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Is it true that adjoint of a normal operator A can be written as a polynomial in A?

Selvakumar A
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1 Answers1

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The trick appears to be this. Suppose the operator $A$ acts on a finite dimensional space. Then $A$ being normal means both $A$ and $A^*$ can be simultaneously diagonalised.

That is we can write $D_2=UA^* U^*$ and $D_1=UA U^*$, with $D_1$ and $D_2$ diagonal and U unitary. Then we can use lagrange interpolation to find a polynomial $p$ such that $p(D_1)=D_2$.

$D_1$ and $D_2$ will have the same number of distinct and equal entries as $(D_2)^*=D_1$ from the above. Supposing that $p(X)=a_nx^n+\ldots+a_1x+a_0$, then $$D_2=p(D_1)=a_n(D_1)^n+\ldots+a_1D_1+a_0=a_nU(A)^n U^*+\ldots+a_1UAU^*+a_0=Up(A)U^*$$

Thus $A^*=U^*D_2U=p(A)$, since $UU^*=1$.

Note your question is point $7.$ on the wiki for normal matrices of which there was the above hint.

snulty
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  • Interesting idea. But something more seems to be need it seems to me. You cannot transform any diagonal matrix in any other. Equal entries will stay equal. – quid Jul 27 '16 at 00:37
  • @quid $(D_2)^=D_1$ is true I believe? So that would suggest that each diagonal matrix will have the same number of distinct entries? I'm used to using $\dagger$ for adjoint of a matrix, but was using $$ for consistency with the link. – snulty Jul 27 '16 at 00:40
  • Yes, this seems correct to me (it follows right away taking the conjugate of $UAU^{\ast}$). Then I agree it is a complete arguemntt;. – quid Jul 27 '16 at 00:43
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    @quid I'll update it to include this! – snulty Jul 27 '16 at 00:44
  • A normal matrix over a finite-dimensional inner product space is not necessarily diagonalizable. – user0 Jun 17 '19 at 17:55
  • @MauriceP are you talking about over finite fields? Over $\mathbb{R}$, over $\mathbb{C}$? – snulty Jun 17 '19 at 18:13
  • See the fourth comment to https://math.stackexchange.com/a/277192/389981. I am interested in this issue because I am trying to solve Exercises 3 and 5 on page 347 of Linear Algebra by Hoffman and Kunze. – user0 Jun 17 '19 at 18:34
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    @MauriceP Yeah I'm not really sure on those questions, since normal operators are diagonalisable over the complex numbers. I don't know how you proceed if you want to avoid complex numbers entirely. For instance everything here uses complex numbers from what I can tell https://math.stackexchange.com/q/675239/128967 – snulty Jun 18 '19 at 12:34
  • I solved my problem for the finite-dimensional case: Temporarily pass to the inner-product space over the field extended to be algebraically closed; in doing so, adjoints will not change because they are unique. Using your answer, it is now easy to prove that if two normal operators commute, then their product is normal. In passing back to the original space, that result will still be true. See https://math.stackexchange.com/q/118424. – user0 Jun 19 '19 at 17:52