# Matrix Diagonalisation eigen?

1. Sep 7, 2011

### quietrain

1. The problem statement, all variables and given/known data

i have matrix A which is diagonalisable

by doing an example on wiki under the section " how do diagonalise a matrix"
http://en.wikipedia.org/wiki/Diagonalizable_matrix

i realise that

A x is not equal to $\lambda$ x , where x are eigen vectors of A , $\lambda$ is eigen values

x A = $\lambda$ x

and i think

A x = x $\lambda$

QUESTION 1)
if this is so, why do they always write the hamiltonian as H |x> = E |x> ??? shouldn't it be H |x> = |x> E ?

if i remembered correctly, for matrix multiplication AB =/= BA right?

but i read wiki and it says something like (i can't remember the exact phrasing)

"it is equal if both A and B are diagonalisable matrix , and are both n by n matrix. "

QUESTION 2)
also, for PT A P = $\lambda$

if i want to "bring over" the P, is it like this

A = P $\lambda$ PT

but why is it like this?

thanks!

2. Sep 7, 2011

### niklaus

You are right that matrices don't commute, however the multiplication of a matrix with a scalar (real or complex number) always commutes. So λx = xλ. What you write above is also not correct. Ax is not at all the same as xA, the wiki page is correct.

Question 2):

this works for so called orthogonal matrices where P^T * P = P*P^T = 1 (the identity matrix). It is not true for general matrices.

3. Sep 7, 2011

### HallsofIvy

Staff Emeritus
You do not have to choose orthonormal vectors for eigenvalues but you can. If you do then the matrix P is "orthogonal"- in particular $P^T= P^{-1}$. Without P being orthogonal, that is not true but it is still true that $P^{-1}AP= D$ so, multiplying on the left by P and on the right by $P^{-1}$,
$$P(P^{-1}AP)P^{-1}= PDP^{-1}$$
$$(PP^{-1})A(PP^{-1})= IAI= A= PDP^{-1}$$

In the case that P is "orthogonal", You can replace $P^{-1}$ with $P^T$.

4. Sep 7, 2011

### quietrain

oh shucks ... for question 1 i realise i multiplied the P ,matrix of eigen vectors, with the diagonal matrix eigen values

but in fact i think they are talking about the individual eigen value and eigen functions , not the combined P.

so yup i get it thanks everyolne!