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3

No. Counterexample: $$ \pmatrix{0&1\\0&0} $$ is not similar to any symmetric matrix. On the other hand, every diagonalizable matrix with real eigenvalues is similar to a symmetric matrix.


3

First, what do you mean by "the eigenvalues of $A$ plus the eigenvalues of $B$"? What is true (under the condition $AB = BA$) is that each eigenvalue of $A+B$ is the sum of an eigenvalue of $A$ and an eigenvalue of $B$. It is not true in general that each sum of an eigenvalue of $A$ and an eigenvalue of $B$ is an eigenvalue of $A+B$. If $A$ and $B$ ...


2

You have correctly stated that for any symmetric $A$, there exist rank $1$ matrices $v_iv_i^T$ such that $$ A = \sum \lambda _i v_i v_i^T $$ However, it is impossible to select a fixed set $\{v_1v_1^T,\dots,v_nv_n^T\}$ such that every $A$, there exists a choice of $\lambda_i$ such that $A$ has the above form. That is, there is no basis for the set of ...


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I accidentally saw this paper today on the internet concerning the case $n=4$: Jeremy Levick, Rajesh Pereira and David W. Kribs (2015), The four-dimensional Perfect-Mirsky Conjecture, Journal: Proc. Amer. Math. Soc., 143: 1951-1956. Abstract: We verify the Perfect-Mirsky Conjecture on the structure of the set of eigenvalues for all $n\times n$ ...


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The linear subspace of symmetric matrices is actually of dimension $n(n+1)/2$. The linear subspace of diagoanl matrices is of dimension $n$. The similar transformation (spectral decomposition) maps $n(n+1)/2$ space to $n$ space. You have confused 2 different spaces with a single space.


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Even if it might be to late for answering. The following work provides a characterization of the convergence of the spectrum with respect to the Hausdorff metric. The continuity is satisfied whenever the norms of all polynomials (up to degree 2) of the operator behave continuous. One could do much better: In particular quantitative estimates are provided. ...


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The Spectral Mapping Theorem allows you to more easily compute the spectrum of some operators. If you know that you can write an operator $A$ as $A=f(a)$ for $f\in hol(a)$ and $a\in\mathcal{A}$, where you already know the spectrum of $a$, you can compute the spectrum of $A$, since $\sigma(A)=f(\sigma(a))$.


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If $\rho$ is a bounded measurable function on $[a,b]$, then $E(t)=\int_{a}^{t}\rho(u)du$ is a function of bounded variation and, for any continuous function $g$, $$ \int_{a}^{b}g(t)dE(t) = \int_{a}^{b}g(t)\frac{dE}{dt}dt = \int_{a}^{b}g(t)\rho(t)dt. $$ In your case, $$ \|E(t)x\|^{2}=\int_{-\infty}^{t}|x(u)|^{2}du $$ The above easily ...


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Since $\rho(A)<1$, there is an induced norm $N(.)$ s.t. $N(A)<1$. Let $\mu=N(A)$; thus $N(A^k)\leq \mu^k$. Since the norms are equivalent, there is a fixed $C$ s.t. $||.||\leq CN(.)$. Finally $||A^k||\leq CN(A^k)\leq C\mu^k$. EDIT. If $E$ is a Banach space and $A\in L(E)$ is bounded, then one has the same result. According to Gelfand, there is ...


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The spectral radius is the biggest eigenvalue of a matrix. There is a linked norm called the spectral norm, which is infact the square root of the biggest eigenvalue of the matrix $A^*A$. (So not linked to its own spectral radius, but to the spectral radius of $A^*A$) Norms are always convex. Due to triangle inequality and to the homogeneity. Edit: ...



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