# Relation af a linear operator to its adjoint in a finite-dimensional vector-space

I'm self-studying a bit of linear algebra and got stuck with the following problem:

Let $\tau$ be a normal operator on a complex finite-dimensional inner product space V or a self-adjoint operator on a real finite-dimensional inner-product space. I want to show that $\tau^*=p(\tau)$, where $p(x)\in\mathbb{C}[x]$.

This looks to me like using minimal or characteristic polynomials, but I don't manage to prove it. Then my idea was to use the orthogonal spectral resolution of $\tau$, i.e. $\tau=\lambda_{1}\rho_{1}$+...+$\lambda_{k}\rho_{k}$, where $im(\rho_{i})=E_{\lambda_{i}}$ ($\lambda_{i}$ eigenvalues, $E_{\lambda_{i}}$ corresponding Eigenspace), and $ker(\rho_{i})=\bigodot_{j\neq i} E_{\lambda_{j}}$. Then using properties of the adjoint I find. $\tau^*=\overline{\lambda_{1}}\rho_{1}$+...+$\overline{\lambda_{k}}\rho_{k}$.And then? Thank you for any suggestion.

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## 2 Answers

Then, you are done, because given $k$ complex numbers $\lambda_1,\ldots,\lambda_k$, you can always find a polynomial $p$ with $p(\lambda_j)=\overline{\lambda_j}$. And $p(\tau)=p(\lambda_j)\rho_j$ (this is where you use that the projections $\rho_j$ are pairwise orthogonal, i.e. $\rho_i\rho_j=0$ if $i\ne j$).

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Thank you! Is there a proof using characteristic polynomials? – Lucien Sep 5 '12 at 14:40
I don't know. I cannot imagine how to use characteristic polynomials to show that two operators are equal. – Martin Argerami Sep 5 '12 at 15:14
Characteristic polynomials are often over-hyped. The convenient expression in terms of determinant is a red herring. – paul garrett Sep 5 '12 at 15:54

When A is normal, you can use Lagrange's interpolation formula to construct a polynomial P such that $\overline{\lambda_j} = P(\lambda_j)$, where $\lambda_j$ are the eigenvalues of A.

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Thank you for this observation. – Lucien Sep 5 '12 at 17:38
@Lucien:You are welcome. – Mhenni Benghorbal Sep 5 '12 at 18:02