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Diagonalisation of matrices

  • Thread starter Benny
  • Start date
  • #1
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0
Hi I'm wondering if the 'order' in which vectors are taken is important in the process of matrix diagonalisation. To clarify what I mean here is an example.

[tex]
A = \left[ {\begin{array}{*{20}c}
7 & { - 2} \\
{15} & { - 4} \\
\end{array}} \right]
[/tex]


I need to diagonalise matrix A. So I need a matrix D such that [tex]D = P^{ - 1} AP[/tex].

I calculate the eigenvalues for A, and got bases for the eigenspace associated with each of the eigenvalues. Following the procedure in my book I took the union of the two(it turned out that there are two bases) bases which I found to be: {(2,5),(1,3)}.

So [tex]P = \left[ {\begin{array}{*{20}c}
2 & 1 \\
5 & 3 \\
\end{array}} \right] \Rightarrow P^{ - 1} = \left[ {\begin{array}{*{20}c}
3 & { - 1} \\
{ - 5} & 2 \\
\end{array}} \right][/tex] where I have formed the matrix P whose columns are the vectors in the set which is the union of the two bases for the eigenspaces.

My calculations yield [tex]D = \left[ {\begin{array}{*{20}c}
2 & 0 \\
0 & 1 \\
\end{array}} \right][/tex].

The answer is [tex]D = \left[ {\begin{array}{*{20}c}
1 & 0 \\
0 & 2 \\
\end{array}} \right][/tex].

I'm not sure where my error is. I've checked the matrix multiplication for D and also PP^-1 = I.
 

Answers and Replies

  • #2
matt grime
Science Advisor
Homework Helper
9,395
3
there is no canonical diagonal form. if the basis of e-vectors is u,v for you, then in the book they've simply got v,u in the reverse order.

they differ by a poermutaion of the basis (try multiplying yours by the matrix with 1's on the off diagonal places and zero on the diagonals (a self inverse symmetric matrix)
 
  • #3
TD
Homework Helper
1,022
0
The eigenvalues appear in the diagonal matrix in the same order as the corresponding eigenvectors in the columns the P-matrix.
 
  • #4
584
0
Thanks for the help matt grime and TD.
 
  • #5
1
0
PLEASE CHANGE THE ORDER.TAKE
p=
1 2
3 5
you'll get the desired answer within a moment.This change does'nt alter the physical meaning,but the form only.
 

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