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For example, if $$ \eqalign{A &= \pmatrix{1 & t\cr 1 & 1\cr},\ v = \pmatrix{\sqrt{t}\cr 1},\ u =\pmatrix{1\cr \sqrt{t}},\cr \frac{(u^T v)^2}{(u^T u)(v^T v)} &= \frac{4t}{(1+t)^2} \to 0 \ \text{as}\ t …

Upper semicontinuity of the spectrum is the following statement: if $U$ is a neighbourhood of $\text{Sp}(x)$, then there is a neighbourhood $V$ of $x$ such that $\text{Sp}(y) \subset U$ for all $y \in …

In any infinite-dimensional Hilbert space, the only such operators $V$ that work for all $T$ are scalar multiples of the identity. Suppose $V$ is not a scalar multiple of a unitary. Then there are l …

If $\lambda_\max$ is the greatest eigenvalue of $T$, the least eigenvalue of $A$ is between $-\lambda_\max$ and $\max(b_1, b_n) - \lambda_\max$.

Take $M$ with all $M_{ij} = 1$, $M_1 = I$, $M_2 = M-I$. Since the eigenvalues of $M$ are $0$ and $n$, $M_1$ and $M_2$ both have rank $n$ if $n > 1$. In the other direction, if $M_1 = M_2 = M/2$, $\te …

No, and it's not true on Hilbert space either. For example, on $\mathbb C^2$ or $\mathbb R^2$ try $$ A = \pmatrix{0 & 0\cr 1 & 0\cr},\ x = \pmatrix{1\cr -1\cr},\ \lambda = 1$$ The spectrum is $\{0\}$, …

What I would consider the obvious proof uses only the Banach space structure. If $\lambda$ is in the resolvent set, the graph $G(T)$ of $T$ maps in an obvious way to the graph of $(T-\lambda I)^{-1}$ …

By orthonormal, I suppose you mean $\text{tr}(C_i C_j) = 0$ for $i \ne j$, $\text{tr}(C_i^2) = 1$? The estimate you gave is tight, in the sense that it is an equality if, for example, all $w_i = \ep …

That closed form is $c^{1/2} (2n-1)$ when $c$ is a positive real. These eigenvalues must be analytic as functions of $c$ as long as they don't collide or go off to $\infty$, and they don't as long a …

No. For example, the spectral radii of $$ A = \pmatrix{0 & 1 & 1\cr 1 & 0 & 1\cr 0 & 0 & 0\cr},\ A' = \pmatrix{0 & 0 & 3\cr 1 & 0 & 1\cr 0 & 0 & 0\cr}$$ are $1$ and $0$ respectively.

By "$I$ is well-defined", I presume you mean you have a continuous injection $\iota$ of $X$ into $Y$. If $X$ and $Y$ are not isomorphic it will not be a bijection, because there is no continuous line …

The trace is the sum of the diagonal elements, so (when $z$ is real) it's always true for at least one $i$, and the only way it can be true for all of them is that all diagonal elements are equal. F …

If $p(z) = a \prod_{j=1}^d (z - r_j)$ and $q(z) = b \prod_{j=1}^e (z - s_j)$, $r_1 \le r_2 \le \ldots \le r_d$, $s_1 \le s_2 \le \ldots \le s_e$, then let $$ f(z) = \dfrac{p'}{p} - \dfrac{q'}{q} = \su …

Let $f$ be your function that is $0$ in a disk $D_0$ around $0$ and $1$ in a disk $D_1$ around $1$. There is (by Runge) a sequence of polynomials $g_n$ such that $g_n \to f$ uniformly on $D_0 \cup D_ …

If we take $X = E_{1,i}$ (the matrix with $X_{1,i} = 1$ and all other entries $0$) and $Y = E_{j,1}$, $XAY$ has $(1,1)$ entry $A_{ij}$ and all others $0$, and its spectral radius is $A_{ij}$. So $\rh …