# Thread: Diagonalisation of symmetric matricies

1. ## Diagonalisation of symmetric matricies

I have just learnt how to reduce a symmetric matrix to a diagonal one, using the equation below:

$\displaystyle D={P}^{T}AP$
Where A is the original symmetric matrix and D is the diagonal matrix.
P is an orthogonal matrix whose columns consist of the normalised eigenvectors of A.

My problem is how to write the above equation in terms of A.

2. Originally Posted by incblue
I have just learnt how to reduce a symmetric matrix to a diagonal one, using the equation below:

$\displaystyle D={P}^{T}AP$
Where A is the original symmetric matrix and D is the diagonal matrix.
P is an orthogonal matrix whose columns consist of the normalised eigenvectors of A.

My problem is how to write the above equation in terms of A.
You just need to solve for A.

First multiply on the left by $\displaystyle P$ and remember that $\displaystyle PP^T=P^TP=I$

$\displaystyle PD=PP^TAP=AP$ Now multiply on the right by $\displaystyle P^T$

$\displaystyle PDP^T=APP^T=A \iff A=PDP^T$

3. Originally Posted by TheEmptySet
Now multiply on the right by $\displaystyle P^T$

$\displaystyle PDP^T=APP^T=A \iff A=PDP^T$
I am struggling with this step. I thought that you always had to multiply on the left.

4. Originally Posted by incblue
I am struggling with this step. I thought that you always had to multiply on the left.
You can multiply on either side but remember that matrix multiplication does not commute.

For example in the real numbers is we have

$\displaystyle \frac{1}{2}x=5$ if we multiply on the left or the right by 2 we can still solve for x by the communicative property, but with matrices it is not so nice.

$\displaystyle A\mathbf{x}=\mathbf{b}$ If we multiply on the left by the inverse of A we get

$\displaystyle A^{-1}A\mathbf{x}=A^{-1}\mathbf{b} \iff \mathbf{x} =A^{-1}\mathbf{b}$

but if you multiply on the right you get

$\displaystyle A\mathbf{x}{A^{-1}}=\mathbf{b}A^{-1}$

and since multiplication does not commute we cannot simplify.

5. Okay, thanks for the explanation.