Comment by Martin Argerami on If $P$ is a selfadjoint operator on a Hilbert...
Glad I could help. I have included yet another path.
View ArticleComment by Martin Argerami on Conditional expectation on Cross product von...
It's pretty straightforward. I've edited the answer.
View ArticleComment by Martin Argerami on A bounded linear operator from $X ^*\rightarrow...
Now I see what you mean. I'll try to find time to think about this later today.
View ArticleComment by Martin Argerami on Necessity of parallelogram law for unique...
That's great. It will be good to have criticism :)
View ArticleComment by Martin Argerami on Oddity in definition(s) of quasi compact operator
I'm not familiar with this topic, but definition 2 looks really weird to me. You have that $I+2P$, with $P$ rank-one, is quasicompact. But $I+\frac12\,P$, which is again a compact perturbation of the...
View ArticleComment by Martin Argerami on Is this assumption correct in proof of...
You already said it in your comment, so I'm not exactly sure what the question is. I have added some detail to the answer to see if it helps you.
View ArticleComment by Martin Argerami on Why do elements of the Gel'fand spectrum map...
@FelipeDilho: if you mean the Spectral Theorem, then no. This result holds in any unital Banach algebra. And you have to include zero if the Banach algebra is non-unital.
View ArticleComment by Martin Argerami on Why do elements of the Gel'fand spectrum map...
The way I know how to prove this stuff, yes, this is circular. Because the proof that $C^*(A)$ is isomorphic to $C(\sigma(A))$ passes from first using the Gelfand transform to show that $C^*(A)\simeq...
View ArticleComment by Martin Argerami on help showing a property for a weak operator...
$|T|=(T^*T)^{1/2}$, which is a norm limit of polynomials in $T^*T $.
View ArticleComment by Martin Argerami on Injective $\mathbb{C}$-vector space...
Glad I could help. The essence of the problem you had is to distinguish between the complement as a set, and the (non-unique) linear complement.
View ArticleComment by Martin Argerami on Why am I getting a finite integral for infinite...
@JonathanZ: +1, but "principal".
View ArticleComment by Martin Argerami on how to calculate $e^{tA}$ with $tr(A)=0$ and...
You should try it! :D
View ArticleAnswer by Martin Argerami for Matrix norm induced by a positive-definite Matrix
By definition,\begin{align}\|C\|_A^2&=\max\{\|Cx\|_A^2:\ \|x\|_A=1\}\\[0.2cm]&=\max\{(Cx,Cx)_A:\ \|x\|_A=1\}\\[0.2cm]&=\max\{x^TC^TACx:\ x^TAx=1\}\\[0.2cm]\end{align}The equality $x^TAx=1$...
View ArticleAnswer by Martin Argerami for Image of single GNS representation and kernel...
I think the phrase$\pi(A)$ is "an image of the part of $A$ that $\rho$ can see?"is wrong. It implies that if $\rho$ doesn't see some part of $A$, the same is true for $\pi$, and that's not always the...
View ArticleAnswer by Martin Argerami for An alternate proof of Fuglede's theorem
(several years later, a comment made this answer resurface so I'm updating it so that it keeps up with the comments. I still don't know how to finish the proof through the outline suggested by...
View ArticleAnswer by Martin Argerami for Question about Proof of BDF Theorem in Analytic...
The information is there in Chapters 2 and 3, but not in a way that one immediately sees the assertion as phrased in the paragraph you quote. It might be obvious to an expert (not my case), but it...
View ArticleAnswer by Martin Argerami for Why $\lim\limits_{n\to...
You can see the same problem with something like $n^{1/\log n}$. If you think "of $n$ first", then the limit would be infinite. If you think "of $1/\log n$ first", the limit would be $1$. But it's...
View ArticleAnswer by Martin Argerami for What is the preferred terminology for...
If the range of $T$ is closed, then so is the range of $T^*$. In that case, the orthogonal of the kernel of $T$ is simply the range of $T^*$.
View ArticleAnswer by Martin Argerami for Examples of an inner product on $\mathscr{l}^1$
The obvious $\ell^2$ product$$x\cdot y=\sum_nx_ny_n$$works because every element in $\ell^1$ is in $\ell^2$. Of course this gives you the $2$-norm, and $\ell^1$ is not complete with respect to this...
View ArticleAnswer by Martin Argerami for $\int_0^\lambda z de_z+\lambda p$ is a positive...
Fix $u\in M'$. Let $0=a_0<a_1<\ldots<a_m=\lambda$ be a partition....
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