Question

Let λ be an eigenvalue of TεB(X,X) ε ,X is a comlex Banach space.If p is a polynomial Show that P(λ) is an eigenvalue of P(T).

Answer 1

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Let $\lambda$ be an eigenvalue of $T\in B(X,X)$, where $X$ is a complex Banach space. If $P$ is a polynomial, show that $P(\lambda)$ is an eigenvalue of $P(T)$.

Solution. First of all prove the following fact: if $\lambda$ is an eigenvalue of $T$ and $\mu$ is an eigenvalue of $S$, then $a\lambda+b\mu$ is an eigenvalue of $aT+bS$ for all $a,b\in\mathbb{C}$. Indeed, for any $x\in X$ we have

$(aT+bS)x=aTx+bSx=a\lambda x+b\mu x=(a\lambda+b\mu)x.$

Furthermore, show that $\lambda^{n}$ is an eigenvalue of $T^{n}$, if $\lambda$ is an eigenvalue of $T$ (here $n$ is a nonnegative integer). Use the induction on $n$. The case $n=0$ is trivial: $T^{n}=I$ and $\lambda^{n}=\lambda^{0}=1$, which is the only eigenvalue of $I$. Suppose $n\geq 1$ and $\lambda^{n-1}$ is an eigenvalue of $T^{n-1}$. Prove that $\lambda^{n}$ is an eigenvalue of $T^{n}$. Let $x$ be an eigenvector of $T^{n-1}$, corresponding to $\lambda^{n-1}$. Then

$T^{n}x=T(T^{n-1}x)=T(\lambda^{n-1}x)=\lambda^{n-1}Tx=\lambda^{n-1}\cdot\lambda x=\lambda^{n}x,$

therefore $x$ is an eigenvector of $T^{n}$, corresponding to $\lambda^{n}$.

Now consider an arbitrary polynomial $P(x)=a_{0}+a_{1}x+\ldots a_{n}x^{n}$ and an eigenvalue $\lambda$ of $T$. Prove that $P(\lambda)=a_{0}+a_{1}\lambda+\cdots+a_{n}\lambda_{n}$ is an eigenvalue of $P(T)=a_{0}I+a_{1}T+\cdots+a_{n}T^{n}$. Indeed, as was shown above, $\lambda^{k}$ is an eigenvalue of $T^{k}$ for all $k=0,\ldots,n$. Then $\sum_{k=0}^{n}a_{k}\lambda^{k}$ is an eigenvalue of $\sum_{k=0}^{n}a_{k}T^{k}=P(T)$. ∎

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