# Linear algebra rank question

1. Nov 14, 2006

### indigogirl

Linear algebra questions (rank, generalized eigenspaces)

Hi,

This seems to be an easy question on rank, but somehow I can't get it.

Let U be a linear operator on a finite-dimensional vector space V. Prove:

If rank(U^m)=rank(U^m+1) for some posiive integer m, then rank(U^m)=rank(U^k) for any positive integer k>=m.

It's in the section introducing Jordan canonical forms, so I assume the proof involves that. I tried induction and got to 'dim(U^m+p+1)<=rank(U^m) where p>1,' but I'm not sure how useful that is.

Also, I'm having trouble with this problem (not rank question, but i can't edit the title)

Let T be a linear operator on a finite-dimensional vector space V whose characteristic polynomical splits. Suppose B is Jordan basis for T, and let lambda be an eigenvalue of T. Let B'=B union K(lambda). Prove that B' is a basis for K(lambda). (K(lambda) is the generalized eigenspace corresponding to lambda)

I'd definitely appreciate some help with this!

Last edited: Nov 14, 2006
2. Nov 14, 2006

### JasonRox

This is how induction should work here.

The first step is given, so we know it is true for n=1, where n is a natural number.

The hypothesis, so assume rank(U^m)=rank(U^(m+n)).

Now, show for rank(U^m)=rank(U^(m+n+1)).

But rank(U^((m+n)+1)) = rank(U^(v+1)) *v=m+n

But, it is given that from some positive integer v, that we have...

rank(U^(v+1)) = rank(U^v)

So, we substitute that into above...

But rank(U^((m+n)+1)) = rank(U^(v+1)) = rank(U^v) = rank(U^(m+n))

But the hypothesis tells us that...

rank(U^(m+n)) = rank(U^m)

So, substitute that in above, and we get...

But rank(U^((m+n)+1)) = rank(U^(v+1)) = rank(U^v) = rank(U^(m+n)) = rank(U^m)

So...

rank(U^((m+n)+1)) = rank(U^m)

And we are done.

No mention of Canonical Forms is mentionned. I know nothing about them. :grumpy:

3. Nov 14, 2006

### indigogirl

I don't think this proof works:

You can't say,

rank(U^((m+n)+1)) = rank(U^(v+1)) *v=m+n
= rank(U^v)

because what you're trying to PROVE is that rank(U^(v+1))=rank(U^v)

It might be clearer to explain using numbers.

Let m=3, then from the problem, rank(U^3)=rank(U^3+1)=rank(U^4).

In the assumption step, we assume rank(U^3)=rank(U^3+1)=rank(U^3+2)...=rank(U^3+n) where n is any fixed integer... let's say n=4

But, you don't know what happens for the n+1th case. You can't say that
rank(U^(3+4)+1)=rank(U^(3+4) because you only know that the assumption step works until n=4... not anything over... you have to PROVE that rank(U^(3+4)+1)=rank(U^(3+4).

To say this in another way, I think you're making the mistake of assuming that v=any integer (greater than m) so you're saying rank(U^any integer greater than m)=rank(U^m) which is exactly what you're trying to prove.

4. Nov 14, 2006

### JasonRox

I see what you mean, but I simply used what we are given.

We are trying to prove rank(U^(3+5)=rank(U^(3), and I just used the rank(U^(3+4)+1)=rank(U^(3+4) property, which we are given.

5. Nov 14, 2006

### indigogirl

Let me explain this better. Using what we're given, we have:

rank(U^((m+n)+1)) = rank(U^(v+1)) for v=m+n

rank(U^v)=rank(U^(m+n))=rank(U^m)=rank(U^m+1)

but the last statement in no way implies that rank(U^m+1)=rank(U^v+1)=rank(U^v)

6. Nov 14, 2006

### indigogirl

Anyways, I found an easier, induction-less proof.

U^(m+1)(V)=U^m(U(V)), and this subspace is included in U^m(V)

If rank(U^m)=rank(U^m+1), then the above 'inclusion' turnsinto an equal sign.

So, U^m(U(V))=U^m(V).

Rewriting U(U^m(V))=U^m(V)

This means that U can be applied to U^m(V) any number of times whithout chanignt eh range of the transformation.

So, U^k(V)=U^m(V) for k>=m

Then, the ranks of the above two are equal.