Mathematics

Questions tagged [functional-calculus]

Functional calculus allows the evaluation of a function applied to a linear operator or a matrix. The function could be a polynomial, a holomorphic function, a continuous function or a measurable function defined on the spectrum of an operator or a Banach algebra. Functional calculus is a basic and powerful tool in the spectral theory of operators and operator algebras and is part of functional analysis.

213 questions
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Cauchy's integral formula for Cayley-Hamilton Theorem

I'm just working through Conway's book on complex analysis and I stumbled across this lovely exercise: Use Cauchy's Integral Formula to prove the Cayley-Hamilton Theorem: If $A$ is an $n \times n$ matrix over $\mathbb C$ and $f(z) = \det(z-A)$ is…
Sam
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Why do zeta regularization and path integrals agree on functional determinants?

When looking up the functional determinant on Wikipedia, a reader is treated to two possible definitions of the functional determinant, and their agreement is trivial in finite dimensions. The first definition is based on zeta function…
anon
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A property of exponential of operators

Let $X$ be a Banach space. $A\in B(X)$ is a bounded operator. we can define $e^{tA}$ by $$e^{tA}=\sum_{k=0}^{+\infty}\frac{t^kA^k}{k!}$$ I am interested in this property: If $x\in X$, such that the function $t\mapsto e^{tA}x$ is bounded on…
user165633
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The matrix logarithm is well-defined - but how can we *algebraically* see that it is inverse to the exponential, as a finite polynomial?

This question is inspired by this which I saw earlier today. I started writing my answer, to share the insight that the matrix logarithm can be defined on matrices that do not have unit norm using an alternative technique. Now, Sangchul has posted a…
FShrike
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Isolated point in spectrum

"Any isolated point in the spectrum of a self-adjoint operator must be an eigenvalue". Is there an easy way to see this? The spectral theorem tells us that any self-adjoint operator is unitarily equivalent to a multiplication operator in $L^2$ for…
Ida
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Why do we introduce the continuous functional calculus for self-adjoint operators?

The question is, as the title already says: Why do we introduce the continuous functional calculus for self-adjoint operators? To be more specific, why self-adjoint operators? It is introduced (in short) as follows: Let $T$ be a linear bounded…
Denis
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Searching two matrix A and B, such that exp(A+B)=exp(A)exp(B) but AB is not equal to BA.

We know that if two matrix $A$ and $B$ commutes then $\exp(A+B)=\exp(A)\exp(B)$. I am trying to find two matrix that does not commute but $\exp(A+B)=\exp(A)\exp(B)$ is true for them. Can anybody give exact example. Thanks
Fin8ish
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Exponential of the Laplacian operator as diffusion equation

Let $u$ be a function on a domain $\Omega$ with some fixed boundary condition. I have recently seen a notation $e^{\tau \Delta}u$ as meaning the the time evolution of $u$ by diffusion for a time $\tau$. I'm curious where this notation comes from,…
rviertel
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$A$ self-adjoint then $A^2$ self-adjoint

Let $H$ be a Hilbert space. Let $A:D(A)\to H$ be a self-adjoint operator. When can we say $A^2:D(A^2) \to H$ is self-adjoint? Is it enough to define $D(A^2)=\{u\in D(A) | Au\in D(A)\}$ ? And can you please tell me a book where this exact result is…
Alucard-o Ming
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Meromorphic symbol pseudodifferential operator

Context: For some function $f\in\mathcal{M}(\mathbb C)$ (meromorphic function on $\mathbb C$ ), I am interested in linear operators $T_f$ that act on functions of the form $g_a:x\mapsto \exp(ax)$ in the following way: $$T_fg_a(x)=f(a)g_a(x).$$ Here…
Kolakoski54
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Spectral measure and measurable functional calculus

I am learning about measurable functional calculus at the moment. I have learned that given a operator $T \in B(H)$ that is self adjoint, there exists a measurable functional calculus $\overline{\Phi}:B_b(\sigma(T)) \rightarrow B(H)$ , i.e. a map…
NTc5
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If an element in a Banach algebra is anihilated by an analytic function then it must be algebraic.

Let $A$ be a Banach algebra, let $a\in A$ and suppose $f(a)=0$ , where $f$ is an analytic function defined on an open set $U$ containing $\sigma(a)$ . Prove that $a$ is algebraic in the sense that $p(a)=0$ for some polynomial $p$ . PS: I have just…
Ruy
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Strong continuity of the Borel functional calculus

I have sometimes heard that the Borel functional calculus maps bounded pointwise convergent sequences of Borel functions to strongly convergent sequences of operators. I gather "sequence" is important here, due to the measure theory aspect, we can't…
Jeff
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If $a$ and $b$ commute in a $C^*$-algebra and $a$ is normal, then $f(a)$ and $b$ commute for any continuous $f$

I'm trying to find a way to demonstrate the following: Let $(A,*,\|\cdot\|)$ be a unital $C^*$-algebra. If $a,b\in A$ commute and $a\in A$ is normal (i.e. $a^*a=aa^*$), then for every continuous function $f:$Sp$(a)\to\mathbb{C}$, $f(a)$ and $b$…
user75206
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Almost commuting elements and functional calculus

Let $A$ be an unitial $C^*$ -algebra, $X \subseteq \mathbb{C}$ a compact set and $f : X \to \mathbb{C}$ a continuous function. Prove that for any $\epsilon > 0$ there is a $\delta > 0$ such that whenever $x,y \in A$ with $x$ normal, $\sigma (x)…
Student 00
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