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In other words, a (real) Banach space $X$ is Hilbert irreducible iff it has no $2$-dimensional subspace isometric to $\mathbb R^2$ with the Euclidean norm. In $M_n(\mathbb R)$, the subspace $Y$ consi …

Even one positive definite operator on an infinite-dimensional Hilbert space need not have any eigenvectors at all: it might have continuous spectrum. The more general statement is the Spectral Theor …

Here's a counterexample (subject perhaps to what you consider "natural"). Take a separable Hilbert space with orthonormal basis $\{u_n : n = 1, 2, \ldots\}$ and the operator $A$ defined by $$ A u_n = …

I think the point is this. It is impossible in general to decide computationally whether two computable real numbers $\alpha$ and $\beta$ are equal. If in fact they are not equal, by computing suff …

Suppose $G$ is abelian. Presumably the measure being used is Haar measure. $T_f$ is unitarily equivalent (via the Fourier transform) to multiplication by $\widehat{f}$ on $L^2(\widehat{G})$. This h …

Consider $H = L^2[0,\infty)$ with $(,) = \langle,\rangle$, $A: H \to H$ the shift operator $A f(t) = f(t+1)$, so that $$ A^* f(t) = \cases{f(t-1) & if $t \ge 1$\cr 0 & otherwise\cr …

Yes, of course. By definition of the adjoint operator, $\{[-A^* y, y]: y \in \mathscr D(A^*)\}$ is the orthogonal complement in $H \oplus H$ of the graph $\{[x, Ax]: x \in \mathscr D(A)\}$ of $A$. …