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My question is regarding an old post from the site, namely:

$T$ is normal if and only if exist polynomial $p$ s.t $T^{*}=p(T)$

My question is : How do you prove that such a polynomial has $\deg< \dim V$? It seems weird to me that the polynomial can not have degree equal to $\dim V.$

José Carlos Santos
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MathBro
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2 Answers2

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If $T$ is normal on a complex inner product space $X$ of finite dimension $N$, then $T$ has an orthonormal basis of eigenvectors, which must also be a basis of eigenvectors for $T^*$ because $$ \|(T-\lambda I)x\|=\|(T^*-\overline{\lambda}I)x\|,\;\;\; x\in V,\;\lambda\in\mathbb{C}. $$ If $\{\lambda_1,\lambda_2,\cdots,\lambda_n\}$ is the set of distinct eigenvalues of $T$, then $$ P_j = \frac{1}{\Pi_{k\ne j}(\lambda_j-\lambda_k)}\Pi_{k\ne j}(T-\lambda_k I) $$ is the orthogonal projection onto $\mathcal{N}(T-\lambda_j I)=\mathcal{N}(T^*-\overline{\lambda_j}I)$. Therefore, $$ T=\sum_{j}\lambda_j P_j,\;\;\; T^*=\sum_{j}\overline{\lambda_j}P_j. $$ In particular, $T^*$ is a polynomial in $T$. (Likewise, $T$ is a polynomial in $T^*$.) The order of the polynomial is less than $N=\dim(X)$.

Disintegrating By Parts
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  • I am not quite following what you are doing. I have trouble understanding how you are concluding that "The order of the polynomial is less than =dim()" – MathBro Oct 17 '20 at 20:10
  • @MathBro $P_j$ is a product of $N-1$ linear factors with $T$ and $I$, where $N$ is the number of distinct eigenvalues. – Disintegrating By Parts Oct 17 '20 at 20:16
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Note that if $T$ is nilotent, there is nothing to do as $T=\mathbf 0$. Now assume $T\neq \mathbf 0$.
Further suppose you have $d\geq n$ and
$p\big(T\big) = a_0 I + a_1 T + .... +a_{n-1}T^{n-1}+ a_nT^n + ... + a_dT^d$

by Caley Hamilton $T^n$ may be written as a linear combination of $\{I, T,...,T^{n-1}\}$, and so may $T^{n+1}$, ...., and so may $T^d$

so you can re-write $p$ to have degree $\leq n-1$.

user8675309
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