# Diagonalisation of matrices

1. Sep 21, 2005

### 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.

2. Sep 21, 2005

### 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)

3. Sep 21, 2005

### TD

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

4. Sep 21, 2005

### Benny

Thanks for the help matt grime and TD.

5. Jul 2, 2010