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### Hermitian Operators and the Spectral Theorem

I understand that in a finite-dimensional vector space $V$, a diagonalizable linear operator $T: V \to V$ decomposes $V$ into a direct sum of its invariant eigenspaces, on each of which it restricts ...
809 views

### When is the restriction of a normal operator not normal?

I was proving the spectral theorem for normal operators on finite-dimensional complex vector spaces today during a test, when I arrived at the point in which If $T\in\operatorname{End}(V)$ is ...
• 903
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### Adjoint of an integral operator

I'm reading through a text about integral operators and I've come across the following theorem: Let $k:\mathbb{R}^2\rightarrow\mathbb{C}$ be a kernel, $T:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ ...
997 views

### Adjoint of a nonlinear operator?

For any linear operator $A$, the adjoint $A^*$ is defined as a linear operator that satisfies $$\langle v, Au\rangle = \langle A^*v, u\rangle$$ Moreover, one has that $(A^{*})^{*} = A$ (proof here). ...
• 1,596
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### Help Showing that the Adjoint Operator $T^*$ is Surjective if and only if $T$ is Injective

Let $T\in L(V,W)$,where $L(V,W)$ denotes a linear map from a vector space $V$ to vector space $W$. I want to prove that $T$ is injective iff $T^*$ is surjective, where $T^*$ is the adjoint of $T$. I ...
1k views

### inner product and adjoint operator

This is a problem I found in Schaum's Outlines: Linear Algebra, and I was wondering if someone knew how to solve it. I began using integration by parts, but that approach did not lead to any ...
• 323
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### Why are self-adjoint operators important?

I am learning about self-adjoint and normal operators. So far, they have come up in the Spectral theorem, which says self-adjoint operators have an eigenvalue basis and a corresponding diagonal ...
• 1,716
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### Schur decomposition of a matrix with distinct eigenvalues is almost unique

Let $M\in \mathbb C^{n,n}$ have $n$ distinct eigenvalues, and let $U_1, U_2$ be two Schur-forms of $M$. Show that if $U_1, U_2$ have equal diagonals, there is a hermitian diagonal matrix $Q$ such that ...
• 7,500
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### Adjoint of multiplication by $z$ in a Hilbert Space (Bergman space)

I am learning Hilbert space theory from Halmos' "Introduction to Hilbert space and the theory of spectral multiplicity". While talking about understanding adjoints (p. 39), he calls special ...
• 3,287
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### Exponential Law for based spaces

I realize most people work in "convenient categories" where this is not an issue. In most topology books there is a proof of the fact that there is a natural homeomorphism of function spaces (with ...
• 1,486
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I hope this is enough of a question (and less of a start of a discussion) to be allowed here. I am trying to get my head around the ideas of inverse and adjoint operators. To keep it simple, let's ...
• 502
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### Self adjoint operators on a Hilbert space

Let $H$ be a Hilbert space and let $T\in \mathcal{B}(H)$ such that $T$ is self-adjoint. I want to show that if $T$ is non-zero, then $T^n\neq 0$ for all $n\in \mathbb{N}$. Suppose $n$ be the least ...
• 4,754
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### For positive self adjoint $T$, show $|\langle Tx,y\rangle|^2 \le \langle Tx,x\rangle \langle Ty,y\rangle$

As in title, $T$ is a positive self adjoint, bounded linear operator on a Hilbert Space $X$ and I'd like to show $$|\langle Tx,y\rangle|^2 \le \langle Tx,x\rangle \langle Ty,y\rangle$$ Self adjoint ...
• 598
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### Invertibility in a finite-dimensional inner product space

Let $T$ be an invertible linear operator on a finite-dimensional inner product space. I just want a hint as to how I should prove that $T^{*}$ is also invertible and $( T^{-1} )^{*} = ( T^{*} )^{-1}$. ...
• 321
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### Spectral gap and Poincaré inequality

Consider the PDE $$\partial_t u = L u$$ where $L = \Delta + \nabla V \cdot \nabla$ is a self-adjoint operator. I read that if $L$ has a spectral gap $\lambda > 0$ then "[convergence of the ...
• 1,716
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### Derivative of inner product via adjoint operator vs. complex derivatives

Dear math enthusiasts, I need to take the derivative of an inner product involving an operator on one side and I'd like to do this via the adjoint operator. However, it seems I'm doing something ...
• 2,530
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### Find adjoint operator of an operator T

I would like to find the adjoint operator of $$T\colon L^2([0,1])\to H^1([0,1]),\quad x\mapsto\int\limits_0^t x(s)\, ds.$$ Here $H^1([0,1])$ is the Sobolev space $W^{1,2}([0,1])$. I tried to find ...
204 views

### Why is the Compactness of an Operator so important? What is the use of compact operators in Mathematics?

Compact Operators have been the major topic in our Operator Theory course for the past few weeks. All the theorems which tell us whether a operator is compact or not are clear to me, but I still don't ...
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• 633
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Let $(M,g)$ be a Riemannian manifold and let $E \to M$ be a real vector bundle over $M$. Let $d_A = d+A$ be a covariant derivative on $E$. It is an $\mathbb R$-linear map $d_A \colon \Gamma(E) \to \... • 8,346 5 votes 1 answer 139 views ### Show properties of the linear operator$L((a_n)_n)=\left(\frac{1}{n}*a_n\right)_n$Let$L\colon \ell^p \rightarrow \ell^p$such that$L((a_n)_n)= \left(\frac{1}{n}*a_n\right)_n$. 1) Determine$L'$(adjoint operator),$\ker(L)$,$\ker(L')$,$\operatorname{rg}(L)$,$\operatorname{rg}(...
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In abstract Hodge theory there is the following lemma: Let $H$ be a Hilbert space and $A \in \mathcal{C}(H)$ a densely defined, closed operator (so possibly unbounded) and $A^*$ its adjoint operator. ...
### Prove or disprove: $\operatorname{Adj} (A)$ is diagonlizable $\implies A$ is diagonalizable
For $2X2$: $$A:\\ \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ $$\operatorname{Adj}(A):\\ \begin{bmatrix} d & -c \\ -b & a \end{bmatrix}$$ So the statement is true. The ...