# Homework Help: Linear Algebra - Eigenvalues / Eigenvectors

1. Mar 6, 2008

### steelphantom

I have two problems here; one I think I almost have but I'm stuck, and the other I'm pretty much stumped on.

1. The problem statement, all variables and given/known data
Suppose V is a complex vector space and T is in L(V). Prove that T has an invariant subspace of dimension j for each j = 1, ... dim(V).

2. Relevant equations

3. The attempt at a solution
I know that every operator on a finite-dimensional, nonzero complex vector space has an eigenvalue. I think we also know that V has subspaces of dimension j for each j = 1, ... dim(V). Each of these subspaces is complex, so each of them has at least one eigenvalue, so T(v) = $$\lambda$$v for each subpace, which makes T invariant on that subspace. Am I totally off the wall here, or am I close?

1. The problem statement, all variables and given/known data
Suppose n is a positive integer and T in L(Fn) is defined by T(x1, ... , xn) = (x1 + ... + xn, ... , x1 + ... + xn). Find the characteristic polynomial of T.

2. Relevant equations
Not really an equation, but the notation "F" represents either R or C.

3. The attempt at a solution
With respect to the standard basis B, the matrix M(T, B) is an n x n matrix consisting of all 1s. To find the characteristic polynomial, I need to find det(I*x - M(T, B)), but I am having trouble doing that. Any tips on how to calculate this determinant? Or is there perhaps a simpler approach? I know the eigenvalues are 0 and n.

Last edited: Mar 6, 2008
2. Mar 6, 2008

### HallsofIvy

?? T is a linear transformation. It doesn't have "subspaces". There may be subspaces that have some property in relation to T. I rather suspect, from what you say below, thinking of T as a linear transformation ON the subspace that you are referring to invariant subspaces- which is what you are trying to prove!

Looks like a problem posted just this morning! Same advice: try n= 2, n= 3, n= 4 and see what happens you should see a fairly simple pattern. In any case, you know that you find the eigenvalues by solving the equation with the characteristic polynomial set to 0 don't you? That means that the "characteristic polynomial" is just $(x- \lambda_1)(x- \lambda_2)\cdot\cdot\cdot$.

3. Mar 6, 2008

### steelphantom

Duh... Stupid mistake. I meant V has subspaces of dimension j for each j = 1, ... dim(V). This is true, correct? Now does the rest of my logic make sense?

4. Mar 6, 2008

### Dick

For the second one, how do you know the eigenvalues are 0 and n? If you know that then (as Halls said) the characteristic polynomial must be (x-0)^j*(x-n)^k where j+k=n. Now you just need to find the multiplicity of each eigenvalue to fix j and k.

5. Mar 6, 2008

### steelphantom

This was from our homework last week. We were given the same T and asked to find its eigenvalues and eigenvectors.

Edit: Ok, I've found the characteristic polynomial. Now could anyone tell me if my revised approach to the first one is correct?

Last edited: Mar 6, 2008
6. Mar 6, 2008

### Dick

Nope. The approach to the first problem is off the wall. Sure, V has subspaces of all dimensions<=n. But you need INVARIANT subspaces. Having an eigenvector trivially gives you a one dimensional invariant subspace, but it's not clear where to go from there. Here's a hint. Suppose the matrix of T is upper triangular. Does that make it clear what the invariant subspaces are? The second part of the hint is that all nxn complex matrices can be written in upper triangular form by a similarity transformation. Did you prove that?

Last edited: Mar 6, 2008
7. Mar 6, 2008

### steelphantom

We proved that all n x n complex matrices can be written in upper triangular form. Are the invariant subspaces Im(T - $$\lambda$$I) ?

8. Mar 6, 2008

### Dick

Nooo. If T is in upper triangular form with respect to the basis x1,x2,...xn define V_k to be all linear combinations of x1,...,xk. V_k is a subspace. Can you show V_k is invariant? Just imagine the matrix multiplication.