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6

$$ A(A + I)^{-1} = (A + I - I)(A + I)^{-1}\\ = (A + I)(A + I)^{-1} - I(A + I)^{-1}\\ = (A + I)^{-1}(A+I) - (A + I)^{-1}I\\ = (A + I)^{-1}(A+I-I)\\ = (A + I)^{-1}A $$ since $(A + I)$ commutes with its own inverse, and $I$ commutes with anything.


5

Assuming this is an operator on $\mathbb{R}^2$ with the standard inner product $\left< x, y \right> = x \cdot y$, then yes, this is the correct reasoning. In general, you need to specify an inner product to know if an operator is self-adjoint, because self-adjointness means $\left< x, A y \right> = \left< A x, y \right>$. In this case, ...


4

Because $A$ is normal, then $\mathcal{N}(A)=\mathcal{N}(A^{\star})$ is invariant under $A$ and $A^{\star}$, as is its orthogonal complement $$ \overline{\mathcal{R}(A)}=\mathcal{N}(A)^{\perp}=\mathcal{N}(A^{\star})^{\perp}=\overline{\mathcal{R}(A^{\star})}. $$ That allows you to reduce to the case where $\mathcal{N}(A)=\{0\}$, and where the ranges ...


3

See theorem 1.2, page 4 of this file http://www.stat-athens.aueb.gr/gr/master/sumschool/files/Kravvaritis.pdf


3

No, the operator is not self adjoint. Your reasoning is correct. Note that the definition of the adjoint of an operator is not simply that the adjoint of $A$ is $A^T$. However, in the context of real matrices taken with respect to the standard dot product or with respect to the standard dual basis, this is what "adjoint" comes to mean. An operator is ...


3

let $f\in C[0,1]$ and $M=\max_{[0,1]\times [0,1]} k$. Then $$\left|\int_0^1 k(x,y) f(y) dy\right| \le M \int_0^1 |f|\le M |f|_{C[0,1]}$$ Now take $\sup$ over $x\in [0,1]$. (Note that the map in question is obviously linear)


2

It should be closed. Let $T:V\to W$ be a bounded linear operator between normed linear spaces. We have that $$ \ker(T) = \{x\in X: Tx =0\} = T^{-1}(\{0\}). $$ Since bounded operators are continuous and $\{0\}$ is a closed set in the norm topology, we see that the kernel is closed. The kernel is not open unless $T$ is the zero operator. Assume $T\neq 0$. We ...


2

Your reasoning is correct, for a continuous (bounded) everywhere defined operator $A$ on a Banach (in particular on a Hilbert) space, the denseness of the range of $\lambda\mathbb{I} - A$ together with the boundedness of the inverse already implies the surjectivity of $\lambda\mathbb{I} - A$, and an equivalent definition of the resolvent set in this setting ...


2

Yes, you can even get $\operatorname{ran}(TS) = \ker S $, without taking the closure on the left. Let $A:\ker T\to H$ be any isomorphism; composing it with a projection $H\to \ker T$ we get a linear operator $B:H\to H$ such that $B(\ker T) = H $. Define $$S = T^\perp + TBT^\perp,\quad \text{ where } T^\perp = I-T$$ I leave it for you to check that ...


1

A closed densely-defined linear operator on a Hilbert space can have empty spectrum. For example, let $H=L^{2}[0,1]$ and let $A=\frac{d}{dx}$ on the domain consisting of absolutely continuous $f \in L^{2}$ for which $f(0)=0$ and $f' \in L^{2}$. To show that $A$ has no spectrum, it is enough to prove that the resolvent $R(\lambda)=(A-\lambda I)^{-1}$ exists ...


1

Nearly every Physical system has a linear regime where if you superimpose two small causes, the resulting effect is the superposition of the two separate causes, and if you double a small cause, then the effect is doubled. The state of the system is characterized by a collection of numbers (a vector) and the principle of superposition is stated in terms of ...


1

In fact, for any scalar product, $A$ is never self-adjoint. Let $f$ be a non-degenerate symmetric bilinear form on $\mathbb{R}^2$ and $M=\begin{pmatrix}a&b\\b&c\end{pmatrix}$ be its matrix in the canonical basis. The adjoint $\tilde{A}$ of $A$ is defined by $f(x,\tilde{A}y)=f(Ax,y)$, that is $M\tilde{A}=A^TM$. Thus $A$ is self-adjoint iff $MA=A^TM$. ...


1

Every normed linear space is Hausdorff, hence singletons are closed sets in a normed linear space. $T:V\to W$ is bounded iff it is continuous, Now $\{0\}$ is closed in $W$ and $T$ is continuous. So, by continuity of $T$, $$ T^{-1}(\{0\})$$ is closed. Hence $$\ker(T) = \{x\in X: Tx =0\}$$ is closed.


1

Another way, first we show linearity: $K(f+g)=\int_0^1 k(x,y)f(x,y)dy+ \int_0^1 k(x,y)g(x,y)dy=\int_0^1 k(x,y)(f(x,y)+g(x,y))dy $ , by linearity of the integral. Then we use 1st countability of $C[0,1]$ (since it is a metric space), and show sequential continuity, which is equivalent to continuity: Assume $f_n \rightarrow f$ in $C[0,1]$ , so that $Sup ...


1

This is only a check! The spectrum is neglible: $$0\leq\lambda_\mathbb{C}(\sigma(H))\leq\lambda_\mathbb{C}(\mathbb{R})=0$$ By functional calculus: ...


1

Let $r>$ be such that: $ A \subset B_r(0) $. With $B_r(0)$ the ball of radius $r$ centered in $0$. Suppose $T(A)$ is not bounded. Then for each $n \in N$ there exists a $x_n \in A$ such that: $||Tx_n|| > n+1 $. The sequence $\frac{x_n}{n+1}$ converges to $0$ because $A$ is bounded. But clearly $|| T \left( \frac{x_n}{n+1} \right)|| >1$. This ...



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