For questions about adjoints, in the category-theoretic or inner-product-space sense, as well as about adjugate matrices.

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A connection between a matrix norm and a related matrix's largest eigen-value

I have been asked to prove that for $A\in M_n(\mathbb{C})$, with $||A||:=\sup_{x\in\mathbb{C}^n,|x|=1}|Ax|$, $$||A||=\sqrt{\lambda}$$ where $\lambda$ is the eigen value of largest modulus of $A^*A$. ...
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1answer
25 views

adjoint representations

I am trying to work out the adjoint representations of $$H=\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right), X = \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} ...
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46 views

Adjoint of an adjoint linear map

My question is as it says in the title really. I've been reading Nakahara's book on geometry and topology in physics and I'm slightly stuck on a part concerning adjoint mappings between vector ...
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1answer
52 views

A functor that has both left and right adjoints

What can we say about a functor that has both left and right adjoints? I vaguely recall hearing that it is then an equivalence of category. Is it true? If not, then under what conditions it is true? ...
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21 views

Proof of inverse matrix element explicit formula [duplicate]

There is a matrix: A, and exists an inverse matrix: A^-1 which elements are b. (b)ij = adj(Aji) / det(A) What is the proof of this equation?
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40 views

Existence of adjoint of the inverse

Let $H$ be a Hilbert space over $\mathbb{F}$ and $V$ be an inner product space over $\mathbb{F}$. Let $T:H\rightarrow V$ be a bounded linear bijection. If $V$ is a Hilbert space, then the open ...
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2answers
125 views

Question on adjoint functors

Can someone provide me an enlightenment on the following three statements? (I stumbled on them at the part dealing injective modules in a text of homological algebra.) 1) Let $F \dashv G \colon ...
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2answers
90 views

Tensor-Hom Adjunction In Monoidal Categories?

Is there a generalization of the tensor-hom adjunction to monoidal categories, or is it a special property of $\mathsf{Mod}$-$R$?
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1answer
46 views

Adjunctions via Reflections and the Axiom of Choice

I have met two ways of defining adjunctions: via the triangle identities, and via reflections. Proposition 3.1.2 Let $F:\mathsf A \rightarrow \mathsf B$ be a functor and $B$ an object of $\mathsf ...
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2answers
69 views

Prove that $T^*$ is injective iff $ImT$ Is dense

Let X,Y be two normed spaces, and $T:X\rightarrow Y$ a bounded linear operator. prove that the adjoint operator $T^*$ ($T^*f(x)=f(Tx)$ is injective iff $ImT$ is dense any help would be great guys. I ...
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23 views

Change of base - Hermitic matrices

This exercise comes from a university exam (http://www.ubacs.com.ar/foro/viewtopic.php?f=67&t=3079, link in spanish). I'll copy it in english for everyone. It's #3: We define in $C^{n×n}$ the ...
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34 views

Adjoint linear operators and inner products question; why does $\langle T(x),T(x)\rangle =\langle T^*T(x),x\rangle $?

I have seen this multiple times in my textbook; $\langle T(x),T(x)\rangle=\langle T^*T(x),x\rangle$; why is this true? I know the definition of adjoint is if $\langle x,T(y)\rangle=\langle ...
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1answer
57 views

Kernel of adjoint and orthogonal complement images

Alright, suppose we are given $V$, a finite dimensional inner product space, and a linear map, $T:V \rightarrow V$, with its corresponding adjoint, $T^\star :V \rightarrow V$. I want to show: ...
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2answers
53 views

Sequences or 'chains' of adjoint functors [duplicate]

Suppose we have (some categories and some functors such that) $F_1$ is left adjoint to $G_1$, $G_1$ left adjoint to $F_2$, $F_2$ left adjoint to $G_2$. Will $F_1$ then be equal to $F_2$ (and $G_1$ to ...
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1answer
32 views

Green's operator, differential forms

In "Foundations of Differential Manifolds and Lie Groups" by Frank Warner on page 225 there is defined Green's operator: $G: E^p(M) \rightarrow (H^p)^{\perp}$ by setting $G(\alpha)$ to equal the ...
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2answers
171 views

More than one pair of “nice” adjoint functors between different concrete categories

Though adjoint functors provide a universal description for many concrete mathematical constructions, these constructions usually revolve around finding a single "canonical" way to transform one type ...
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1answer
54 views

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 ...
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13 views

Adjoint of Polynomial?

I understand adjoint of matrices. Say I have a transformation to the standard basis [1,x,$x^2$]. Given: T(p(x)) = p(x+i) Find T*(p(x)) for all. How do I apply the adjoint transformation to the ...
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1answer
31 views

Is it always the case that a free construction satisfies this universal property?

this might be a stupid question, but I'm not sure if this is true (at least in some class of cases). Let $F : \mathcal{C} \rightarrow \mathcal{D}$ be left adjoint to an inclusion $\mathcal{D} ...
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2answers
38 views

Finding the determinant of a matrix given the adjoint

My attempt: Knowing that $$A(AdjA) = IdetA$$ I took the determinant on both sides: $$det(A)det(AdjA) = det(det(A))$$ So, $$det(A)det(AdjA) = (det(A))^3$$ $$det(AdjA) = (det(A))^2$$ $$det(A) = ...
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2answers
82 views

Proving that $\chi_{T^*}=\overline{\chi_T}$ and $m_{T^*}=\overline{m_T}$ (characteristic and minimal polynomials of adjoint map)?

For a linear map $T:V\to V$ where $V$ is a finite dimensional inner product space over $\mathbb{C}$, I know the result $\chi_{T^*}=\overline{\chi_T}$ (where $T^*$ is the adjoint map for $T$). My ...
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1answer
48 views

Adjoint of a matrix and inverse of a matrix

As everyone know that we can use a matrix $A$ to represent an operator $T$. The adjoint of a matrix $A$ is denoted as $A^*$, which takes complex conjugate of $A$ and then transpose. My problem ...
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1answer
18 views

Show $\sigma(T)=\sigma{(\overline{T^{*}})}$

Let $T \in B(H)$ be a bounded operator. Is $\sigma(T)=\sigma{(\overline{T^{*}})}$ true for $T$? $\textbf{TRY-}$ I have proved it is true for normal operator but could not do it for bounded ...
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33 views

Product of two positive compact, self adjoint operators

If we have two positive compact , self adjoint operators; $A$, $B$. Is the product $AB$ a positive operator?
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178 views

If a linear operator has an adjoint operator, it is bounded

This is a question I'm struggling with for a while: Let $H$ be a Hilber space. Let $T,S: H\rightarrow H$ be linear operators (not neccessarily bounded) such that for every $x,y\in H$: $\langle ...
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0answers
15 views

Product of self adjoint transformations

If $A$ and $B$ are linear transformations such that $A$ and $AB$ are self-adjoint and such that $\ker (A) \subset \ker (B)$, then does there always exist a self-adjoint transformation $C$ such that ...
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55 views

Adjoint Operator of a Compact Operator

In the proof of the fact that the ad-joint operator $T^*$ of a compact operator $T$ defined on a separable, infinite dimensional Hilbert space $\mathcal H$ is also compact, I read that "$\|P_nT - T\| ...
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2answers
34 views

Existence Adjoint Operator: Boundedness?

Context This would make the check on the GNS construction much more simple. Problem Given a Hilbert space $\mathcal{H}$. Consider a merely linear operator $A:\mathcal{H}\to\mathcal{H}$. Suppose ...
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1answer
48 views

Diagonalisability of Self-Adjoint Operators for Non-Symmetric Metrics

Let $V$ be a finite dimensional vector space and $(\cdot,\cdot)$ a non-degenerate bilinear form. When $(\cdot,\cdot)$ is symmetric, every self-adjoint operator on $V$ is diagonalisable. What happens ...
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60 views

Linear self-adjoint operator

Prove that there does not exist a linear self-adjoint operator T on R3 with the standard Euclidean scalar product such that T((1, 2, 3)) = (3, 2, 1) and T((4, 5, 6)) = (4, 5, 6). Where do I begin? ...
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1answer
58 views

Right-adjoint to the forgetful functor $R-\mathbf{Alg} \to \mathbf{CRing}$

Does the forgetful functor $U: R-\mathbf{Alg} \to \mathbf{CRing}$ have a right-adjoint? I checked that it commutes with finite colimits but I couldn't guess any other candidate than the tensor ...
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3answers
187 views

Connection between categorical notion of adjunction and dual space/adjoint in vector spaces

I'm an economist, not a mathematician. I've been trying to make sense of some concepts in functional analysis: dual, bidual, adjoint, natural mapping. The definitions of these notions come out of ...
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82 views

Adjoint functors and inclusions

Background / Motivation: Consider the functor $S \colon \mathrm{Mod} \to \mathrm{Comm.Alg}$ which sends a module to the symmetric algebra over that module. Let $M$ be a $k$-module and let $R ...
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21 views

Adjoint of Complex Matrix

Assume A is a matrix with complex entries. Prove that all diagonal elements of the matrix (A*)A are real where A* is the adjoint of A. Where do I start here? Thanks in advanced!
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1answer
34 views

Connection between adjoint of a matrix and adjoint of an operator

Let $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ with $$T(x,y) = \left[ \begin{array}{ccc} 1x+2y \\ 3x+4y \end{array} \right] $$ The matrix representation of $T$ is $$ A= \left[ \begin{array}{ccc} 1 ...
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1answer
55 views

Using the property $A\mathrm{adj}(A) = \det(A)I_n$, prove that $\det(\mathrm{adj}(A^3)) = (\det(A))^{3n}-3$

Suppose that $A$ is invertible $n \times n$ matrices. Then using the property $A\mathrm{adj}(A) = \det(A)I_n$, prove that $\det(\mathrm{adj}(A^3)) = (\det(A))^{3n}-3$ I started with. ...
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4answers
63 views

Let $A$ be a $3\times3$ matrix. Given $\mathrm{adj}(A)$, find $\det(A)$.

Let $A$ be a $3\times3$ matrix such that $$\mathrm{adj}(A) = \begin{pmatrix}3 & -12 & -1 \\ 0 & 3 & 0 \\ -3 & -12 & 2\end{pmatrix}.$$Find the value of $\det(A)$. I know that ...
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2answers
53 views

Let A be an invertible nxn matrix. Prove that $\det(\operatorname{adj}(A^{-1})) = (\det(A))^{1-n}$

Let $A$ be an invertible $n\times n$ matrix. Prove that $\det(\operatorname{adj}(A^{-1})) = (\det(A))^{1-n}$ I tried starting with $A^{-1} = 1/\det(A) \cdot \operatorname{adj}(A)$ I tried everything ...
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1answer
29 views

Adjoint functors of sheaves and stalks

Let $X$ and $Y$ be topological spaces and $F:Sh(X)\to Sh(Y)$, $G:Sh(Y)\to Sh(X)$ be functors between the categories of sheaves over the respective topological spaces. It seems like a very important ...
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0answers
19 views

Matrix of Adjoint operator (Hermitian conjugate)

Can someone tell me what I have to do? Operator $D$ of the matrix $D_f=\begin{bmatrix} 2&1\\ 2&0\end{bmatrix}$ with basis $f_1=(1,1), f_2=(0,1)$ of vector space $\mathbb R^{2\times 2}$ with a ...
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2answers
91 views

Understanding the significance of a functor being full/faithful, especially with adjoints

I'm working through "Basic Category Theory" by Tom Leinster and am trying to get clarity on how to reason about things... one thing I'm not sure about is how to think about what a functor being ...
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41 views

Why is the Galerkin-Method not optimal for non-self-adjoint equations

often i read phrases that explain the bad behavior of standard Galerkin-FEM for convection dominated problems by the equations beeing non-self-adjoint. Examples: Zienkiewicz, The Finite Element ...
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1answer
44 views

What is the map $\Sigma K(G,n) \to K(G,n+1)$?

Since $\Omega K(G, n+1)$ is a $K(G,n)$, we have a CW approximation/homotopy equivalence $K(G,n) \xrightarrow{\sim} \Omega K(G,n+1)$. The adjoint of this map is a map $\Sigma K(G,n) \to K(G,n+1)$. ...
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38 views

$AB*\text{adjoint}(BA)=I$

$AB*\text{adj}(BA)=I$ Prove: $1$. $|AB|=1$ $2$. $AB=BA$ As for $2$. what I have menage is $AB*AB^{-1}=AB^{-1}*AB=AB*\text{adj}$(BA)=I$ \rightarrow BA=AB$ How do I solve $1$. and is $2$. is ...
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1answer
40 views

Compact Operator on Hilbert Space

How do I show that the range of $\lambda I-T$ is all of $H$ (Hilbert Space) if and only if the null-space $\bar\lambda I-T^{\ast}$ is trivial? Thanks!
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1answer
73 views

Adjoint functors for the power set monad

There is the power set functor, $T$, which gives raise to a monad: For a set $X$, we set $TX:=\mathcal P(X)$ and for $f:X\to Y$, we set $T(f):=S\mapsto f(S)$, where $f(S)$ denotes the direct image. ...
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63 views

Proof of Fisher-Cochran's theorem

$dim(E)=n$ We have $u_1, u_2..., u_p$ self-adjoint operators which belong to $E$ $(i)$ : $rk(u_1)+...+rk(u_p)=n$ $(ii)$ : $q_1(x)+...q_p(x)=x.x$ with $q_i$ the quadratic form $q_i(x)=u_i(x).x$ for ...
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3answers
52 views

Inner product space over generalized number systems

Apologies for the lengthy setup, but I want to make sure I am clear on how I am using the notation, and what I mean by the phrase "generalized number system". Define a generalized number system $G$ ...
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1answer
47 views

self-adjoint operator without eigenvalues?

I have a self-adjoint operator $d$ which acts on vector fields defined on $\mathbb{R}^n$. I am interested on its eigenvalues. That is, I study the equation $d(X)-\lambda X=0$. I have found that if ...
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1answer
134 views

Right-adjoint to the inverse image functor

Let $X$ be a set. We can turn $\mathcal P(X)$ (the power set of $X$) into a category by taking inclusion maps as morphisms. Now consider a function $f : X \to Y$, which induces the functor $f^{-1} : ...