How Does the Rank Stability of Linear Operators Influence Their Powers?

In summary: So, rank(U^k)=rank(U^m) for k>=mIn summary, Linear algebra questions (rank, generalized eigenspaces) can involve proving that if rank(U^m)=rank(U^m+1) for some positive integer m, then rank(U^m)=rank(U^k) for any positive integer k>=m. This can be done using induction or by showing that the range of the transformation remains the same when U is applied to U^m(V), resulting in rank(U^m)=rank(U^k) for k>=m. Additionally, understanding Jordan canonical forms can be useful in solving these types of problems.
  • #1
indigogirl
9
0
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!
 
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  • #2
indigogirl said:
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.

I'd definitely appreciate some help with this!

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.
 
  • #3
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
indigogirl said:
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.

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
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
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.
 

Related to How Does the Rank Stability of Linear Operators Influence Their Powers?

1. What is the rank of a matrix in linear algebra?

The rank of a matrix is the number of linearly independent rows or columns in the matrix. In other words, it is the maximum number of rows or columns that are not linearly dependent on each other.

2. How is the rank of a matrix related to its dimensions?

The rank of a matrix is always less than or equal to the number of rows or columns in the matrix. In other words, the rank cannot exceed the smaller dimension of the matrix.

3. Can a matrix have a rank of zero?

Yes, a matrix can have a rank of zero if all of its entries are equal to zero. This means that all of the rows or columns are linearly dependent on each other and do not contribute any new information to the matrix.

4. What is the significance of the rank in linear algebra?

The rank of a matrix is important because it determines the number of solutions to a system of linear equations. A matrix with full rank has a unique solution, while a matrix with a lower rank may have infinitely many solutions or no solutions at all.

5. How is the rank of a matrix calculated?

The rank of a matrix can be calculated by performing row reduction operations on the matrix and counting the number of non-zero rows in the reduced matrix. This is also known as finding the echelon form of the matrix.

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