Finding X of (X^-1)AX = B

  • Thread starter smerhej
  • Start date
  • #1
20
0

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- [itex]\lambda[/itex]I)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..?
 

Answers and Replies

  • #2
20
0
Also in case the notation is different elsewhere, lambda = eigenvalue.
 
Last edited:
  • #3
761
13
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.
 

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