## Linear algebra – Scope of the null space of A

I apologize for the brief nature of this question, but I do not think it has been clarified in a previous post on this topic – Find a Covering Set for Null Space.

When we say that the null space of the matrix $$A$$ is equal to the duration of a set of vectors:

$$N (A) = span ( v with v_1, vec v_2,$$ $$… vec v_n)$$

Are we actually saying that:

$$A ( vec v_1) + A ( vec v_2) +$$ $$… A ( vec v_n) = vec 0$$

## abstract algebra – Let $phi: mathbb {Z} _7 times mathbb {Z} _7$ be a homomorphism such that $phi ^ 5 = text {id}$. Show that $phi$ is the identity.

Let $$phi: mathbb {Z} _7 times mathbb {Z} _7 to mathbb {Z} _7 times mathbb {Z} _7$$ to be a homomorphism such as $$phi ^ 5 = text {id}$$. CA watch $$phi$$ is the identity.

My attempt: Since $$ker ( phi) subset ker ( phi ^ 2) subset cdots subset ker ( phi ^ 5) = 0$$it follows that $$phi$$ is injective. Since $$| mathbb {Z} times mathbb {Z} |$$ is finished, $$phi$$ is automatically an isomorphism.

then $$phi$$ should map a subgroup to a subgroup. Consider $$phi ( mathbb {Z} _7 times 0)$$, so he should be associated with $$mathbb {Z} _7 times 0$$ or $$0 times mathbb {Z} _7$$. Without loss of generality, we consider the first case. In particular, $$phi ( 1.0)$$ should be a generator for $$mathbb {Z} _7 times 0$$. assume $$phi (1,0) = (a, 0)$$. then $$(1,0) = phi ^ 5 (1,0) = phi ^ 4 (a, 0) = phi ^ 3 (a ^ 2,0) = cdots = (a ^ 5,0)$$. So, $$a ^ 5 = 7k + 1$$ for some people $$k in mathbb {Z}$$. By testing $$a in mathbb {Z} _7$$, we see that the only possibility is that $$a = 1$$. The discussion for $$phi (0,1)$$ Is similar.

Could someone help me take a look at my attempt? Is my approach reasonable and is there a better way to solve this problem?

## Agalgebraic geometry – Which endomorphisms of Tate's algebra are "algebraic"?

For an abelian variety $$A$$ on a field $$k$$ with different characteristic of $$ell$$ and the Galois group $$G = Gal ( overline k / k)$$there is always an injective map of the form:
$$mathbb Q_ ell otimes End_k (A) to End_G (V_ ell (A))$$
or $$V_ ell$$ is the rational module of Tate. In specific cases, we know that the map is a bijection ($$k$$ = finite fields, numeric fields).

Suppose now that we take the direct limit on both sides to increase $$k$$. Then the left side is simply $$mathbb Q_ ell otimes End _ { overline k} (A)$$ and the map is in $$End (V_ ell (A))$$ but it is (almost?) never a bijection.

For example, for non-super singular curves or in characteristic curves $$0$$ and $$A$$ elliptic curve, the right side is one or two dimensions, while the right side is four-dimensional.

Can we classify or characterize the endomorphisms of $$V_ ell (A)$$ which are "algebraic", ie come from the map above for some $$k$$?

## reference query – These two natural structures $A_ infty$ -structures on the realization of an isomorphic cosimplective commutative algebra?

Given a cosimplicial commutative algebra $$A bullet$$ on a characteristic zero field, there are two ways to produce a $$A_ infty$$-structure on its realization $$| A ^ bullet | : = int ^ Delta C ^ * ( Delta ^ bullet) otimes A ^ bullet$$, or $$C ^ * (-)$$ is the simplicial complex cochain, that is to say $$| A ^ bullet |$$ is the cochain complex with $$| A ^ bullet | ^ n = A ^ n$$ and differential the alternate sum of coface cards:

• The Alexander-Whitney map equips the realization functor with a lax monoidal structure, so that it sends the monoid $$A bullet$$ in simplified groups to a monoid in cochain complexes, that is to say a differential graded algebra. Explicitly, the product of $$a_p in A p$$ and $$a_q in A ^ q$$ is (to sign) $$f_ {p, q} (a_p) b_ {p, q} (a_q)$$, or $$f_ {p, q}: A ^ p to A ^ {p + q}$$ and $$b_ {p, q}: A ^ q to A ^ {p + q}$$ are the cofaces "front and back". Note that this does not use the commutativity of $$A bullet$$, and the product is not commutative in general. Call it (dga, and so in particular) $$A_ infty$$-algebra $$| A ^ bullet | _ {CG}$$.
• The coend $$Omega ^ * (A ^ bullet): = int ^ Delta Omega_P ^ * ( Delta ^ bullet) otimes A ^ bullet$$, or $$Omega ^ * _ P ( Delta ^ n) = k (t_0, points, t_n, mathrm dt_0, points, mathrm dt_n) / (t_0 + points + t_n-1, mathrm dt_0 + points + mathrm dt_n)$$ carries a natural (commutative) dga structure. As explained by Cheng and Getzler, there is a simplicial retraction $$Omega_P ^ * ( Delta ^ bullet) rightleftarrows C ^ * ( Delta ^ bullet)$$, giving rise to a retraction $$Omega ^ * (A ^ bullet) rightleftarrows | A ^ bullet |$$ along which this structure can be transferred to a $$A_ infty$$-structure on $$| A_ bullet |$$ (in fact, even a $$C_ infty$$-structure). The operations are given by sums on trees. Call it $$A_ infty$$-algebra $$| A ^ bullet | _ {CG}$$.

Obviously, these two structures are very different: for example, the $$2$$The secondary functioning of the first is associative, but not commutative, while that of the second is commutative, but not associated.

There is a special case where I know that these $$A_ infty$$ the algebras are equivalent: Si $$M$$ is a smooth variety and $$A ^ bullet = operatorname {Sing} ^ bullet (M)$$ is the commutative algebra cosimplicial functions on smooth simplices, there is a zigzag
$$| A ^ bullet | _ {CG} rightarrow Omega ^ * (A ^ bullet) rightarrow int ^ Delta Omega ^ * ( Delta ^ bullet) otimes A ^ bullet leftarrow Omega ^ * (M) rightarrow | A ^ bullet | _ {AW}$$
where are the cards, in the order:

• the canonical $$A_ infty$$-morphism produced by the transfer of homotopy
• the inclusion (simplex) of polynomial forms in smooth forms
• the map $$omega mapsto ( sigma mapsto sigma ^ * omega)$$
• an explicit $$A_ infty$$-isomorphism obtained by Chen's iterated integrals, see for example here.

Is there a natural $$A_ infty$$-isomorphism between $$| A ^ bullet | _ {AW}$$ and $$| A ^ bullet | _ {CG}$$? Can it be reasonably explicit? If so, does it extend the construction above to $$A ^ bullet = operatorname {Sing} ^ bullet (M)$$, or is there at least one natural (and explicit?) $$A_ infty$$-Homotopy between them?

## Operator Theory – Meaning of Affiliate to a von Neumann Algebra

Thank you for contributing an answer to MathOverflow!

• Please make sure to respond to the question. Provide details and share your research!

But to avoid

• Make statements based on the opinion; save them with references or personal experience.

Use MathJax to format equations. MathJax reference.

## Abstract algebra – How to apply the theory of "homology" to the Goldbach conjecture using the result of Helfgott (ternary Goldbach true)?

The result of Helfgott: every odd integer $$n geq 7$$ is the sum of $$3$$ (odd) bonuses. I saw somewhere in the newspaper the mention of odd prime numbers. It means that every single integer $$n geq 10$$ is the sum of 4 odd prime numbers (you just added a $$3$$ to each odd number).

Let's first assume something bold, to simplify the discussion, that for each integer $$x geq 13$$, the even number $$2x$$ can be expressed as the sum of $$4$$ distinct prime numbers: $$2x = p + q + r + s$$. To handle all cases, I'm sure it's possible, because it would take either an enumeration of the small set of cases, or a combinatorial argument. Reason is $$x geq 13$$ is because: $$3 + 5 + 7 + 11 = 26 = 2 cdot 13$$ is the smallest even number which is the sum of $$4$$ distinct odd primes.

If you are not familiar with the basics of simplicial homology but you know what the free $$Bbb {Z}$$-module on a set of symbols is, then google "introduction to simplicial homology". You basically need limit cards: $$partial_n: C_n to C_ {n-1}$$ who take each element of the module $$C_n$$ at its "limit" in $$C_ {n-1}$$, its border being represented as a formal element in the free module, whatever the border map, it must be a $$Bbb {Z}$$homomorphism -module and satisfy also: $$partial_ {n-1} circ partial_n = 0$$.

So, according to Helfgott's result (and the net distinction hypothesis), each integer $$x geq 13$$ is such that the even number $$2x = p + q + r + s$$, or $${p, q, r, s }$$ are distinct odd prime numbers. As a result, there is $${4 choose 2} = 6$$ ways of forming even smaller numbers (by adding two of the prime numbers involved in the sum of $$x$$) $$2x_i, i in {1, dots, 6 }$$ such as $$2x_i + 2x_j = 2x$$ or equivalent $$x_i + x_j = x$$. Now, there is another problem with duplicates as $$x_i = x_j$$ is a possibility. But for now, suppose that $$x_i neq x_j$$, for reasons of discussion. So, to recap, everyone $$x_i = u + v$$ or $$u, v in {p, q, r, s }$$.

So, if we associate each $$x_i$$ with the official symbol $$bar {x_i}$$ which might not be an integer (for the moment) then we will call $$bar {x_i}$$ the faces of the simplex $$bar {x}$$. But $$bar {x}$$ could also be a simplex face of another integer! It is therefore possible to obtain a long-chain complex eventually.

So we have at least $$Delta = { bar {x}, bar {x_i}: i in {1, points, 6 } }$$ like a complex. This does not necessarily correspond to an abstract simplicial complex in which $$forall sigma in Delta,$$ if $$tau subset sigma$$then $$tau in Delta$$. Since for one, we have not defined the $$bar {x_i}$$ to be sets – these are just formal symbols.

Another approach would be?

remember that $$sum limits_ {i = 1} ^ 6 alpha_i x_i = x$$ or $$sum limits_ {i = 1} ^ 6 alpha_i = 2, alpha_i in {0, 1$$. It's a bit like a "convex combination". Anway, as you can see, I took elements of simplicial homology and tried to adapt it. Do you have any ideas on how to proceed?

## group theory gr. – What are the compacts Aut (A) of an algebra A (G), G finite, which contains the identity?

If we have an algebra A on a finite group G, then if G is non-abelian, we can have a non-trivial set of compact automorphisms of A that map the elements of G to an isomorphic set to G. it only has to contain the identity, but I have not checked that.

For example, the field algebra $$C$$ more than $$S_3$$ has such a set of automorphisms that form an isomorphic group to $$SO (3)$$. When G is dihedral of size $$2p$$, the A (G) have compact automorphisms meeting the above requirement with at most $$3 (p-1) / 2$$ generators is $$p$$ is strange, or $$3 (p-2) / 2$$ if $$p$$ is same.

Has anyone classified this?

Thank you very much to Nik Weaver for explaining the language with which to ask the question.

## algebra precalculus – Real Roots of $1+ frac {x} {1!} + frac {x ^ 2} {2!} + frac {x ^ 3} {3!} + cdots + frac {x ^ n} {n!}$

Let $$Q_n (x)$$ to be the degree $$n$$ polynomial
$$1+ frac {x} {1!} + Frac {x ^ 2} {2!} + Frac {x ^ 3} {3!} + Cots + frac {x ^ n} {n !}$$
How many real roots the equation $$Q_n (x) = 0$$ to have?

My attempt:

It's obvious that $$Q_n (x)$$ will have all its actual roots in the negative part of the actual line if there are any. In addition, we notice that if $$n$$ is odd, then there is at least one real root by the complex conjugate root theorem. So I suppose there is exactly one root for $$n$$ strange and there is no root for $$n$$ even.

However, I do not know how to analyze $$Q_n (x)$$. All I can do is take derivatives, which does not provide more useful information. Any clue is appreciated! Thank you.

## linear algebra – Retrieves the translational component of a particular matrix multiplication

Let's say that I have a rotation matrix R 4x and a 4×4 translation matrix T. If I multiply the matrices in this order T * R, the translation component of T will not be affected. But if you multiply them in the R * T order, it will be affected by R.'s rotation values.

How can I describe the position vector of the resulting matrix from the multiplication R * T, in terms of the rotational components of R and the translational component of the matrix T? Can I describe it as the scalar product of the respective orientation component of the matrix R and the translation component of the matrix T?

## abstract algebra – $I$ is a free R-module if and only if $I = Ra$

Problem: Let $$I$$ to be a non-zero ideal of a commutative ring $$R$$ with identity. Prove it
$$I$$ is a free R-module if and only if $$I = Ra$$ for some people $$a in R$$ it's not a zero
divider.

My idea is:
($$Rightarrow$$) Let $$I$$ to be generated by $${a_1, … a_n } (n> 1)$$, then I have $$a_1x_1 + … + a_nx_n = 0$$with $$x_1 = a_2, x_2 = -a_1, …$$($$n$$ is same); $$x_1 = a_2, x_2 = -a_1-a_3, x_3 = a_2, x_4 = a_5, x_5 = -a_4, …$$ ($$n$$ is odd).
So, $$a_1 = … = a_n = 0$$. Contradiction. So, $$n = 1$$.

Is it good? Help me! Thank you!