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

learn more… | top users | synonyms

1
vote
0answers
27 views

When is $adj(A)$ nilpotent?

Are there any conditions regarding the $adj(A)$ being nilpotent for some square matrix A?
1
vote
1answer
28 views

Clever way to prove $\langle A,X\rangle=x^TAx$ with $X=xx^T$, $A\in S^n$?

How to prove $\langle A,X\rangle=x^TAx$ with $X=xx^T$, $A\in S^n$? (inner product of matrices) $xx^T$ is rank one. The following is one way to prove it: $$\langle A,X\rangle=\text {tr}(AX)$$ ...
0
votes
3answers
40 views

Is every invertible matrix $A$ an Adjugate matrix of some other matrix $B$? If so, is $B$ unique?

Is it true in general, true for a specific field ($\mathbb R$/$\mathbb C$) or false? could it be that $A=adj(B),A=adj(C)$ but $B\not=C$?
0
votes
0answers
15 views

Rank of adjoint of a matrix [duplicate]

I need to prove these 3 statements, and I don't know how to start... A is an nxn matrix: 1) if rank(A) = n then rank(adj(A)) = n 2) if rank(A) = n-1 then rank(adj(A)) = 1 2) if rank(A) < n-1 ...
0
votes
0answers
22 views

Finding the adjoint of the left translations semigroup on $L^p (\Bbb R)$

If $t \mapsto T_l (t)$ is the left translation operator by $t$ on $L^p (\Bbb R)$ given by $\Big( T_l (t) (f) \Big) (s) = f (t + s)$, find the adjoint of the left translations semigroup. Note that on ...
3
votes
1answer
72 views

Operator $f \mapsto u(f)$ solution of non-homogeneous Laplace equation is compact and self-adjoint

Let $u : L^2_0(D) \to L^2_0(D): = \lbrace f \in L^2 : \int_D f = 0 \rbrace $ be the linear operator which associates $f$ to $u(f)$ the solution of $$ \begin{cases} \Delta u = f & \text{in } D \\ ...
1
vote
1answer
30 views

determinants of matrix with adjoints of order 2

Let $A$ be a square matrix of order $2$ with $\lvert A \rvert\ne 0$ such that $\big\lvert A+\lvert A \rvert \operatorname{adj} (A)\big\rvert=0$, then the value of $$\big\lvert A-\lvert A \rvert ...
1
vote
2answers
58 views

Does every invertible matrix A has a matrix B such that A=Adj(B)?

I'm trying to understand if it's always true, always true over $\mathbb C$ or never true. I know that if $A$ is invertible, than there exists $A^{-1}$. $$A=\frac{1}{det (A^{-1})}Adj(A^{-1})$$ So I ...
1
vote
0answers
14 views

Singular values of the differentation operator

I was trying to solve this exercise but I can't get to answer given in book: Find the singular values of the operator D over $R_2[x]$ (That is polynomials with degree equal or less than 2) define as ...
0
votes
0answers
20 views

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 ...
1
vote
1answer
26 views

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 ...
0
votes
1answer
20 views

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 ...
1
vote
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 ...
2
votes
2answers
44 views

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 ...
3
votes
2answers
84 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 ...
2
votes
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}> = ...
1
vote
1answer
34 views

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 ...
1
vote
1answer
28 views

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 ...
0
votes
2answers
36 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 ...
0
votes
2answers
16 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 ...
1
vote
0answers
33 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 ...
1
vote
1answer
46 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 ...
2
votes
1answer
23 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 ...
0
votes
2answers
44 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 ...
0
votes
0answers
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 ...
2
votes
1answer
34 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 ...
3
votes
0answers
20 views

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 ...
1
vote
0answers
27 views

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, ...
3
votes
1answer
34 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 ...
2
votes
1answer
71 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 ...
0
votes
1answer
46 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 ...
0
votes
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 $$\langle ku,u\rangle = \lambda\langle u,u\rangle$$
3
votes
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( ...
0
votes
0answers
11 views

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 ...
1
vote
1answer
58 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} = ...
0
votes
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. ...
1
vote
1answer
46 views

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 ...
1
vote
1answer
41 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 ...
0
votes
2answers
31 views

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 ...
2
votes
1answer
25 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, ...
0
votes
0answers
21 views

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.
-1
votes
1answer
31 views

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 ...
0
votes
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 $Q$. ...
1
vote
1answer
31 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 ...
1
vote
1answer
29 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 ...
2
votes
2answers
44 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 ...
1
vote
2answers
39 views

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 ...
0
votes
1answer
28 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?
2
votes
3answers
88 views

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 ...
1
vote
1answer
33 views

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 ...