# Tag Info

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 $$T x = \lim_{N} T\sum_{n=1}^{N}\langle x,e_n\... 0 \newcommand{\Reals}{\mathbf{R}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}An invertible linear operator f on \Reals^{n} maps the unit ball to an ellipsoid whether or not f is symmetric (or even diagonalizable): One strategy is to write the unit ball as the locus of a quadratic inequality and perform a linear change of variables, concluding ... 1 Suppose that T-\lambda I is invertible, then it has trivial kernel and is bounded. Particularly you can solve the equation$$(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.$$Note that \lambda=\alpha_m is a serious problem here. ... 0 The two vertices with acute angles are at distance \sqrt{10} in the first, \sqrt{2} in the second. EDIT: There are many other differences. For example, the longest edge of the first polygon has length 2, of the second 2 \sqrt{2}. 2 The spectrum of an operator, in this case an unbounded operator, greatly depends on the domain of definition and the surrounding Banach or Hilbert space. For example on the space C^2[0,1] of twice differentiable functions with the usual Banach norm, the Laplacian is a bounded operator, and hence its spectrum is bounded. On the other hand, if we take the ... 2 In the finite dimensional case, we get a similar argument to work: the max is continuous (as a function of V), and the space of k-dimensional vector spaces is compact, so the minimum is reached. In the infinite dimensional case, the main difference is that \mathcal{G} (k, \infty) is no longer compact. It is still a metric space, and the application ... 0 Turns out that the specific property does not hold. There can be many counterexamples. 0 Since \Delta is formally self-adjoint on C^\infty(M), eigenspaces corresponding to distinct eigenvalues are orthogonal to each other, i.e., for all j \neq i, \mathcal{P}_i(M,g) \subset \mathcal{P}_j(M,g)^\perp \cap C^\infty(M). Hence, for all j \neq i, since V_j \subset \mathcal{P}_j(M,g)^\perp \cap C^\infty(M), it follows that \mathcal{P}_i(M,g)... 2 For the first part: If V has dimension k then we have$$\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.$$This is a maximum ... 0 Yes, you can consider it a standard (but quite tedious) linear algebra exercise for any fixed n,k (in your example, n=k=3). More generally, the article you linked provides the following answer on page 3. Although it's not really clear what you mean by "finding a basis". H^k is the orthogonal complement of all polynomials of the form |x|^2 P(x), ... 1 It is clear that 0 is in the spectrum of K, since K is compact. Let us address if it is an eigenvalue: if$$ 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,...