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Jun
1
comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$
@tomasz Great, thanks for your comment, I misinterpreted the question.
Jun
1
comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$
I don't know the answer off the top of my head and right now I don't have time to think. I'll get back to you.
Jun
1
comment Proof that if f is function, continuous on an interval I then f(I) is also an interval
Just a minor thing: If you have $f(\alpha)<\lambda< f(\beta)$ you can't have $\lambda=f(\alpha)=f(\beta)$. You'd have to slightly rephrase this. And I'd change $x_\lambda \in I$ into $x_\lambda \in (a,b)$ just to be more precise. But it's not necessary. Otherwise this looks good to me!
Jun
1
comment Proof that if f is function, continuous on an interval I then f(I) is also an interval
@Bardo That's great. If you like you can post it then people on this site can check it for you.
Jun
1
comment Necessary conditions for $A=K+\operatorname{Ker}(\phi)$
One sufficient condition would be the conditions necessary to apply the splitting lemma. The splitting lemma tells you that if you have a short exact sequence: $$ 0 \to \mathrm{Ker}(\phi) \xrightarrow{i} A \xrightarrow{\phi} K \to 0$$ then $A$ is isomorphic to the direct sum of $K$ and $\mathrm{Ker}(\phi)$ if there exists an algebra homomorphism $t:K \to A$ such that $\phi \circ t$ is the identity on $K$.
Jun
1
comment Proof that if f is function, continuous on an interval I then f(I) is also an interval
@Bardo No, I agree with you, as stated it doesn't make much sense. But it's clear to me what it's trying to do: it wants to show that $f(I)$ is an interval by applying the intermediate value theorem. It completely fails to mention what lambda is and that is must be between $\alpha$ and $\beta$. On top of that, the English is broken. Maybe it would be helpful to you if (in addition) you could get yourself a better book from the library.
Jun
1
comment Please verify this definition of locally convex vector space
@luka5z If you found this answer helpful then you might consider upvoting it (in addition to accepting).
Jun
1
comment Please verify this definition of locally convex vector space
I think so. ${}$
Jun
1
comment Please verify this definition of locally convex vector space
No, I think it's because a norm is a seminorm. Then the family consists of just one thing: the norm. As for your definition: No, I think the definition states that the topology on the space is induced by the separating family of seminorms. But maybe that's what you mean?
Jun
1
comment Please verify this definition of locally convex vector space
Sorry, I misread your question. Note that every normed space is locally convex.
Jun
1
comment Please verify this definition of locally convex vector space
Can you include the definition of locally convex vector space in your question please?
Jun
1
comment $V$ is open , then $V=\{x\in \mathbb R:f(x)>0\}$ for some continuous function $f$
@Etienne Too bad : ) Thank you for your comment!
May
31
comment $V$ is open , then $V=\{x\in \mathbb R:f(x)>0\}$ for some continuous function $f$
@Etienne Would it also work to take the convolution of the characteristic function of $V$ with a bump function?
May
31
comment Compact Operator <=> Separable Range
@Freeze_S Compact operators have much nicer properties than arbitrary operators. You can think of them as behaving a bit like operators on finite dimensional vector spaces.
May
31
comment Compact Operator <=> Separable Range
It's true that compact implies separable range. The other direction I suspect does not hold. Why not try to find a counterexample?
May
28
comment How to write this as one matrix?
What is the notation $(\cdot, \cdot)$?
May
28
comment Show that $Z_i\sim N(\mu_i,V_{ii})$
Why the downvote?
May
28
comment Show that $E(Z)=\mu, Cov(Z)=V$
Why the downvote?
May
27
comment Prove or disprove: If L : V → U and M : U → V are linear mappings such that (M ◦ L)(x) = x for all x ∈ V, then M is onto.
I think you're absolutely right, it's obvious and there is nothing to prove. Given any $x$ in $V$, $Lx$ maps to it.
May
27
comment Math Major: How to read textbooks in better style or method ? And how to select best books?
This is the second time that I felt frustrated that my reading is far too slow and I read your answer and then didn't feel so bad anymore. Wonderful.