0
$\begingroup$

In my textbook, a theorem states that:

"Theorem 6.17. Let T be a linear operator on a finite-dimensional real inner product space V. Then T is self-adjoint if and only if there exists an orthonormal basis $\beta$ for V consisting of eigenvectors of T."

My understanding of this theorem is that T is self-adjoint iff T is diagonalisable. Is my understanding correct? If yes, that doesn't sound very true because there are many diagonalisable matrices but not self-adjoint. For example, $ \left(\begin{matrix} 1 & -1 & 1 \\ -1 & 1 & 1 \\ -1 & -1 & 3 \end{matrix}\right) $ is diagonalisable but not self-adjoint.

$\endgroup$
9
  • 1
    $\begingroup$ True, but they stop being eigenvectors in the process. $\endgroup$
    – user700480
    Mar 18, 2023 at 10:47
  • 2
    $\begingroup$ Why would they be? Let's say $v_1,v_2,\ldots v_n$ are all eigenvectors for different eigenvalues. Orthogonalization goes like: $w_1=v_1$ (so far so good!), $w_2=v_2-\frac{\langle v_2, v_1\rangle}{\langle v_1, v_1\rangle}v_1$ and, if $\langle v_2, v_1\rangle\ne 0$, this is a linear combination of $v_1, v_2$ which is not in the span of either of those two vectors alone, i.e. itself cannot be an eigenvector. And let's not even involve $w_3, w_4$ etc. $\endgroup$
    – user700480
    Mar 18, 2023 at 10:56
  • 1
    $\begingroup$ You can, of course, orthonormalize eigenvectors for the same eigenvalue, because any linear combination of eigenvectors for the same eigenvalue (unless it is zero) is again an eigenvector for the same eigenvalue. But it does not work across different eigenvalues. $\endgroup$
    – user700480
    Mar 18, 2023 at 10:59
  • 1
    $\begingroup$ Correct. And that is what your original statement says as well. It will have it iff the the corresponding operator is self-adjoint (which is to say that its matrix in any orthonormal basis is symmetric). $\endgroup$
    – user700480
    Mar 18, 2023 at 11:22
  • 1
    $\begingroup$ Got it. Now I realise my fundamental misconception. Thanks buddy! $\endgroup$
    – Khanh
    Mar 18, 2023 at 11:24

1 Answer 1

1
$\begingroup$

Orthogonalization of a set of eigenvectors will not generally yield a set of eigenvectors.

Suppose $v$ and $w$ are eigenvectors for $T$, relative to distinct eigenvalues $\lambda$ and $\mu$.

For simplicity, we can assume the vectors have norm $1$, so the orthogonalization process gives $$ w'=w-(v,w)v $$

Can $w'$ be an eigenvector relative to some eigenvalue $\nu$? Let's check: \begin{align} & \nu w'=\nu w-\nu(v,w)v \\ & Tw'=Tw-(v,w)Tv=\mu w-\lambda(v,w)v \end{align} Now we need $$ (\nu-\mu)w+(v,w)(\lambda-\nu)v=0 $$ Since $v$ and $w$ are linearly independent, being eigenvectors relative to distinct eigenvalues, this imposes $$ \nu=\mu,\qquad (v,w)(\lambda-\nu)=0 $$ and this is only possible for $(v,w)=0$, that is, the two eigenvectors are orthogonal to begin with.

Thus we see that the existence of an orthogonal basis of eigenvectors requires that eigenvectors relative to different eigenvalues are orthogonal.

$\endgroup$
6
  • $\begingroup$ Normal matrices have their eigenvectors orthogonal to each other. However, this is still short of being self-adjoint? $\endgroup$
    – Khanh
    Mar 18, 2023 at 12:48
  • $\begingroup$ @Khanh Normal real matrices are self-adjoint exactly because they're diagonalizable with an orthogonal matrix. $\endgroup$
    – egreg
    Mar 18, 2023 at 13:49
  • $\begingroup$ Hi @egreg, do you mean all normal real matrices are self-adjoint? (0 -1 // 1 0) is a normal real matrix but it doesn't look self-adjoint to me. Where do I get it wrong? $\endgroup$
    – Khanh
    Mar 19, 2023 at 3:01
  • $\begingroup$ @Khanh Sorry, I should have written “diagonalizable normal”. $\endgroup$
    – egreg
    Mar 19, 2023 at 8:30
  • $\begingroup$ (0 -1 // 1 0) is diagonalisable over the complex field. So you meant "diagonalisable normal real matrices"? $\endgroup$
    – Khanh
    Mar 19, 2023 at 8:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .