# Generalized Eigenspace and JOrdan Form

1. May 25, 2010

### hitmeoff

1. The problem statement, all variables and given/known data
For each linear operator T, find a basis for each generalized eigenspace of T consisting of a union of disjoint cycles of generalized eigenvectors. The find a Jordan canonical form J of T.

a) T is the linear operator on P2(R) defined by T(f(x)) = 2f(x) - f '(x)

2. Relevant equations

3. The attempt at a solution

OK, so I know the matrix rep of this transformation on the standard basis {1, x, x2}:
T(1) = 2
T(x) = -1 + 2x
T(x2) = -2x + 2x2

$$[T]_{\beta}$$ =
2 -1 0
0 2 -2
0 0 2 and the eigenvalue of this matrix is $$\lambda$$ = 2 with a multiplicity of 3.

I know the JOrdan form will be :

2 1 0
0 2 1
0 0 2 , just not sure how to get the basis or how to get the J from the [T]

2. May 25, 2010

### HallsofIvy

Staff Emeritus
Any eigenvector corresponding to eigenvalue 2 must satisfy
$$\begin{bmatrix}2 & -1 & 0 \\0 & 2 & -2 \\ 0 & 0 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}$$
$$= \begin{bmatrix}2a- b & 2b- 2c & 2c\end{bmatrix}=\begin{bmatrix} 2x \\ 2y \\ 2z\end{bmatrix}$$

So we have 2a- b= 2a, 2b- 2c= 2b, and 2c= 2c. The first says that b= 0 and the second that c= 0. All eigenvectors are multiples of $$\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$$.

The characteristic equation for this operator is $(\lambda- 2)^3= 0$ and, since a matrix always satifies its own characteristic equation, we must have $(T- 2I)^3v= 0$ for all vectors v. Of course if Tv= 2v, that is if v is an eigenvector, the (T- 2I)v= 0 so $(T- 2I)^3v= 0$. But the eigenvectors for this T only span a one-dimensional subspace.

We need to find v such that (T-I)v is not 0 but is in that eigen-space: then we would have $(T- I)^2((T-I)v)= (T-I)^2v= 0$- a "generalized eigenvector".

That is, we look for x, y, z such that
$$\begin{bmatrix}0 & -1 & 0 \\ 0 & 0 & -2 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}$$$$= \begin{bmatrix} -y \\ -2z \\ 0 \end{bmatrix}= \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$$.

It is easy to see that
$$\begin{bmatrix}0 \\ -1 \\ 0\end{bmatrix}$$
works.

But now we need a vector such that neither (T- 2I)v= 0 nor $(T- 2I)^2v= 0$ but such that $(T- 2I)^3v= 0$. Now we need $(T- 2I)^2v$ to be an eigenvector which will be the case if
$$\begin{bmatrix}0 & -1 & 0 \\ 0 & 0 & -2 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}$$$$= \begin{bmatrix}-y \\ -2z \\ 0\end{bmatrix}= \begin{bmatrix}0 \\ -1 \\ 0\end{bmatrix}$$

That gives
$$\begin{bmatrix}0 \\ 0 \\ \frac{1}{2}\end{bmatrix}$$

Those three vectors form the basis. If P is the matrix having those vectors as columns, then $P^{-1}TP$ will be in Jordan Normal Form.

Last edited: May 25, 2010