# Thread: matrix diagonalisation - real and complex matrices - some observations

1. I think the problem with your computation is that, while A is real, its eigenvalues are not necessarily real (we have not assumed that A is symmetric). Hence, in order to test for symmetry, you should be computing complex conjugate transposes (I use daggers for that operation); that is, what you must really show is that

$\displaystyle A^{\dagger}=(P^{\dagger}DP)^{\dagger}=P^{\dagger}D ^{\dagger}P=P^{\dagger}DP=A,$ but it is not necessarily true that

$\displaystyle D^{\dagger}=D.$

You can prove that every normal matrix is diagonalizable, and it is also true that not every normal matrix is symmetric.

2. Originally Posted by Ackbeet
... while A is real, its eigenvalues are not necessarily real (we have not assumed that A is symmetric).
That is right. But look at my previous post, I included in the hyphotesis $\displaystyle D\in\mathbb{R}^{n\times n}$" . That is, "ortoghonally diagonalizable on $\displaystyle \mathbb{R}$".

Fernando Revilla

3. Ok, but I would claim that's an additional hypothesis that's not implied by being orthogonally diagonalizable. Not all orthogonally diagonalizable real matrices have real eigenvalues (which is what your D consists of).

4. Originally Posted by Ackbeet
Ok, but I would claim that's an additional hypothesis that's not implied by being orthogonally diagonalizable. Not all orthogonally diagonalizable real matrices have real eigenvalues (which is what your D consists of).
The following solve the question:

Definition Consider $\displaystyle A\in \mathbb{R}^{n\times n}$ . We say that $\displaystyle A$ is orthogonally diagonalizable if and only if there exists an orthogonal matrix $\displaystyle Q\in \mathbb{R}^{n\times n}$ such that $\displaystyle Q^tAP=D\in \mathbb{R}^{n\times n}$ with $\displaystyle D$ diagonal.

I've just used that definition, which implies I've not added any hypothesis. For your comments I see that you are using another definition, I know which one, and I accept it. So, you and me are right and we are not talking about a mathematical problem but about a convention problem .

Fernando Revilla

5. Sure, fine. You can define things that way. But then you can't have normal matrices to be orthogonally diagonalizable.

6. Originally Posted by Ackbeet
Sure, fine. You can define things that way.

Actually, that is a standard definition.

But then you can't have normal matrices to be orthogonally diagonalizable.

We can, $\displaystyle A\in\mathbb{R}^{n\times n}$ symmetric implies $\displaystyle A$ normal.

Fernando Revilla

7. Originally Posted by FernandoRevilla
We can, $\displaystyle A\in\mathbb{R}^{n\times n}$ symmetric implies $\displaystyle A$ normal.Fernando Revilla
I meant the other direction: you'll have some normal matrices that are not symmetric, which you won't be able to orthogonally diagonalize.

8. Originally Posted by Ackbeet
I meant the other direction: you'll have some normal matrices that are not symmetric, which you won't be able to orthogonally diagonalize.
Right.

Fernando Revilla

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