# 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.

• Eigenvectors for different eigenvalues are always orthogonal. But eigenvectors for the same eigenvalue need not be. Aug 16, 2020 at 12:33
• @AnginaSeng So is the proof in the lecture note wrong? Aug 16, 2020 at 12:34

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.

• Yes I know he means $A$ has a set of eigenvectors and they are orthogonal.Except the mistake in the statement,I also wonder if his proof itself is wrong. Because he proved $T^{T}=T^{-1}$ directly from $\left(T^{-1} \Lambda T\right)^{T}=T^{T} \Lambda T^{-T}$.Is this wrong？ Aug 16, 2020 at 12:47
• @Tree23 Is that wrong? I don't see why it's right. But it's easy to see that $T$ is orthogonal. Because as above we have $v_j\cdot v_k=0$, for $j\ne k$, so the columns of $T$ are orthonormal. Aug 16, 2020 at 12:59
• There is something wrong in the pdf in the proof of what is there item 2) because there is a $v$ disappearing into nothingness (turning scalars into vectors) when reading left to right. Also I cannot guess what is intended. I recommend using David C. Ullrich's proof instead. Aug 16, 2020 at 22:24

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.