Hi people,

Can someone tell me when a matrix actually becomes diagonizable??

and when eigenvalue is considered distinct and not distinct?

Thanks.

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- Jun 7th 2013, 09:34 PMn22exams 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 PMchiroRe: 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 AMHallsofIvyRe: exams very soon -help-Diagonalization
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.