New answers tagged

1

I'm going to revise the notation to make things easier to follow. Say $X^*$ is the space of bounded linear functionals on $X$, and similarly for $Y^*$. I'm going to write $x$ and $y$ for elements of $X$ and $Y$ and I'm going to write $x^*$ and $y^*$ for elements of $X^*$ and $Y^*$. Define the adjoint $T^*:Y^*\to X^*$ as usual: $$T^*y^*=y^*T.$$We can use ...


0

We use the Closed Graph Theorem. Assume $x_n \to x$ and $Tx_n \to y$. We need to prove that $Tx = y$. By Hahn Banach this is true if $f(y) = f(Tx)$ for all $f \in Y^*$. But $f(Tx) = \lim_n f(Tx_n) = f(y)$. So we are done. (In case $X,Y$ are Banach.)


1

The proof follows from the following theorem from basic calculus: Let $\{a_n\}_{n\in\mathbb N}$ be a sequence of nonnegative numbers with the property that $a_{n+m}\leq a_n \cdot a_m$ for all $n,m\in\mathbb N$. Then the limit $\lim\limits_{n\to\infty}\sqrt[n]{a_n}$ exists and it is equal to $\inf\limits_{n\in\mathbb N}\sqrt[n]{a_n}.$ Theorem about ...


2

Geometrically we have just one operator, namely $A$. This is the map that moves the points of our space $V:={\mathbb R}^n$ around. In order to deal with this $A$ in a computational way we have to choose a basis for $V$. At the outset we would use the standard basis $(e_i)_{1\leq i\leq n}$ of ${\mathbb R}^n$, and with respect to this basis the operator $A$ ...


7

I would start from the logical definition of the matrix factorial, without assuming that we want to cover all properties that we know from factorial in set of reals. We define standard factorial as $1 \cdot (1+1) \cdot (1+1+1) \cdot ... \cdot (1+1+...+1+1)$ So first let us define $[n]!$ using the same logic replacing 1 with identity matrix. The obvious way ...


0

EDIT (see below) Following Robert Israel's idea here's another example, which is even Hilbertian: on the space $L^2(0,1)$ consider, again, the polynomials and the linear operator $T$ defined as $$T(x^{2k})=0\quad > T(x^{2k+1})=x^{2k+1}, $$ which agrees with the identity on the dense subspace consisting of odd-degree polynomials and is ...


1

The first part is a notational device, to be able to describe a change of basis operator $B$, by defining its action on the (standard?) basis $\{e_1,\ldots,e_n\}$: $$ Be_j = u_j, \;\; j=1,\ldots,n $$ Then we can related the matrix representation of any other linear operator $A$ on $\mathbb{R}^n$ with respect to the standard basis and the representation of ...


2

Travis' answer is very nice. It would be good to mention that (almost) any matrix function can be made into a power-series expansion, which eventually involves the values of the function on the eigenvalues of the matrix multiplied by the eigenvectors. In other words the matrix function is completely characterised by the values it takes on the eigenvalues ...


1

Consider the Banach space $X = C[0,1]$ of continuous functions on $[0,1]$, with the operator $A$ of multiplication by $x$ (i.e. $Af(x) = x f(x)$). This has spectrum $[0,1]$. Let $E$ be the subspace of $X$ consisting of polynomials. This is invariant under $A$. However, $A - \lambda I$ is never surjective as an operator from $E$ to $E$, e.g. there is no ...


-1

$X=R^2=Vect\{e_1,e_2\}$, $A(e_1)=e_1, A(e_2)=2e_2$, $E=R^2-Re_2$. The spectrum of $A$ restricted to $E$ is $1$ and the spectrum of $A$ is $\{1,2\}$.


0

At first, let us review three points: (1) Since $\phi$ is pure then $s(\phi)$, the support of $\phi$ is a minimal projection in $M$. (2) Let $z_{\pi}$ be the support of $\pi$ (since $\ker\pi$ forms a $w^*$-closed 2-sided ideal then there is central projection $q\in M$ with $\ker\pi=Mq$. Support of $\pi$ is defined by $1-q$. Therefore the restriction of $\pi$ ...


0

One may check that: No sub-basic open set in the strong operator topology is an open set in the weak operator topology.


2

Let $\def\norm#1{\left\|#1\right\|_{L^2}}f \in L^2(0,\infty)$. We have \begin{align*} \norm{Tf}^2 &= \def\abs#1{\left|#1\right|}\int_0^\infty \frac 1{x^2} \abs{\int_0^x f(y)\, dy}^2 \, dx\\ &\le \int_0^\infty \frac 1{x^2}\left(\int_0^x y^{1/4}y^{-1/4}\abs{f(y)}\, dy\right)^2 \, dx \\ &\le \int_0^\infty \frac 1{x^2} ...


0

After some fiddling, I feel I have a satisfactory answer. Define $\omega := [2,\infty) \cap \sigma(T)$, and $P := E(\omega)$. This is a self-adjoint projector by definition of $E$. By symbolic calculus, we have $$((TP - 2P)x,x) = ((T-2I)Px,x) = \int_{\sigma(T)} (\lambda - 2)\chi_{\omega} d(E_\lambda x,x).$$ We know $(E_\lambda x,x)$ is a positive measure ...


0

For any $x\in\mathbb H $ we have $Ax=\lim A_nx$, with $A_nx\in\text {ran }\,A_j $. Because a Hilbert space is metric, we can do this with sequences. Thus $$ \text {ran}\,A\subset \text {cl}\left (\bigcup_n\text {ran}A_j\right). $$


12

I don't have enough reputation points to comment on Travis' answer, but his numerical result is incorrect. Using Julia I get A = [1 3;2 1] EVD = eigfact(A) V = EVD[:vectors] g = gamma(EVD[:values]) gammaA = V * diagm(g) * inv(V) factA = A * gammaA 3.62744 8.84231 5.89488 3.62744 As long as cond(V) isn't too terrible, I've found the above procedure ...


0

\begin{align} TP-2P&=(T-2I)P,\\ 2(I-P)-T(I-P)&=-(T-2I)(I-P) \end{align} $P=E[2,\infty)$ does the job because $I-P=E(-\infty,2]$.


0

The form domain of $A$ is $(A+I)^{-1/2}$ because $A$ is positive. This is because $$ (Ax,x)+(x,x) = ((A+I)x,x) = ((A+I)^{1/2}x,(A+I)^{1/2}x)=\|(A+I)^{1/2}x\|^2. $$ Because all of the form domains are the same, then $(A_n+I)^{1/2}(A+I)^{-1/2}$ must be a bounded operator for all $n$. Therefore, on the domain of $A$, the following is bounded: $$ ...


3

Let $T=\frac{1}{i}\frac{d}{dt}$ be defined on the domain $\mathcal{D}(T)$ consisting of all absolutely continuous functions $f \in L^2[0,1]$ for which $f(0)=0=f(1)$. More precisely, $f \in \mathcal{D}(T)\subset L^2[0,1]$ is an equivalence class of functions equal a.e. with one element $\tilde{f}$ of the equivalence class that is absolutely continuous on ...


4

Consider the space $H = L^2([0,1],\mu)$, where $\mu$ is the Lebesgue measure. Define $T$ to be the multiplication with the identity function, i.e. $$(Tf)(x) = x\cdot f(x).$$ Since the identity function is bounded, $T$ is bounded ($\lVert T\rVert \leqslant 1$), and since it is real-valued, $T$ is self-adjoint. Clearly $T$ has no eigenvalues, since $$(T - ...


167

For any holomorphic function $G$, we can define a corresponding matrix function $\bar{G}$ via (a formal version of) the Cauchy Integral Formula: We set $$\bar{G}(B) := \frac{1}{2 \pi i} \oint_C G(z) (z I - B)^{-1} dz ,$$ where $C$ is an (arbitrary) anticlockwise curve that encloses the eigenvalues of the (square) matrix $B$. Note that the condition on $C$ ...


32

The gamma function is analytic. Use the power series of it. EDIT: already done: Some properties of Gamma and Beta matrix functions (maybe paywalled).


1

Take $H= \mathscr l^2(\mathbb N)$. Let $S_n$ be the switch $$S_n\left(\sum_k x_k e_k\right):=\sum_{k\neq0,n}x_k e_k+x_0e_n+x_ne_0$$ This is a bounded linear operator and $S_n^2=id$. Let $f(n)$ be a function that rises monotonously and unboundedly with $n$, but that satisfies $\lim_n \frac{f(n)}{\sqrt n}=0$. Let $\pi_k$ be the orthogonal projection onto the ...


1

It occurred to me that your problem has to do with a creation and annihilation operator, according to \begin{eqnarray*} X &=&U+V \\ a^{\ast } &=&V,\;a=U \end{eqnarray*} see below. Let $\mathcal{H}=l^{2}$ \ with elements $u=u_{1},u_{2},\cdots $ and let $K$ be defined by \begin{eqnarray*} \mathcal{D}(K) ...


1

It is $n-1$. Note that due to the structure of the root spaces, you have nontrivial invariant subspaces of any dimension up to $n-1$. In fact $inv(A)$ is itself $n-1$ for any matrix $A$, for the same reason. Changed: replaced "of any (complex) dimension" by "of any dimension" (I was assuming that the OP wanted $A$ to have only real entries).


1

For every normed space $X$, the dual $X^*$ is $1$-complemented in $X^{***}$. Indeed, let $i:X\to X^{**}$ be the canonical embedding; then its adjoint $i^*$ is a projection of norm $1$ of $X^{***}$ to $X^*$. Simply put, it takes a functional $\phi:X^{**}\to \mathbb{C}$ and composes it with $i$. In particular, the above applies to $\mathbb{B}(\mathbb{H})$, ...


2

No for question 1: for example, if $A=0$, then $A+B$ will still have $n-1$ zero eigenvalues. No for question 2, in general, even if the answer to 1 was "yes": note that eigenvalues vary continuously with the entries of the matrix, and so, for almost all values of the entries, the eigenvectors will vary continuously. Added: No for question 3: for example, ...


2

Suppose there exists such a norm $\lVert \cdot \rVert$ on $C(\mathbb{R},\mathbb{R})$. Then for each $a \in \mathbb{R}$, there is a constant $c_a > 0$ such that $$\lvert f(a) \rvert = \lvert T_a(f) \rvert \le c_a \lVert f \rVert$$ for all $f \in C(\mathbb{R},\mathbb{R})$. This is a contradiction because there exists continuous functions $f : \mathbb{R} \to ...


1

I have studied the whole book of Rosenberg and I can asure you that it is a very good choice to start with, I have studied some parts of the books of Gilkey and Lawson-Michelson too and I can asure you they are quite advanced, it would be hard to begin with them. I guess the book of Lawson-Michelson is better for you because it covers the classical proof of ...


1

Consider the definition of $f(U)$ given by $$ f(U)x = \frac{1}{2\pi}\int_{0}^{2\pi}f(e^{i\theta})\left(\sum_{n=-\infty}^{\infty}e^{-in\theta}U^{n}x\right)d\theta $$ If $f(z)=z^{n}$, then $f(U)=U^{n}$ for $n=0,\pm 1,\pm 2,\pm 3,\cdots$. The inner sum is a vector function of $\theta$: $$ S(\theta)x=\sum_{n=-\infty}^{\infty}e^{-in\theta}U^nx $$ The $k$ ...


0

But there is such a norm. Just take the supremum norm. Added: I am assuming that $C(\mathbb R,\mathbb R)$ is the set of bounded continuous functions.


2

Let $A$ be a finite-dimensional Banach algebra. The group $K_0(A)$ only depends on the underlying ring structure of $A$. In what follows, $A$ can be any finite-dimensional algebra over an arbitrary field $k$, well $A\neq 0$. The map $\{\text{finitely generated projective $A$-modules}\}\mapsto \mathbb Z$ given by $P\mapsto\dim_kP$ is well defined (being $A$ ...


0

Using induction, one may see that $(S^*)^nTS^n=T$ for all natural number $n$ (since $S^*S$=idenity). It is routinely to be checked that $wot-\lim (S^*)^nTS^n=0$ which means that $T=0$.


0

This is like Lemma 5.1.4 in Kadison & Ringrose. The key is to use the weak-operator compactness of the unit ball of $\mathcal{B}(\mathbb{H})$, Theorem 5.1.3 (p. 306). You then use the general topological fact that compactness is equivalent to "every net has a convergent subnet." Alas, I'm not sure I can shed much light on the proof of compactness ... ...


3

Well, no answer has been given, so here's one: We know $\{\cos nx:n=0,1,\dots \}\cup \{\sin nx:n=1,2,\dots \}$ is an orthogonal basis of $L^2[-\pi,\pi].$ Suppose $g\in L^2[-\pi,\pi]$ is odd. Then $$\int_{-\pi}^\pi g(x)\cos nx\, dx = 0, n=0,1,\dots $$ Thus such a $g$ can be written uniquely as $$ \tag 1 g(x)=\sum_{n=1}^{\infty} b_n \sin nx,$$ the sum ...


1

Define $T:L^1(\Bbb R)\to L^1(\Bbb R)$ by $Tf = gf$, where $$g(t)=\frac{1}{1+t^2}.$$Functions with compact support are dense in $L^1$, hence $T(L^1)$ is dense in $L^1$. But you can easily find an example showing that $T(L^1)\ne L^1$; hence $T(L^1)$ is not closed in $L^1$.


2

Clearly, $\|A_n\|=\|B_n\|=1$, since $\|A_n\|=\|B_n\|\le 1$ and $\|A_ne_n\|=1$, $\|B_ne_1\|=1$. However, for every $x\in \ell_2$, $$ A_nx,\,B_nx\to 0, $$ which means that $A_n,B_n\to 0$, weakly!


1

If $A$ is an operator on a Hilbert space $\mathcal H$, the norm of $A$ is defined as $$\|A\| := \sup_{\|x\|=1}\|Ax\|. $$ So if $x\in\ell^2$ with $\|x\|=1$, then $$\|A_n x\|=\|x_n e_1\| \leqslant \|x_n\|\|e_1\|\leqslant 1, $$ which implies that $\|A\|\leqslant 1$. Since $A_n e_1 = e_1$ and $\|e_1\|=1$, indeed $\|A_n\|=1$. Computing the norm of $B_n$ follows ...


2

As Gro-Tsen commented, the fact that $\Gamma$ is an isometry implies its image is closed because $\mathfrak{U}$ is a complete metric space (since it is closed in $C(X)$). Any subset of a metric space that is complete as a metric space is automatically closed, since a sequence that converged to a point outside the set would be a Cauchy sequence with no limit ...


1

Suppose that $\langle f(y_\alpha):\alpha\in A\rangle$ does not converge to $f(y)$. Then there is a subnet $\langle y_{\alpha_\beta}:\beta\in B\rangle$ of $\langle y_\alpha:\alpha\in A\rangle$ such that $\{f(y_{\alpha_\beta}):\beta\in B\}$ is bounded away from $f(y)$. Now apply the argument given in the book to the net $\langle y_{\alpha_\beta}:\beta\in ...


1

I believe the answer is yes - let's see if I can prove it. First we need to define "bounded". Recall that $S\subset X$ is said to be bounded if for every open set $V$ containing the origin there exists $c>0$ so that $S\subset cV$. Now given two topological vector spaces $X$ and $Y$ we say that $T:X\to Y$ is bounded if $T(S)$ is bounded in $Y$ for every ...


2

Your proposed domain for the adjoint $X^\star$ appears to me to be correct. As defined, $$ (Xf,g) = \sum_{j=0}^{\infty}(\sqrt{j+1}f_{j+1}+\sqrt{j}f_{j-1})\overline{g_j}. $$ By definition of adjoint, $g\in\mathcal{D}(X^{\star})$ iff there exists $h \in \ell^2$ such that the following holds for all $f \in \mathcal{D}(X)$: $$ ...


1

By definition, unitaries preserve the norm. So $$ \|Cy-CSy\|=\|C(y-Sy)\|=\|y-Sy\|. $$ Thus, $CV=V$.


1

You are correct. We can note that our space is isometric to the subspace of $l^2(\mathbb{N})$ comprised of sequences with finitely many non-zero entries, by the isometry $\sum_{n=0}^N a_n z^n\mapsto (a_0, a_1, \ldots, a_N, 0, 0, \ldots)$. (Note that, indeed, $\left\langle \sum_{n=0}^N a_n z^n, \sum_{n=0}^M b_n z^n\right\rangle = ...


1

I'd say the main argument would be that this formulation is not really advantageous. The thing is that for $\dim X < \infty$, $B[0,1]$ and $S$ are compact and $\left\Vert Tx \right\Vert$ is continuous. Hence, there exists some $x_0 \in S$ such that $$\left\Vert T \right\Vert = \left\Vert Tx_0 \right\Vert$$ making the supremum a maximum. The supremum over ...


0

As MaoWao suggests ; Just take the norm of the integrand. The additional factor r(M)+δr(M)+δ comes from the length of the contour line.


1

a) The first task here is to interpret the question appropriately. I guess it should be read as, "Show that for every partial isometry $V\in B(H)$ on a finite-dimensional Hilbert space $H$, the restriction $V|_{(\ker V)^{\perp}}$ can be extended to a unitary on $H$." (The only extension of $V$ to $H$ is $V$ itself, and it is certainly not true that every ...


0

The derivative of $f$ at the point $\gamma(t)$ is a linear map from $\mathbb{R}^n$ to $\mathbb{R}$. This is a linear approximation of $f$ near that point, which you can visualize as a hyperplane in $\mathbb{R}^{n+1}$ that is tangent to the graph of $f$. In multivariable calculus, the gradient of $f$ is essentially the transpose of this linear map. Read ...


2

$t$ is fixed there, so set $\gamma(t)=b$. Thus he's referring to $f'(b)$, which is a linear map from $\Bbb R^n$ to $\Bbb R$.


1

Since $f : E \to \mathbb R$ and $E$ is open in $\mathbb R^n$, we have for each $y\in E$, $f'(y) \in L(\mathbb R^n, \mathbb R)$. In particular, at $y = \gamma(t)$ we have $f'(\gamma(t)) \in L(\mathbb R^n, \mathbb R)$. You might be thinking of $(f\circ \gamma)' (t)$ instead.



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