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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.

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  • $\begingroup$ Eigenvectors for different eigenvalues are always orthogonal. But eigenvectors for the same eigenvalue need not be. $\endgroup$ Aug 16, 2020 at 12:33
  • $\begingroup$ @AnginaSeng So is the proof in the lecture note wrong? $\endgroup$
    – Tree23
    Aug 16, 2020 at 12:34

2 Answers 2

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Tree23
    Aug 16, 2020 at 12:47
  • $\begingroup$ @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. $\endgroup$ Aug 16, 2020 at 12:59
  • $\begingroup$ 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. $\endgroup$
    – Vincent
    Aug 16, 2020 at 22:24
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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.

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