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Let $T$ be an operator on Hilbert space.

Define $\sigma(T)=\lbrace \lambda\in \mathbb{C} | \lambda I - T~\textrm{is not invertible}\rbrace$.

How can I prove that $\sigma(T^n)=\lbrace \lambda^n|\lambda\in \sigma(T)\rbrace$?

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You probably want $\sigma(T) = \{ \lambda \in \mathbb C \;|\; \lambda - T \text{ is not invertible}\}$. –  Alexander Thumm Mar 23 '11 at 14:12
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@user8484: Your $\sigma (T)$ is the *resolvent of $T$*, not its spectrum. The spectrum of $T$ is the set $\sigma (T):=\{ \lambda \in \mathbb{C} |\ \lambda -T \text{ is not invertible}\}$. –  Pacciu Mar 23 '11 at 14:13
    
Right. I changed it. Thanks. –  user8484 Mar 23 '11 at 14:32
    
Your operator must be bounded too... –  Freeze_S Jul 6 at 5:30

1 Answer 1

up vote 4 down vote accepted

Note: Given a polynomial $P(x)$, we can talk about the operator $P(A)$ which is the polynomial applied to $A$. Since $A$ commutes with itself, and the identity commutes with everything, we can factor $P(A)=(A-c_1 I)(A-c_2 I)\cdots (A-c_m I)$ by the fundamental theorem of algebra.

Hint: The polynomial $x-\lambda$ divides $x^n-\lambda^n$. This shows $\{\lambda^n:\ \lambda\in \sigma(A)\} \subset \sigma(A^n)$. To show that each element $r\in\sigma(A^n)$ is of the form $\lambda^n$, $\lambda\in \sigma(A)$, consider $x^n-r$, and factor this over $\mathbb{C}$ as a polynomial. Since the polynomial evaluated at $A$ is non-invertible, at least one factor must non-invertible. Hence $r^{1/n}\zeta_n\in \sigma(A)$ where $\zeta_n$ is some root of unity, and the proof is finished.

Hope that helps,

Edit: More generally, suppose we are given an operator $A$, and that $Q(x)=\frac{p(x)}{q(x)}$ is a rational function whose poles do not lie in $\sigma(A)$. Then $Q(A)$ makes sense as an operator, and $$\sigma(Q(A))=\{Q(\lambda):\ \lambda\in\sigma(A)\}.$$ The proof of this is very similar to the proof of what you want.

Edit 2: Since we are talking about the spectrum, I removed the earlier edit. I'll still mention, the exact same proof works with the directions changed to show that the resolvent is preserved under polynomials.

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Doesn't your proof hold only in the finite dimensional case? I'm in trouble here, because I don't know how to define the characteristic polynomial in the infinite dimensional case. –  Pacciu Mar 23 '11 at 14:23
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@Pacciu: This is fine for infinite dimensions. I am not referring to the characteristic polynomial, but rather polynomials applied to $A$. Given $P(x)$, the operator $P(A)$ makes sense, and the factoring over $\mathbb{C}$ of $P(x)=\prod_{i=1}^n(x-c_i)$ also holds for $P(A)=\prod_{i=1}^n(A-c_i I)$ since $A$ commutes with itself and the identity. Now, the non-invertibility of a factor implies that of the whole polynomial, which is what we use in the above argument. –  Eric Naslund Mar 23 '11 at 14:29
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On the other hand, the proof (and result) does depend on the scalar field having all roots of unity. The statement does not hold for a 90-degree rotation in the plane for instance: its real spectrum is empty but the real spectrum of its square is not. –  lhf Mar 23 '11 at 14:34
    
Ok, got it! :-D Thank you. –  Pacciu Mar 23 '11 at 14:39
    
@Eric: Be careful about what you use: in your answer, noninvertibility of the polynomial implies noninvertibility of at least one factor (right) in your comment, noninvertibility of a factor implies noninvertibility of the polynomial (only right here, in general wrong) –  Freeze_S Jul 6 at 5:44

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