1
$\begingroup$

In this wiki page the spectral theorem for finite dimension case, the positive definiteness of inner product is assumed. However in the following proof of spectral theorem, it seems to me that this condition is only used in proving the eigenvalues are real.

$\lambda_1\langle{e_1},{e_1}\rangle = \langle A(e_1),e_1\rangle=\langle e_1,A(e_1)\rangle=\bar{\lambda}\langle{e_1},{e_1}\rangle$

Positive definiteness of inner product implies $\langle{e_1},{e_1}\rangle\neq0$ then $\lambda_1=\bar{\lambda}$.

It seems to me the rest of the proof dose not use positive definiteness of inner product.

I wonder if positive definiteness is just a sufficient condition for spectral theorem or not.

Any help would be much appreciate.

$\endgroup$
0
$\begingroup$

I think I have found where we need the definiteness.

In the proof of spectral theorem we use the fact that the orthogonal complement of an eigenvector $\mathrm{w}$ is an invariant subspace of the symmetry operator.

Here is the point :
We need definiteness to assert that the linear space $V$ is indeed an direct sum $V=\mathrm{w}\bigoplus \mathrm{w}^{\perp}$ so as to proceed our proof by induction.

For example, let $V=\mathbf{C}^2$ with $\langle X,Y\rangle=x_1y_1-x_2y_2$. The scalar product is indefinite. Now suppose the symmetry operator $A$ have an eigenvector $\mathrm{w} =(1,i)$, then $\mathrm{w}^{\perp}=\mathrm{w}$. We cannot have an orthogonal basis consists of eigenvectors for this case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.