- #1

looth

- 5

- 0

Is there a theorem which classify the diagonalizable matrices??

If so, could someone please kindly tell me which journal is it.

Thanks

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter looth
- Start date

- #1

looth

- 5

- 0

Is there a theorem which classify the diagonalizable matrices??

If so, could someone please kindly tell me which journal is it.

Thanks

- #2

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 971

- #3

looth

- 5

- 0

HallsofIvy said:there exist a basis for the vector space consisting of eigenvectors of the matrix.

Thanks for your reply. For me, that's so called characterization.

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

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 971

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

- #5

looth

- 5

- 0

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

regards

looth

Last edited:

- #6

Dr Avalanchez

- 18

- 0

* 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

matt grime

Science Advisor

Homework Helper

- 9,426

- 4

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

looth

- 5

- 0

HallsofIvy said:

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

Share:

- Last Post

- Replies
- 10

- Views
- 562

- Replies
- 52

- Views
- 1K

MHB
Matrices

- Last Post

- Replies
- 1

- Views
- 366

MHB
Matrices Proof

- Last Post

- Replies
- 5

- Views
- 755

- Last Post

- Replies
- 3

- Views
- 950

- Last Post

- Replies
- 11

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 315

- Replies
- 39

- Views
- 1K

- Replies
- 24

- Views
- 688

- Replies
- 4

- Views
- 985