# Classification of diagonalizable matrices

1. Sep 27, 2005

### looth

Hi,
Is there a theorem which classify the diagonalizable matrices??
If so, could someone please kindly tell me which journal is it.
Thanks

2. Sep 27, 2005

### HallsofIvy

Staff Emeritus
I'm not sure you would call it a "classification" but a matrix is diagonalizable if and only if there exist a "complete set of eigenvectors". That is, if there exist a basis for the vector space consisting of eigenvectors of the matrix.

3. Sep 29, 2005

### looth

Maybe I should put it in this way. Normal matrices are diagonalizable nand there also exist some non-normal, diagonalizable matrices. My question is: Are there any necessarly and sufficent conditions that characterize those subclasses of matrices??

thanks & regards
looth

4. Sep 29, 2005

### HallsofIvy

Staff Emeritus
A sufficient condition for real matrices is that they be symmetric.

A sufficient condition for linear transforms in general is that the be "self-adjoint" (which reduces to being symmetric for real matrices).

5. Sep 30, 2005

### looth

Self-adjoint matrix is a special type of normal matrix. What I'm interested is the characterization for those non-normal, diogonalizable matrices.

regards
looth

Last edited: Sep 30, 2005
6. Oct 1, 2005

### Dr Avalanchez

I'm still not sure what you mean, but these are some criteria I know of:
* the eigenvectors are a basis for the vectorspace
* all the algebraic multplicities equal the geometrical multiplicities (the vectorspace is a direct sum of its eigenspaces)
* the characteristic polynom has no multiple roots

7. Oct 1, 2005

### matt grime

It should be pointed out that those are not equivalent criteria (the last one is not implied by either of the first two, which are equivalent).

The only true characterization is (for an nxn matric) to have n eigen vectors. That is it, exactly. Nothing more, nothing less. These matrices are in bijection with C^n (assuming we're working over C here) modulo the relation x~y iff there is a permutation of the coordinates of x that gives y. That might be an interesting space to look at...

8. Oct 6, 2005

### looth

What I mean is a theorem like this:

Let M be a diagonalizable matrix. Then precisely (or maybe either) one of the following holds:
1. M is normal.
2. .....
3. ....
etc.

I only manage to state one of the case because
so far the only "nice" necessary condition (for a matrix to be diagonalizable) that I know is being normal matrix.

Regards
looth