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Show there exists $\lambda \in \sigma(A)$ such that $|\lambda - \lambda_0|<\epsilon$

It suffices to use the fact that $$\|(\lambda_0-A)^{-1}\|=\frac1{d(\lambda_0,\sigma(A))}$$ for $\lambda_0\in\rho(A),$ where $d(\lambda_0,\sigma(A))=\inf_{\lambda\in\sigma(A)}|\lambda-\lambda_0|$. This ...
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Prove that $L_0 = (-i)^m \frac{d^m u}{dx^m}$ is a positive operator

When $m$ is even, let $m=2n$ and it follows $$\langle L_0u,u\rangle=(-1)^n\langle u,u^{(2n)}\rangle=(-1)^{2n}\langle u^{(n)},u^{(n)}\rangle=\|u^{(n)}\|^2\geq0.$$ This equals zero iff $u^{(n)}\equiv0$, ...
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A question on two self-adjoint operators with same spectrum

A self-adjoint operator $T$ admitting a cyclic vector is, as you said, necessarily the multiplication operator on $L^2(\sigma(T),\mu)$ given by $$ T(f)x=xf(x). $$ So, once $\sigma(T)$ is fixed, ...
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Why does from $\|Tf_n\| \rightarrow 0$ follow that $\lambda \in [0,1]$ are approximate eigenvalues?

There is a typo in the first part. Actually, $\|(T-\lambda)f_n\|^{2}=2n\int_{\lambda-\frac 1 n}^{\lambda+\frac 1 n} (x-\lambda)^{2}dx =\frac 4 {3n^{2}} \to 0$.
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Is a holomorphic Function of a matrix invertible?

There is this theorem in spectral theory that says (if $g$ is continuous and defined on the spectrum of linear bounded $A$) $$ \sigma(g(A)) = g(\sigma(A)). $$ $\sigma$ of course denotes the spectrum. ...
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Understanding the Spectral Theorem for (Bounded) Operators

Your $A$ can be written as $\sum \lambda_i P_i,$ where $P_i$ is the projection onto the eigenspace of $\lambda_i.$ A sum is a finite version of an integral.
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1 vote
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Non-commutative spectrum in Banach Algebras

You have that $T^*T=I$, the identity operator, and it is trivial to check that $\sigma(I)=\{1\}$. Meanwhile, $TT^*=I-P$, where $P$ is the projection onto the first coordinate. Then $\sigma(TT^*)=\{0,1\...
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Show there exists $\lambda \in \sigma(A)$ such that $|\lambda - \lambda_0|<\epsilon$

Just to add two comments: The result holds for any normal operator $A$. If $A$ is normal, so is $(A-\lambda_0)^{-1}$ (if $\sigma_0\not\in \sigma(A)$), and $\|(A-\lambda_0)^{-1}\|$ is equal to its ...
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