# If the eigenvalues of a matrix are real, the matrix is diagonalizable?

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.

• Spoiler: all symmetric matrices are diagonalizable – Exodd Mar 12 '17 at 23:24
• I know. This whole section is on that. They want us to prove it. I can't just answer the problem by saying, "this symmetric matrix is diagonalizable because symmetric matrices are diagonalizable." – dagny Mar 12 '17 at 23:26
• @dagny You should tell us exactly what they are asking you to show: one could answer this question in many levels of abstraction, from "all symmetric matrices are diagonalizable" to explicitly providing a diagonalization of your "specific symmetric matrix." – angryavian Mar 12 '17 at 23:30
• @angryavian Okay, I edited it – dagny Mar 12 '17 at 23:36

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.