Diagonalize matrix

I have the matrix [2 3;3 2] and wish to diagnoalize it.

First I find the eigenvalues of : -1 and 5 and put them into a matrix as:

P=[-1 0;0 5]

Then find the eigenvectors for each value and put them into a matrix
D= [1 1; -1 1]

Then find the inverse of p

-1/5 [-1 0; 0 5]

and then what?

You've diagonalised it. It may be better to call your diagonal matrix of eigenvalues $D$ and your invertible matrix of eigenvectors $P$

Then, calling your matrix $A$, you have $A=PDP^{-1}$

You may want to check that $AP=PD$ and that $P$ is invertible (e.g. has determinant 0).

By the way, it isn't the inverse of the diagonal matrix of eigenvalues you require, instead the invertible matrix of eigenvectors.

* non zero

Just to elaborate on what TheEd said (because understanding the concept means it's easier to remember):

The idea of a diagonalised matrix is that it takes a two particular vectors (the eigenvectors) and scales them.

Now, as it is, the matrix $\begin{pmatrix} 2 & 3 \\ 3 & 2 \end{pmatrix}$ tells you that it sends $\begin{pmatrix}1 \\ 0 \end{pmatrix}$ to $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and $\begin{pmatrix}0 \\ 1 \end{pmatrix}$ to $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$. However, you also know that it sends $\begin{pmatrix}1 \\ 1 \end{pmatrix}$ to $5 \times \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and $\begin{pmatrix}-1 \\ 1 \end{pmatrix}$ to $-1 \times \begin{pmatrix} -1 \\ 1 \end{pmatrix}$, so what you want to get is a matrix which:

1. Sends $\begin{pmatrix}1 \\ 0 \end{pmatrix}$ to $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and $\begin{pmatrix}0 \\ 1 \end{pmatrix}$ to $\begin{pmatrix} -1 \\ 1 \end{pmatrix}$ (i.e. it sends the basis vectors to the eigenvectors)
2. Scales these two vectors appropriately
3. Takes you back to where you started

Since these both do the same thing to both the basis vectors, they must be the same thing, so they're equal.

This is why when you diagonalise you get something in the form $A=P^{-1}DP$, where $D$ is diagonal: the matrix $P$ sends your basis vectors to the eigenvectors of the matrix, then $D$ scales them, and finally $P^{-1}$ takes you back to where you were.

Thank you both. So D is actually the diagonalized matrix?

The question I am given asks for the modal matrix is this the P? And then it asks me to find the product : PAP-1 so is this then giving me D?

Thank you both. So D is actually the diagonalized matrix?

The question I am given asks for the modal matrix is this the P? And then it asks me to find the product : PAP-1 so is this then giving me D?

My lecture notes give me D = P^-1 A P

So does that mean that A = PDP^-1 which is backwards of what is written above

This is also correct (actually the way I learnt it and how I said it in my first post)

Sorry for being dumb here but how does A=PDP-1 and = P-1 DP when matrix multiplication is dependent on the order.