33,642 reputation
12041
bio website jungenschaft-hohenstaufen.de
location Berlin, Germany
age 33
visits member for 2 years, 11 months
seen 13 hours ago

13h
comment I see some contradiction in the definition of orthogonal vectors
Yes. For any invertible matrix $A$ on $\mathbb R^n$, $\left<x,y\right> := \sum_{i=1}^n (Ax)_i(Ay)_i$ is an inner product.
14h
comment What is the limit of the $n$-th power of the shift operator
@tomb. They seem to talk about the right shift $S\colon (x_1, x_2, \ldots) \to (0, x_1, x_2, \ldots)$ there. For this operator $S^n \to 0$ (weakly operator), but $S^n x \not\to 0$ for any $x \ne 0$, so $S^n \not\to 0$ (strongly operator). This example also shows (as $S = T^*$) that $T \mapsto T^*$ is not strongly operator continuous.
14h
comment What is the limit of the $n$-th power of the shift operator
In what topology? Strongly (and hence weakly) operator, we have $T^n \to 0$. In norm, as @DavidMitra said, $T^n$ does not converge.
20h
comment Statistics - Normal distribution
@user9636 Please add these answers to the question and add some thoughts, why these ought to be the correct ones ...
20h
comment limiting joint distribution
Let all $X_n$ equal wiki's $X$, all $Y_n$ equal wiki's $Y$. Then $(X,Y)$ (the limit) is not jointly normal.
1d
comment Show that $A \cup B = (A$ \ $B ) \cup (A \cap B) \cup (B$ \ $A)$
As we know (as I wrote) $A -B, A \cap B, B-A \subseteq A \cup B$, we have $$ A \cup B \supseteq (A-B)\cup (A\cap B) \cup (B-A) $$ which is one inclusion of the two we need for the equality you stated (namely "$\supseteq$").
1d
comment Show that $A \cup B = (A$ \ $B ) \cup (A \cap B) \cup (B$ \ $A)$
What is $\neg B$ for a set?
2d
comment Determinant of the linear map given by conjugation.
See also math.stackexchange.com/questions/275925/… where the case of conjugation on the whole of $M_n(K)$ is discussed
2d
comment Transformation matrix of a polynomial
You are welcome ...
2d
comment Convergence of an infinite power
Do you also want $w\ne 0$? Otherwise take $z = 2^{-1}$.
2d
comment Why are those objects initial or final obejcts?
@Bryan261 Exactly. Now try the other ones.
2d
comment Why are those objects initial or final obejcts?
Ex falso quodlibet ... so if $P$ is any property, $\forall x \in \emptyset: P(x)$ is (vacuously) true.
2d
comment Rewriting the matrix equation $AX = YB$ as $Y = CX$?
As an addition note that as $\def\vec{\mathop{\rm vec}}\vec(AX) = (\def\Id{{\rm Id}}\Id \otimes A)\vec X$, $\vec(YB) = (B^t \otimes \Id)\vec Y$, that can help you to find $C$ for the vectorization
Jul
20
comment Differentiation in Banach spaces
In your second line ... how do you rearange the product $(A+h)^n$ ... in general $A$ and $h$ won't commute ..
Jul
17
comment To derive commutativeness of any group from the normality of all its subgroups & some other conditions
The converse is not true. See, e. g. here.
Jul
17
comment Is there a (linear) functional $f\in (C[-1,1])^*$ such that $f$ maps even continuous function to $0$ and odd continuous function to its infinite norm
You call a function $g \in C([0,1])$ even or odd iff ...? If with $h$ also $-h$ is odd, then such an $f$ cannot be linear.
Jul
17
comment Weak-star lower semicontinuity in $L^\infty$
@asda Yes. Otherwise weakly$^*$-convergence does not make sense. In the $L^\infty$-case, $X= L^1$, of course.
Jul
17
comment Looking for a approximation/solution to my mortgage calculator function
@user3843310 What are you going to use to "evaluate"?
Jul
17
comment Dual of $l^\infty$ is not $l^1$
@MartinSleziak Thanks. ${}$
Jul
17
comment How prove $A^2=0$,if $AB-BA=A$
@math110 In characteristic two, $-1 = 1$, so: yes. I do not aim to give a counterexample for other characteristic, as there the above proof thows no such examples exist.