I am sure I am making some simple definitional mistake, but I seem to have a 'proof' that all normal operators are self adjoint which I am pretty sure is not true but anyway.

Anyway it goes: Suppose $A\in L(V)$ where $V$ is some finite dimensional vector space. Then, by the complex spectral theorem, we can write $A$ as $A=UDU^{*}$ with $U$ unitary and $D$ diagonal.

Then we have $A^{*}=(UDU^{*})^{*}=(U^{*})^{*}D^{*}U^{*}=UDU^{*}=A $ thereby showing that $A$ is self adjoint.

Any clarification would be appreciated. Thanks


1 Answer 1


For diagonal matrices $D$, one has $D=D^*$ only if $D$ has real entries. Since having real entries is not said in the consequence of the version of the spectral theorem you applied, replacing $D^*$ by $D$ is not justified.

(Also you should cite your hypotheses correctly: you never say you assume $A$ to be normal.)

  • $\begingroup$ Ahh right, thanks. That makes more sense. $\endgroup$
    – user434180
    Jun 22, 2015 at 4:35

You must log in to answer this question.

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