that I would need to account for the linear combination involving T^q terms for q < i. But I think I see now that if T^qu_i is linearly independent then if it's coefficient wasn't 0 then we could rearrange it to show that it was indeed linearly dependent. Is that true? If you have a set of vectors and know that some are linearly independent of all the rest, then the linear dependence of any of the vectors can't involve a linearly independent vector? Thanks so much, I am sure I will keep posting questions the next few weeks.

 

[voko]

You have a linear combination, where some vectors will have a minimum i. Put them on the left side of the equation. The ones on the right will all correspond to higher degrees of the operator. Now apply the operator (m - i) times to the whole thing (that's your idea). What do you get?

The only case when that can't work is when all i = m. But what do we have in that case?