Diagonalizable Matrices & Eigenvalues

[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?

Thanks for your time.

 

[Hurkyl]

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

 

[HallsofIvy]

For example, how many distinct eigenvalues does

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

 

[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:

Theorem 3: Suppose that A is an nXn matrix and that A has n distinct eigenvalues, then A is diagonalizable.

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?

Thanks for your responses.

 

[Hurkyl]

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

 

[HallsofIvy]

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..

Yes- sufficient but not necessary.

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?

Thanks for your responses.

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

 

[Este]

Great! Things are pretty clear now, thanks HallsofIvy!