Jordan Canonical Form: Equivalent Operators A and B?

[psholtz]

Homework Statement

Are the operators specified by the matrices:

$$A = \left[\begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{array}\right]$$

$$B = \left[\begin{array}{ccc} 4 & 1 & -1 \\ -6 & -1 & -3 \\ 2 & 1 & 1 \end{array}\right]$$

equivalent?

Homework Equations

See below.

The Attempt at a Solution

My guess is that the answer is "yes", since the eigenvalues of both matrices is the same.

That is, the spectrum of both matrices is:

$$\sigma(A) = {1,1,2}$$

$$\sigma(B) = {1,1,2}$$

..and for matrix B, the geometric multiplicity of the eigenvalue 1 is 1 (i.e., the eigenspace is of dimension 1, as is the case for matrix A).

However, the answer is in the book, and in the book they claim that in fact the operators are not equivalent.

Is this a typo, or am I missing something?

 

[lippyka]

(I have tried to check with the Jordan Normal Form, but in both cases the matrix is already in Jordan Normal Form).