Diagonalisation of matrices

[Benny]

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.


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



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

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

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

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

[matt grime]

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)

[TD]

The eigenvalues appear in the diagonal matrix in the same order as the corresponding eigenvectors in the columns the P-matrix.

[Benny]

Thanks for the help matt grime and TD.

[ARNAB ROY]

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.