# Diagonalizable Matrices & Eigenvalues!

1. Jan 5, 2009

### Este

Hello,

Is it sufficient to determine that a nXn matrix is not diagonalizable by showing that the number of its distinct eigenvalues is less than n?

2. Jan 5, 2009

### Hurkyl

Staff Emeritus
You should empirically test your hypothesis; try it out on the simplest matrices you can imagine.

3. Jan 5, 2009

### HallsofIvy

Staff Emeritus
For example, how many distinct eigenvalues does

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$$
have?

4. Jan 6, 2009

### Este

Ok I got your point...I think the question should be now, how to check whether a matrix is diagonalizable or not?

A theorem I've encountered in my textbook:
and somewhere else I read this is sufficient to prove a matrix is diagonalizable but not the other way around.. that's why I posted the question..

So now, all I can do is to prove that Matrix x is diagonalizable, but if it's not, I can't tell for sure....what other methods should I be using?

5. Jan 6, 2009

### Hurkyl

Staff Emeritus
The most direct way to tell if a matrix isn't diagonalizable is to try and diagonalize it.

6. Jan 6, 2009

### HallsofIvy

Staff Emeritus
Yes- sufficient but not necessary.

An n by n matrix is diagonalizable if and only if it has n independent eigenvectors. Since eigenvectors corresponding to distinct eigenvalues are always independent, if there are n distinct eigenvalues, then there are n independent eigenvectors and so the matrix is diagonalizable. But even if the eigenvalues are not all distinct, there may still be independent eigenvectors.

$$\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$$
as I pointed out before has the single eigenvalue 1 but is already diagonal because <1, 0, 0>, <0, 1, 0>, and <0, 0, 1> are independent eigenvectors.

$$\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}$$
also has 1 as its only eigenvalue. Now, <1, 0, 0> and <0, 0, 1> are the only eigenvectors so this matrix cannot be diagonalized.

Finally,
$$\begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}$$
also has 1 as its only eigenvalue but now only <1, 0, 0> and multiples of that are eigenvectors.

7. Jan 6, 2009

### Este

Great! Things are pretty clear now, thanks HallsofIvy!

Last edited: Jan 6, 2009