Diagonalisation of matrix

  • #1
NavalChicken
17
0

Homework Statement


This is part of a larger problem of finding [tex] e^A [/tex], where [tex]


A = \left[ \begin{array}{ccc} 0 & 0 & -2 \\ 1 & 2 & 1 \\ 1 & 0 & 3 \end{array} \right]

[/tex]


Homework Equations


-


The Attempt at a Solution




I generally have no problems with diagonalising a matrix, however, in this case I just can't seem to get it to work. I've got the eigenvectors to be [tex] (-1, 0, 1)^T, (0, 1, 0)^T, (-2, 1, 1)^T [/tex]

Once carried through the steps of finding [tex] P, A, P^{-2} [/tex] and finding the diagonal, I come out with

[tex] D = \left[ \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 1 \end{array} \right] [/tex]

I've checked it on wolframalpha and on maple and both confirm that my inverse is correct and following through the multiplication both come out with the same result. Chances are I've just made a silly mistake somewhere. Can someone help me spot it?

Thankyou
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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You mean you did D=P^(-1)AP where P is the matrix whose columns are the eigenvectors, right? I did the same thing and got D=diag(2,2,1). What did you get for P^(-1)?
 
  • #3
NavalChicken
17
0
Sorry, I did mean [tex] P^{-1} [/tex]

Which I had as:

[tex]

\left[ \begin{array}{ccc} 1 & 0 & 2 \\ 1 & 1 & 1 \\ -1 & 0 & -1 \end{array} \right] [/tex]
 
  • #4
Dick
Science Advisor
Homework Helper
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Looks right.

With P:
[tex]


\left[ \begin{array}{ccc} -1 & 0 & -2 \\ 0 & 1 & 1 \\ 1 & 0 & -1 \end{array} \right]
[/tex]

I get P^(-1)AP=diag(2,2,1). I do notice that if I do PAP^(-1) I get your off diagonal result D. That's not the right order.
 
Last edited:
  • #5
NavalChicken
17
0
Oh, yes, it was down to wrong order. Very silly!

Thanks for clearing that up
 

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