# exams very soon -help-Diagonalization

• Jun 7th 2013, 09:34 PM
n22
exams very soon -help-Diagonalization
Hi people,
Can someone tell me when a matrix actually becomes diagonizable??
and when eigenvalue is considered distinct and not distinct?
Thanks.
• Jun 7th 2013, 10:20 PM
chiro
Re: exams very soon -help-Diagonalization
Hey n22.

Hint: Given a characteristic equation, how can you obtain the eigen-values? (Also if you have repeated eigen-values what does this mean for the roots of the characteristic polynomial?)
• Jun 8th 2013, 06:09 AM
HallsofIvy
Re: exams very soon -help-Diagonalization
Quote:

Originally Posted by n22
Hi people,
Can someone tell me when a matrix actually becomes diagonizable??
and when eigenvalue is considered distinct and not distinct?
Thanks.

It doesn't make sense to talk about an eigenvalue being "distinct". Two or more eigenvalues are "distinct" if they are different! That is, after all, what "distinct" means.

An n by n matrix is diagonalizable if and only it has n independent eigenvectors. It can be shown that "distinct" (i. e "different") eigenvalues have eigenvectors that are independent eigenvectors so if an n by n matrix has n distinct eigenvalues, it has n independent eigenvectors and so is diagonalizable. The converse is NOT true. If a matrix has multiple eigenvalues it may or may not be diagonalizable.

For example if $\displaystyle A= \begin{bmatrix}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{bmatrix}$
$\displaystyle B= \begin{bmatrix}3 & 1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{bmatrix}$ and
$\displaystyle C= \begin{bmatrix}3 & 1 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3\end{bmatrix}$
Then each of A, B, and C have "3" as a triple eigenvalue. A has three independent eigenvectors, B has two independent eigenvectors, and C has only one independent eigenvector.
A can be "diagonalize" (clearly it is diagonal) but B and C cannot.