For questions about adjoint operators in inner product spaces. For adjoint functors from category theory, use the tag (adjoint-functors).

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Showing that $Im(L^*)=(Ker\: L)^\perp \space \:\mathrm and \:\:Ker(L^*)=(Im\: L)^\perp$

Let $V,W$ be finite-dimensional euclidean or unitarian Spaces and $L: V \to W$ a linear map. I have to show the following: $$Im(L^*)=(Ker\: L)^\perp \space \:\mathrm {and} \:\:Ker(L^*)=(Im\: L)^\perp ...
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1answer
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How to find a diagonalizable linear map T such that any subspace W is T-invariant but not T*-invariant? [on hold]

Let V be a finite-dimensional complex inner product space. Let $W$ be a subspace of $V$ not equal to $\{0\}$ or $V$. Construct a linear operator $T$ on $V$ such that (i) $T$ is diagonalizable (ii) ...
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1answer
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If $(U,〈\;⋅\;,\;⋅\;〉),H$ are Hilbert spaces, $W\in U$, $Y\in H$, $Z\in L(U, H)$ and $f\in L(H,L(H,\mathbb R))$, then $〈Y,fZW〉=〈ZW,fY〉$

Let $(U,\langle\;\cdot\;,\;\cdot\;\rangle)$ and $H$ be Hilbert spaces $W\in U$, $Y\in H$ and $Z\in\mathfrak L(U, H)$$^1$ $f\in\mathfrak L\left(H,\mathfrak L\left(H,\mathbb R\right)\right)$ How ...
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1answer
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Prove that $A = G^{4}$ for some self-adjoint matrix $G$

Let $A\in M_{n}(\mathbb{C})$, $n\geq 2$. Let $P_A(x) = (x-\lambda_1)....(x-\lambda_n)$ be characteristic polynomial of $A$ and all $\lambda_i$ are positive reals. Show that if $A$ is normal, then $A ...
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1answer
24 views

When is this matrix unitary

If we have the matrix $$U=\begin{bmatrix} a & b & \frac{1}{\sqrt{2}} \\ c & 0 & 0 \\ d & e &\frac{-1}{\sqrt{2}} \end{bmatrix}$$ what are the ...
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2answers
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Self adjoint and symmetric operator

I am wondering whether for an operator defined on a real Hilbert space to be positive we need to show that it is self-adjoint at first. It seems to me that they are two different property and can be ...
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60 views

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 ...
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1answer
20 views

Does this inner product manipulation make sense?

Suppose A is a normal matrix over $M_n(\mathbb{C})$, with diagonalization $A = PDP^*$. Consider the inner product $<A\mathbf{v}, \mathbf{v}>$ $<A\mathbf{v}, \mathbf{v}> = ...
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1answer
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Compute $ad_X$, $ad_Y$, and $ad_Z$ relative to a basis

For a lie algebra $\mathbb{g} $ we can define the adjoint representation as: $ ad: \mathbb{g} \rightarrow End(\mathbb{g}) $ as the map such that $ad_x(y)=[x, y] $ for all $\in \mathbb{g} $ I am ...
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1answer
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Is it true that when $X$ is a positive matrix and $Y$ is a self adjoint matrix then $XY$ is positive??

Let $X_{n \times n}$ be a positive matrix i.e $<Xy,y> \ge 0$ for all $y \in \mathbb{C^n}$ and $Y_{n \times n}$ be a self adjoint matrix. Show that $XY$ is positive or find a counterexample. So ...
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2answers
35 views

Proving an complex operator is self-adjoint

Let V be a finite dim complex inner product space. Let T be a linear operator on V. Prove that T is self-adjoint if and only if $\langle T\alpha,\alpha\rangle$ is real $\forall \alpha \in V$. The ...
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2answers
15 views

Proving Kernel Equality of a Linear Transformation and Adjoint

Let $T$ be a linear transformation in an inner product space $V$. Determine if the following it true or false: $$Ker (T)= Ker (T^*T)$$ Where $*$ donates the adjoint operator. Would it help proving ...
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0answers
31 views

Is the adjoint of an isometry $S$ the inverse of $S$?

Let $S$ be an isometry with adjoint $S^\ast$. Prove that $S^\ast S=SS^\ast=I$. Here is what I have so far: Since $S$ is isometry, there is an orthonormal basis of $e_j$'s such that ...
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1answer
44 views

Hodge decomposition thm. Why $\Delta(E^p)=d\delta(E^p)\oplus\delta d(E^p)=d(E^{p-1})\oplus\delta(E^{p+1})?$

$\DeclareMathOperator{Img}{Im}$ In Warner's "Foundations of differentiable manifolds and Lie groups" we read that $$E^p\stackrel{(1)}{=}\Delta(E^p)\oplus H^p\stackrel{(2)}{=}d\delta(E^p)\oplus\delta ...
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1answer
21 views

Are all adjoints lattice homomorphisms?

Obviously something must be wrong in the following reasoning proving that any linear operator $T:X\to Y$ between Banach lattices has a lattice homomorphic adjoint: $\forall a,b\in E':$ $$T'(a\wedge ...
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2answers
39 views

self-adjoint and orthonormal basis

Suppose $F=\mathbb{R}$. Let $A: V\to V$ (where $V$ is a finite dimensional inner product space over $F$) so that $A=A^*$ ("self-adjoint"), then there exists an orthonormal basis of eigenvectors and ...
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34 views

How do I show that if is normal, then if T is a projection then it must be an orthogonal projection?

T is an operator on a finite dimensional inner product space. How do I show that if T is a projection, then it must be an orthogonal projection? I know I have to use the fact that $T^*T=TT^*$, and I ...
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1answer
30 views

Exponential of a self-adjoint operator

Let $\mathcal{H}$ be an Hilbert space. Firstly, I shall define some notions as their definitions may vary: A spectral resolution is a function $E:\mathbb{R}\to\mathcal{L}(\mathcal{H})$ (the space ...
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Finding two adjoints, and showing boundedness of operators

Let $H = l_2$ and consider the following operators: $T,S:H \to H$ $Tx = (0,x_1,x_2,\ldots)$ and $Sx = (x_2,x_3,x_4,\ldots)$ Show they are bounded, and find the adjoint of both: For $T$, I have ...
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adjoint method for computing derivatives

I am curious if anyone has heard of this problem before: Suppose that $u(x,p)$ is a function of $x$ and $p$. These arguments need not be scalars. Let $u(x,p)$ satisfy some differential equation, ...
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1answer
33 views

The adjoint of a linear isometry on normed space is surjective

I come across with the following question: Let $X$ and $Y$ be normed spaces over the same field and $T:X \rightarrow Y$ be a linear isometry, with $\|Tx\|=\|x\| \, \forall x \in X$, then the adjoint ...
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1answer
53 views

Prove that $R(T^{*})^\perp =N(T)$

Let V be an inner product space, with T being a linear operator on V. How do I prove that $R(T^{*})^\perp =N(T)$? I tried setting $x\in R(T^{*})$ and $Ty\in N(T)$, and set up an inner product = 0 ...
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1answer
32 views

Is there any way to find an explicit formula for the adjoint of a linear transformation?

I know that the definition of the adjoint of a linear transformation is defined to be $\langle T(x), y \rangle = \langle x, T^{*}(y) \rangle$ but is there any way to find an explicit formula for ...
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1answer
25 views

Eigen values and self adjoin operator

Can someone give me a clue about how to solve the b part ? All I know is the self adjoint formula $$\langle ku,u\rangle = \lambda\langle u,u\rangle$$
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1answer
52 views

Spectrum $\sigma(T)$ of $T:l^1 \to l^1$ given by $T((a_j))=\left( \sum_{j=2}^{\infty} a_j \right) e_1 + \sum_{j=2}^{\infty} a_{j-1} e_j$

I'm considering the bounded linear operator $T$ on $l^1$ (the space of all absolutely convergent complex sequences) given by (with $e_k=(\delta_{kj})_{j=1,2,...}$) $$T((a_j))=\left( ...
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Let $Q$ be a positive-definite, non-singular, self adjoint linear transformation. Then if $P$ is self adjoint, $Pe_i = \lambda_i Qe_i$

I need to show what's written in the question title, for some set of linear independent vectors $\{e_1,\dots,e_n\}$ and scalars $\lambda_1 \dots \lambda_n$. My hypothesis is that the vectors to use ...
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1answer
51 views

Proving that the eigenvalues of $T^*T$ is non-negative

Suppose $T^*$ is the adjoint of $T.$ I want to prove that $T^*T$ has only non-negative real eigenvalues. I proved that $T^*T$ is self-adjoint (because $\overline{(\overline{A^T}A)^T} = ...
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2answers
68 views

Relating ${\rm ad}(\phi(X))$ with $\phi \circ\,{\rm ad}(X) \circ \phi^{-1}$.

The exercise is to prove that $\phi: {\frak g}\to {\frak g}$ is an automorphism if and only if ${\rm ad}(\phi(X)) = \phi\circ {\rm ad}(X)\circ \phi^{-1}$. Here $\frak g$ is a Lie algebra, of course. ...
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1answer
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Adjoint action on Lie algebra su(2) ($A \in SU(2), X \in \mathfrak{su(2)} \Rightarrow AXA^{-1}\in \mathfrak{su(2)}$)

I am trying to understand ho $SU(2)/\{\pm I\} \cong SO(3)$ (see: how to show $SU(2)/\mathbb{Z}_2\cong SO(3)$) but i am not sure about the adjoint action. In especially, as I understand, the adjoint ...
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1answer
40 views

Spectrum and resolvent of an operator

So for the operator $A:l_2(\Bbb C)\to l_2(\Bbb C)$ defined as: $$A(x_1,x_2,\cdots,x_m,x_{m+1},x_{m+2},\cdots) = (x_1,x_2,\cdots,x_m,0,0,\cdots)$$ We can find the adjoint operator $A^*$ by looking ...
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Trying to understand adjoint operators and the dual space of $l_2(\Bbb C)$

Consider the operator: $A: l_2(\Bbb C) \to l_2(\Bbb C)$ $$A(x_1,x_2,\cdots, x_m,x_{m+1},x_{m+1}, \cdots) = (x_1,x_2,\cdots,x_m,0,0,\cdots)$$ I.e. it's the identity map on the first $m$ values, and ...
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1answer
24 views

Is $f$ unitary if $f\in L(V,V)$ such that $a<\frac{\|f^n(v)\|}{\|v\|}<b$, with $0<a<1<b$, for all $n\in \mathbb{N}$ and nonzero $v$

Let $V$ be a complex inner product space. Is $f$ unitary if $f\in L(V,V)$ such that $a<\frac{\|f^n(v)\|}{\|v\|}<b$, with $0<a<1<b$, for all $n\in \mathbb{N}$ and nonzero $v$? If not, ...
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Self adjoint systems of ordinary differential equations

Could someone direct me to a reasource on self adjoint systems of ODEs? In particular I would like to know how to test a system for being self adjoint.
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Exact Equations and Integrating Factor

I am really having difficulty with this problem. I understand part a, as I did d/dx[x^2y'+(x^2'-2x)y]=0 and then took the integral of both sides to get lny==x+2lnx+C. What I don't understand is how ...
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2answers
47 views

Show that $0$ is the only eigenvalue of $Q$ given that $Q^*=5Q$

$Q$ is a linear operator from $V \to V$ with $V$ being a finite dimensional complex inner-product-space. Given: $Q^*=5Q$, $Q^*$ being the adjoint. Show that $0$ is the only eigenvalue of $Q$. ...
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1answer
30 views

Show that $A$ is self-adjoint if and only if $(Ax,x)$ is real for all x

Let $A:X\rightarrow X$ be a linear transformation, where $X$ is a finite-dimensional complex inner product space. Show that $A$ is self-adjoint if and only if $(Ax,x)\in\mathbb{R}$ for all $x\in ...
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1answer
28 views

adjoint operator of the partial trace map

Could someone explain to me, what is the adjoint map of the partial trace map the (tensored with the identity map), or why does the following equality hold? $Tr(C_A\cdot Tr_{B} D_{AB})=Tr((C_A\otimes ...
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2answers
41 views

Selfadjoint Endomorphism

Question: Let $p >1$ be an integer, let $G = \mathbb{Z}/(p)$, and let $V = \mathbb{C}^G$, which is an inner product space over $\mathbb{C}$ with inner product defined by $\langle f, g\rangle ...
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Every finite dimensional Self-Adjoint Linear Map has an orthonormal basis of eigenvectors Proof

What I understand: By dimension formula $dim V$ $-$ $1$= $dim \langle v \rangle ^ \perp$ since $V$ is the direct sum of $U$ and $U ^ \perp$ if $U$ is a subspace of $V$. dim $\langle v \rangle$ is ...
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1answer
24 views

Adjoint linear transformations

Is it the case that the adjoint of a matrix is just its transpose- the definition is based off inner products. Does it matter whether it is a complex or real inner product?
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3answers
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T compact if and only if T*T is compact.

I have an operator $T \in B(\mathcal{H})$. I need to prove that T is comapct if and only if $T^*T$ is compact. One way is ok, because if A or B is comapct then AB is compact, so I get at once that if ...
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1answer
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Adjoint Transformation

Let $V$ be a finitely-generated inner product space and let $\alpha,\beta_1,\beta_2 \in \text{End}(V )$ satisfy $\alpha^* \alpha \beta_1 =\alpha^* \alpha \beta_2$ . Show that $\alpha \beta_1 = \alpha ...
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1answer
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Let $T:\ell_1 \to c_0$ be linear operator defined as $x_n \to \sum_{k\geqslant n} x_k$. Then $T \in B(\ell_1, c_0)$.

I'm solving some exercises for my Functional Analysis Exam. Here is one on which I am stuck: Let $T: l_1 \to c_0$ be linear operator defined as $x_n \to \sum_{k\geqslant n} x_k$. Show that $T \in ...
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2answers
54 views

Is intersection of a dense subspace and a closed subspace of a Hilbert space also Dense?

I have a Hilbert space $H$ and a closed operator $T$ defined on its domain $D(T)$ which is dense in H. Also $M = \text{range} \ T^n$, for some $n$, is given to be closed. Consider the restriction of ...
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1answer
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If the only self adjoint operators from $\mathcal H$ to $\mathcal H$ are $0$, what comment we can make on $\mathcal H$?

If $\mathcal H$ is a Hilbert space over $\mathbb C $ or $\mathbb R$. The only self adjoint operators from $\mathcal H$ to $\mathcal H$ are $0$ and $I$. --------------------------------(1) Then ...
2
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1answer
35 views

Properties of adjoint matrix in a finite dimensional inner product space

let $V$ be a finite dimensional inner product space. Let $T$ be a linear operator on $V$. Prove that there exists an invertible linear operator $U$ such that $U^{-1}TT^*U = T^*T $ where $T^*$ is ...
2
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1answer
69 views

Operator $T^*$ surjective iff $T$ topologically injective

Let $T : E \to F$ operator between Banach spaces, and $T^* : F^* \to E^*$ - adjoint operator. I want to proof next proposition: $T$ is topologically injective (or equivalently: injective with closed ...
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1answer
58 views

Suppose that $T$ is a normal operator on $V$. Show that $\|T(v+w)\|=10$.

Suppose $T$ is a normal operator on $V$. Suppose also that $v, w \in V$ satisfy the equations $$ \| v \|= \| w \| =2, Tv = 3v, Tw = 4w.$$ Show that $\| T(w+v) \| = 10.$ I thought this problem would ...
0
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1answer
30 views

Determining adjoint operator between spaces with different inner products

I'm given a space of $X \in R^{mxn}$ with inner product $\langle X_1, X_2 \rangle = tr(X_1^T X_2)$ and another space of random vectors $Y \in RV^m$ with inner product $\langle y_1,y_2 \rangle = ...
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3answers
127 views

$T$ is surjective if and only if the adjoint $T^*$ is an isomorphism (onto its image)

I am trying to prove the following statements: Let $X$ and $Y$ be normed spaces (not necessarily complete) Let $T\in L(X,Y)$ (meaning $T:X\to Y$ is a bounded linear map). Let $T^*:Y^*\to X^*$ denote ...