All Questions
Tagged with operator-theory linear-transformations
206
questions
-1
votes
1
answer
27
views
Stability of Subspaces under a Linear Map in Direct Sum Decomposition
Consider the vector spaces $D_1$, $D_2$, $D$ and $X$ such that $D\subset X$ and $D=D_1\oplus D_2$.
Furthermore, suppose that $L:X\longrightarrow D$ is a linear map such that $D_1$ is stable under $L$...
0
votes
1
answer
47
views
We are interested in finding for which $\lambda$ the operator $A - \lambda I$ is not surjective.
We are in the space $X = C[1/2, b]$ for some $b < 1$. We are interested in finding for which $\lambda$ the operator $A - \lambda I$ is not surjective. The operator $A: X \to X$ is given as $(Af)(x) ...
-2
votes
1
answer
117
views
Thus $MV - VM = V^2$. So the spectrum of $V^2$ is $\sigma(V^2) = (\sigma(V))^2=0$. Why??
I am curious if this statement holds (it doesn't make much sense to me, but it was written in solutions in this form):
$\sigma(A)=0\implies\sigma(A^2)=(\sigma(A))^2=0.$
Can anybody explain to me why ...
4
votes
1
answer
72
views
Non-trivial closed maps on Banach spaces
Let's call a map closed if it takes any closed subset to a closed subset.
I'm wondering if there are some "standard" examples of closed (in this sense) linear maps on a Banach space, other ...
0
votes
1
answer
39
views
A bilinear operator is continuous iff verifying $|| \phi (v;w)|| \leq M ||v|| ||w|| $
First I know that this question has all ready be asked for exemple here but for bilinear operator with only one variable here I want to show it for linear operator with two variables.
Question:
Prove ...
0
votes
0
answers
37
views
Adjoint of an operator on scaled Euclidean spaces
For $N\in \mathbb N$, equip $\mathbb C^N$ with the inner product $\langle\mathbf x,\mathbf y\rangle_N := N^{-1}\sum_i \overline x_i y_i$. Let $A$ be an $N\times M$ complex matrix. As an linear ...
3
votes
2
answers
221
views
Every bounded linear operator is an infinitesimal generator
I'm studying the theory of semigroups from Pazy's book. I'm struggling to understand a specific inequality in the proof of a theorem stating that every bounded linear operator is an infinitesimal ...
1
vote
0
answers
25
views
Is the dual of a unitary operator on a Hilbert space always a continous linear operator?
I have encountered this in a course (not functional analysis), but I am wondering whether this is a general fact.
Thank you very much in advance for any input!
1
vote
0
answers
42
views
A question about Composition Operators
Let $(X, \mathcal{B}, \mu)$ be a $\sigma$-finite measure space and $f:X\to X$ be a transformation such that $f(B)\in\mathcal{B}$ and $f^{-1}(B)\in \mathcal{B}$ for every $B\in\mathcal{B}$. Also there ...
1
vote
2
answers
212
views
Prove or disprove that Every bounded linear functional on a Banach space is a quotient map?
Prove or disprove that Every bounded linear functional on a Banach space is a quotient map?
Let $X$ be a Banach space over $\mathbb{K}$ and $f:X\to \mathbb{K}$ be a functional on $X$.
If $f$ is ...
0
votes
1
answer
32
views
Moving a constant operator inside an operator valued integral: Why $A\int_{X}f(t)B(t)dt=\int_Xf(t)AB(t)dt, f:X\to\mathbb{C},A,B$ linear operators
Let $X$ be some non-empty set and $\mathcal{H}$ be a Hilbert space over the complex numbers. Suppose that $A\in \mathcal{B}(H)$ is some fixed bounded operator and $B(t)\in\mathcal{B}(H),t\in\mathbb{R}$...
0
votes
1
answer
53
views
$\lim_{n}||T_n||\not\to 0\implies\sum_{n=0}^\infty T_n$ cannot converge in operator norm topology for self-adjoint $T_n\in\mathcal{B}(\mathcal{H})$
Let $\mathcal{H}$ be a Hilbert space and $\{T_n\}$ be a sequence of self-adjoint bounded linear operators. Take $T:=\sum_{n=0}^\infty T_n$ for now just as an assignment without worrying whether the ...
0
votes
2
answers
231
views
Connection between invariant subspaces and eigenvectors of a linear operator: Showing that if $A,B$ commute then $A$ and $B$ share an eigenvector [duplicate]
I got stuck trying to show that if $A,B$ are two linear operators on a finite dimensional vector space $V$ over the field $\mathbb{F}$, then $A$ and $B$ share an eigenvector. Quick Googling revealed ...
2
votes
2
answers
74
views
Does $x^T T(y) = y^T T(x)$ imply that $T$ is a linear operator?
Let $T:\mathbb{R}^n \rightarrow \mathbb{R}^n$ an operator satysfying:
\begin{equation}
x^T T(y) = y^T T(x) ~~~~\forall (x,y)\in \mathbb{R}^n\times \mathbb{R}^n.
\end{equation}
Does it imply that $T$ ...
0
votes
1
answer
55
views
Linearity of infinite dimensional matrix
The operator $T: \ell^2 \to \ell^2$ is given by the infinite-dimensional matrix with matrix elements
\begin{align*}
t_{kl} = \frac{\vert k - l \vert}{k^2l^2}
\end{align*}
for all $k, l \in \mathbb{N}...