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I'll assume a complex Hilbert space $H$. If $\{ e_n \}_{n=1}^{\infty}$ is an orthnormal basis of $H$ with $Te_n =\lambda_n e_n$, then there is a constant $M$ such that $|\lambda_n| \le M$ for all $n$ because $T$ must be bounded, which ensures the norm converges of all vector sums in this post, such as $$ T x = \lim_{N} T\sum_{n=1}^{N}\langle x,e_n\...


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$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$An invertible linear operator $f$ on $\Reals^{n}$ maps the unit ball to an ellipsoid whether or not $f$ is symmetric (or even diagonalizable): One strategy is to write the unit ball as the locus of a quadratic inequality and perform a linear change of variables, concluding ...


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Suppose that $T-\lambda I$ is invertible, then it has trivial kernel and is bounded. Particularly you can solve the equation $$(T-\lambda I)x = y$$ for any $y\in\ell^p$. Writing $x = (x_m)$ and $y = (y_m)$, we see that $$ (\alpha_m-\lambda)x_m = y_m,$$ i.e. $$x_m = \frac{1}{\alpha_m-\lambda}y_m.$$ Note that $\lambda=\alpha_m$ is a serious problem here. ...


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The two vertices with acute angles are at distance $\sqrt{10}$ in the first, $\sqrt{2}$ in the second. EDIT: There are many other differences. For example, the longest edge of the first polygon has length $2$, of the second $2 \sqrt{2}$.


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The spectrum of an operator, in this case an unbounded operator, greatly depends on the domain of definition and the surrounding Banach or Hilbert space. For example on the space $C^2[0,1]$ of twice differentiable functions with the usual Banach norm, the Laplacian is a bounded operator, and hence its spectrum is bounded. On the other hand, if we take the $...


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In the finite dimensional case, we get a similar argument to work: the max is continuous (as a function of $V$), and the space of $k$-dimensional vector spaces is compact, so the minimum is reached. In the infinite dimensional case, the main difference is that $\mathcal{G} (k, \infty)$ is no longer compact. It is still a metric space, and the application $...


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Turns out that the specific property does not hold. There can be many counterexamples.


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Since $\Delta$ is formally self-adjoint on $C^\infty(M)$, eigenspaces corresponding to distinct eigenvalues are orthogonal to each other, i.e., for all $j \neq i$, $\mathcal{P}_i(M,g) \subset \mathcal{P}_j(M,g)^\perp \cap C^\infty(M)$. Hence, for all $j \neq i$, since $V_j \subset \mathcal{P}_j(M,g)^\perp \cap C^\infty(M)$, it follows that $\mathcal{P}_i(M,g)...


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For the first part: If $V$ has dimension $k$ then we have $$\max_{\substack{x \in V \\ x \neq 0}} \frac{\langle \,x , Ax \rangle}{\langle \, x, x \rangle} = \max_{\substack{x \in V \\ x \neq 0}} \langle \, \frac{x}{\Vert x \Vert} , A \frac{x}{\Vert x \Vert} \rangle = \max_{\substack{x \in V \\ \Vert x \Vert = 1}} \langle \,x , Ax \rangle.$$ This is a maximum ...


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Yes, you can consider it a standard (but quite tedious) linear algebra exercise for any fixed $n,k$ (in your example, $n=k=3$). More generally, the article you linked provides the following answer on page 3. Although it's not really clear what you mean by "finding a basis". $H^k$ is the orthogonal complement of all polynomials of the form $|x|^2 P(x)$, ...


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It is clear that $0$ is in the spectrum of $K$, since $K$ is compact. Let us address if it is an eigenvalue: if $$ Kf=0,$$ this means we have $$\tag{1} 0=t\int_t^1f(s)\,ds+\int_0^ts\,f(s)\,ds. $$ Differentiating (via Lebesgue's Differentiation Theorem), $$\tag{2} 0=\int_t^1 f(s)\,ds-tf(t)+tf(t)=\int_t^1f(s)\,ds,\ \ \ \text{a.e.}. $$ Then, for any $v,t\in[0,...



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