My book says this in an answer to one of the problems (where we're asked to prove that a specific symmetric matrix is diagonalizable)
I know that if there are n distinct eigenvalues of an nxn matrix, then it is diagonalizable…But why would it be true that if the eigenvalues are real, then the matrix is diagonalizable? Couldn't you have an eigenvalue with a multiplicity > 1 that has only one eigenvector, for example, and then not have n linearly independent eigenvectors and not be able to diagonalize the matrix?
-Edit-
The problem in my book is:
Prove that the symmetric matrix is diagonalizable
\begin{bmatrix}0&0&a\\0&a&0\\a&0&0\end{bmatrix}
And they solve it here:
http://calcchat.com/book/Elementary-Linear-Algebra-7e/7/3/7/
I'm confused on where they say, "Since the eigenvalues are real, A is diagonalizable." I don't know how they draw that conclusion. I already know that it's TRUE that symmetric matrices are diagonalizable and have real eigenvalues, but we're supposed to PROVE this is diagonalizable, not just use what we already know.
Best Answer
It doesn't make much sense to compute the eigenvalues without their multiplicity and then say "since the eigenvalues are real, the matrix is diagonalizable" because it doesn't hold in general. It does hold for symmetric matrices but symmetric matrices have real eigenvalues and are diagonalizable so there is no need to calculate anything to deduce your matrix is diagonalizable.
You can say that $A$ is diagonalizable because it is real and symmetric without calculating anything. Alternatively, you can compute the eigenvalues (they will be real) and then compute the geometric multiplicity of each eigenvalue and then conclude $A$ is diagonalizable. It seems that whoever wrote the solution mixed both approaches.