Are eigenvectors of real symmetric matrix all orthogonal? As I learned in linear algebra ,a real symmetric matrix $A$ always has orthogonal eigenvectors so $A$ is orthogonally diagonalizable.But are eigenvectors of real symmetric matrix all orthogonal?
In fact, $A$ is diagonalizable so we can find invertible $P$ and $A=PSP^{-1}=P diag\{\lambda_{1},\cdots,\lambda_{n}\}P^{-1}.$But I cannot prove $P$ is orthogonal.I can only find that $A^{T}=A=PSP^{-1}=(P^{T})^{-1}SP^{T}.$ So $P^{T}PS=SP^{T}P.$This cannot show that $P^{T}P=I_{n}.$
So it this $P$ orthogonal? If not,what is its relationship with the orthogonal eigenvectors?
By the way I came this problem when I was reading a lecture note.http://control.ucsd.edu/mauricio/courses/mae280a/lecture11.pdf
I think his way of proving any symmetric matrix has orthogonal eigenvectors is wrong.
Any help will be thanked.
 A: The theorem in that link saying $A$ "has orthogonal eigenvectors" needs to be stated much more precisely. (There's no such thing as an orthogonal vector, so saying the eigenvectors are orthogonal doesn't quite make sense. A set of vectors is orthogonal or not, and the set of all eigenvectors is not orthogonal.)
It's obviously false to say any two eigenvectors are orthogonal, because if $x$ is an eigenvector then so is $2x$. What's true is that eigenvectors corresponding to different eigenvalues are orthogonal. And this is trivial: Suppose $Ax=ax$, $Ay=by$, $a\ne b$. Then $$a(x\cdot y)=(Ax)\cdot y=x\cdot(Ay)=b(x\cdot y),$$so $x\cdot y=0$.
Is that pdf wrong? There are serious problems with the statement of the theorem. But assuming what he actually means is what I say above, the proof is probably right, since it's so simple.
A: Indeed, you cannot prove that a matrix that diagonalizes $A$ is orthogonal, because it's false.
For instance, take $A=I$ (the identity matrix). Any invertible matrix $P$ diagonalizes $I$, but of course $P$ need not be orthogonal.
If $A$ has $n$ distinct eigenvalues (where $A$ is $n\times n$), then the statement is true, because eigenvectors corresponding to different eigenvalues are orthogonal (see David C. Ullrich answer).
Otherwise you need to take a basis of eigenvectors; then, for each eigenvalue $\lambda$, you take the eigenvectors in the basis corresponding to $\lambda$ and orthogonalize it. Then you get an orthogonal basis of eigenvectors.
And yes, the proof in the lecture notes is wrong: using $A=I$, the argument would prove that every invertible matrix is orthogonal, which is obviously false.
