Finding Invertible Matrices for Equal Eigenvalues

[smerhej]

Homework Statement

Let A =

\begin{bmatrix}
\lambda & a \\
0 & \lambda \\
\end{bmatrix}

and B =

\begin{bmatrix}
\lambda & b \\
0 & \lambda \\
\end{bmatrix}

Assuming that a ≠ 0, and b ≠ 0 ; find a matrix X such that X^-1AX = B.

Homework Equations

(A- \lambdaI)v=0

The Attempt at a Solution

I tried using the following logic: Let B = {v1, v2,...vn}
be the basis of Fⁿ consisting of the columns of X. We know that column j of B is
equal to [Avj ]B, that is, the coordinates of Avj with respect to the basis B.

But because of the two matrices having the exact same eigenvalues, I just end up with a=0, and am unable to actually find an invertible matrix X. Am I misreading the question, the logic, etc..?

 

[smerhej]

Also in case the notation is different elsewhere, lambda = eigenvalue.

 

[dirk_mec1]

Am I missing something because from what I can tell just set up a 2x2 matrix with 4 unknown elements setup 4 equations and find a matrix X which satisfies those four equations.