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Worksheet 5.5 Worksheet: generalized eigenvectors

Let V be a finite-dimensional vector space, and let T:Vβ†’V be a linear operator. Assume that T has all real eigenvalues (alternatively, assume we’re working over the complex numbers). Let A be the matrix of T with respect to some standard basis B0 of V.
Our goal will be to replace the basis B0 with a basis B such that the matrix of T with respect to B is as simple as possible. (Where we agree that the "simplest" possible matrix would be diagonal.)
Recall the following results that we’ve observed so far:
  • The characteristic polynomial cT(x) of T does not depend on the choice of basis.
  • The eigenvalues of T are the roots of this polynomial.
  • The eigenspaces EΞ»(T) are T-invariant subspaces of V.
  • The matrix A can be diagonalized if and only if there is a basis of V consisting of eigenvectors of T.
  • Suppose
    cT(x)=(xβˆ’Ξ»1)m1(xβˆ’Ξ»2)m2β‹―(xβˆ’Ξ»k)mk.
    Then A can be diagonalized if and only if dim⁑EΞ»i(T)=mi for each i=1,…,k.
In the case where A can be diagonalized, we have the direct sum decomposition
V=EΞ»1(T)βŠ•EΞ»2(T)βŠ•β‹―βŠ•EΞ»k(T).
The question is: what do we do if there aren’t enough eigenvectors to form a basis of V? When that happens, the direct sum of all the eigenspaces will not give us all of V.
The idea: replace EΞ»j(T) with a generalized eigenspace GΞ»j(T) whose dimension is mi.
Our candidate: instead of EΞ»(T)=ker⁑(Tβˆ’Ξ»I), we use GΞ»(T)=ker⁑((Tβˆ’Ξ»I)m), where m is the multiplicity of Ξ».

1.

Recall that in class we proved that ker⁑(T) and im(T) are T-invariant subspaces. Let p(x) be any polynomial, and prove that ker⁑(p(T)) and im(p(T)) are also T-invariant.
Hint: first show that p(T)T=Tp(T) for any polynomial T.
Applying the result of Problem 1 to the polynomial p(x)=(xβˆ’Ξ»)m shows that GΞ»(T) is T-invariant. It is possible to show that dim⁑GΞ»(T)=m but I won’t ask you to do that. (A proof is in the book by Nicholson if you really want to see it.)
Instead, we will try to understand what’s going on by exploring an example.
Consider the following matrix.

3.

Find the eigenvectors. What are the dimensions of the eigenspaces? Based on this observation, can A be diagonalized?

4.

Prove that for any nΓ—n matrix A, we have
{0}βŠ†null(A)βŠ†null(A2)βŠ†β‹―βŠ†null(An).
It turns out that at some point, the null spaces stabilize. If null(Ak)=nullAk+1 for some k, then null(Ak)=null(Ak+l) for all lβ‰₯0.

5.

For each eigenvalue found in Worksheet Exercise 5.5.2, compute the nullspace of Aβˆ’Ξ»I, (Aβˆ’Ξ»I)2, (Aβˆ’Ξ»I)3, etc. until you find two consecutive nullspaces that are the same.
By Worksheet Exercise 5.5.4, any vector in null(Aβˆ’Ξ»I)m will also be a vector in null(Aβˆ’Ξ»I)m+1. In particular, at each step, we can find a basis for null(Aβˆ’Ξ»I)m that includes the basis for null(Aβˆ’Ξ»I)mβˆ’1.
For each eigenvalue found in Worksheet Exercise 5.5.2, determine such a basis for the corresponding generalized eigenspace. You will want to list your vectors so that the vectors from the basis of the nullspace for Aβˆ’Ξ»I come first, then the vectors for the basis of the nullspace for (Aβˆ’Ξ»I)2, and so on.

6.

Finally, let’s see how all of this works. Let P be the matrix whose columns consist of the vectors found in Problem 4. What do you get when you compute the matrix Pβˆ’1AP?
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