Linear Algebra - Eigenvalues / Eigenvectors

In summary, the first problem involves proving that for a complex vector space V and a linear transformation T in L(V), T has an invariant subspace of dimension j for each j = 1, ... dim(V). The approach of using the existence of eigenvalues and corresponding eigenvectors is not sufficient, as it only gives one-dimensional invariant subspaces. Instead, the matrix of T can be written in upper triangular form through a similarity transformation, and this makes it clear what the invariant subspaces are. For the second problem, the characteristic polynomial can be found by considering the matrix of T with respect to the standard basis, which consists of all 1s. The characteristic polynomial is then (x-0)^j*(x-n)^
  • #1
steelphantom
159
0
I have two problems here; one I think I almost have but I'm stuck, and the other I'm pretty much stumped on.

Homework Statement


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).

Homework Equations



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) = [tex]\lambda[/tex]v for each subpace, which makes T invariant on that subspace. Am I totally off the wall here, or am I close?

Homework Statement


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.

Homework Equations


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

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.

Thanks for your help, guys!
 
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  • #2
steelphantom said:
I have two problems here; one I think I almost have but I'm stuck, and the other I'm pretty much stumped on.


Homework Statement


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).

Homework Equations



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 T has subspaces of dimension j for each j = 1, ... dim(V).
?? 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!

Each of these subspaces is complex, so each of them has at least one eigenvalue, so T(v) = [tex]\lambda[/tex]v for each subpace, which makes T invariant on that subspace. Am I totally off the wall here, or am I close?

Homework Statement


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.

Homework Equations


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

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.

Thanks for your help, guys!
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 [itex](x- \lambda_1)(x- \lambda_2)\cdot\cdot\cdot[/itex].
 
  • #3
HallsofIvy said:
?? 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!

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
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
Dick said:
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.

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?
 
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  • #6
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?
 
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  • #7
Dick said:
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?

We proved that all n x n complex matrices can be written in upper triangular form. Are the invariant subspaces Im(T - [tex]\lambda[/tex]I) ?
 
  • #8
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.
 

Related to Linear Algebra - Eigenvalues / Eigenvectors

1. What are eigenvalues and eigenvectors?

Eigenvalues and eigenvectors are concepts in linear algebra that are used to understand how a linear transformation affects a vector. Eigenvalues are scalar values that represent how much a vector is stretched or shrunk by the linear transformation. Eigenvectors are the corresponding vectors that are only scaled by the linear transformation, but their direction remains unchanged.

2. How are eigenvalues and eigenvectors calculated?

To calculate eigenvalues and eigenvectors, the linear transformation is represented by a matrix. The eigenvalues are then found by solving the characteristic equation for the matrix. The eigenvectors can be found by solving a system of linear equations using the eigenvalues.

3. What are the applications of eigenvalues and eigenvectors?

Eigenvalues and eigenvectors are used in a variety of fields, including physics, engineering, and computer science. They are used to study and understand the behavior of systems that can be modeled as linear transformations, such as in quantum mechanics, structural analysis, and image processing.

4. Can a matrix have more than one eigenvalue and eigenvector?

Yes, a matrix can have multiple distinct eigenvalues and corresponding eigenvectors. In fact, the number of eigenvalues and eigenvectors is equal to the dimensions of the matrix. However, a matrix can also have repeated eigenvalues and eigenvectors, in which case the eigenspace (the set of all eigenvectors corresponding to a specific eigenvalue) may have a higher dimension than 1.

5. How are eigenvalues and eigenvectors related to diagonalization?

Diagonalization is a process in which a matrix is transformed into a diagonal matrix using its eigenvalues and eigenvectors. This allows for easier computation and analysis of the matrix, as well as revealing important properties of the original matrix. A matrix can only be diagonalized if it has a full set of linearly independent eigenvectors.

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