Articles of adjoint

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, $T:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ be a bounded operator given by $$Tf(y) = \int_{-\infty}^{\infty} k(x,y)f(x)\,dx.$$ Then the adjoint, $T^*$, of $T$ is given by $$T^*f(y) = \int_{-\infty}^{\infty}\overline{k(y,x)}f(x)\,dx.$$ The proof is as follows: $$\begin{align} \langle Tf,g\rangle & \stackrel{\text{def}}{=} \langle […]

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 attention to this example, remarking that “its adjoint is not what at first it might appear to be”: Let $\mathfrak H$ be the set of analytic functions […]

How can we compute the adjoint of the inclusion between two Hilbert spaces?

Let $\mathbb K\in\left\{\mathbb C,\mathbb R\right\}$ $(U,\langle\;\cdot\;,\;\cdot\;\rangle_U)$ and $(V,\langle\;\cdot\;,\;\cdot\;\rangle_V)$ be $\mathbb K$-Hilbert spaces such that $U\subseteq V$ and that the inclusion $\iota:(U,\langle\;\cdot\;,\;\cdot\;\rangle_U)\to(V,\langle\;\cdot\;,\;\cdot\;\rangle_V)$ is Hilbert-Schmidt $C:=\iota^\ast$ denote the adjoint of $\iota$ and $$\langle u,v\rangle_0:=\langle C^{-\frac 12}u,C^{-\frac 12}v\rangle_U\;\;\;\text{for }u,v\in U$$ We can show that $$U=C^{\frac 12}V\;\;\;\text{and}\;\;\;\langle\;\cdot\;,\;\cdot\;\rangle_0=\langle\;\cdot\;,\;\cdot\;\rangle_U\;.$$ How can we compute what the adjoint $C$ of $\iota$ actually is?

Isometry <=> Adjoint left inverse

This question already has an answer here: Isometric <=> Left Inverse Adjoint 2 answers

Prove that $\|T\|=\sup_{\|x\|=1}|\langle x,T(x)\rangle|$.

Let $T$ be a self adjoint bounded linear operator in a Hilbert space $H$. Prove that $$\|T\|=\sup_{\|x\|=1}|\langle x,T(x)\rangle|$$

$\operatorname{Im} A = (\operatorname{ker} A^*)^\perp$

This question already has an answer here: Is the formula $(\text{ker }A)^\perp=\text{im }A^T$ necessarily true? 1 answer

Show $\alpha$ is selfadjoint.

This question already has an answer here: $TT^*=T^2$, show that $T$ is self-adjoint 2 answers

Is it possible to define an inner product such that an arbitrary operator is self adjoint?

Given a vector space $V$ (possibly infinite dimensional) with inner product $(.,.)$. We say an operator $A$ is self adjoint if $(Af,g)=(f,Ag)$. The definition as stated require us to start with an inner product $(.,.)$ in $V$ and check if the operator $A$ satisfies the equality. My question is: If we start with an operator […]

Gradient operator the adjoint of (minus) divergence operator?

Recently I found this statement — the gradient operator is the adjoint of the minus divergence operator — in one of my lecture notes. Knowing only a little about functional analysis, I’m looking for an intuitive interpretation. I already found this topic which has a few good answers, but I’d like to view it from […]

image of adjoint equals orthogonal complement of kernel

Let $T:V\to W$ be a linear map of finite-dimensional spaces. Then $${\rm im}(T^{\textstyle*})=({\rm ker}\,T)^\perp\ .\tag{$*$}$$ I can prove this as follows: $${\rm ker}(T^{\textstyle*})=({\rm im}\,T)^\perp\tag{$**$}$$ is quite easy, and we know $T^{\textstyle*}{}^{\textstyle*}=T$ and $W^{\perp\perp}=W$, so $${\rm im}(T^{\textstyle*})=({\rm im}\,T^{\textstyle*})^{\perp\perp}=({\rm ker}\,T^{\textstyle*}{}^{\textstyle*})^\perp=({\rm ker}\,T)^\perp\ .$$ However, I would be interested in a “direct” proof of $(*)$. It’s fairly easy to […]