untchick04tm

2022-01-04

Prove, that the vector Space Hat (n; F) with the multipliсation

$A\cdot B{\textstyle \phantom{\rule{0.222em}{0ex}}}=AB-BA$
is a F-algebra (algebra over a field F) is such an algbera associative, commutative, untiary?

Joseph Lewis

Beginner2022-01-05Added 43 answers

To prove that the vrctor space Hat(n,f) := {set of $n\times n$ matrices over F} is a F-algebra

Note that Hat(n, F) is said to be F-algebra if for any elements x, y, z$\in$ Hat(n, F) and all elements a, b $\in$ F.

Right distribulity, left distribulity and compatibility with scalars followed.

Note that

1)$(x+y)\cdot z=(x+y)z-z(x+y)$

$=xz+yz-zx-zy$

$=(xz-zx)+(yz-zy)$

$=x\cdot z+y\cdot z$

(Right distribulity)

2)$z\cdot (x+y)$

$=z(x+y)-(x+y)z$

$=(zx-xz)+(zy-yz)$

$=z\cdot x+z\cdot y$

(Left distribulity)

3)$\left(ax\right)\cdot \left(by\right)=axby-byax$

$=ab(xy-yx)$

$=ab(x\cdot y)$

(Compatibility with scalars) So, Hat(n,f) is an F-algebra for x, y, z$\in$ Hat(n, F) and a, b $\in$ F

Note that Hat(n, F) is said to be F-algebra if for any elements x, y, z

Right distribulity, left distribulity and compatibility with scalars followed.

Note that

1)

(Right distribulity)

2)

(Left distribulity)

3)

(Compatibility with scalars) So, Hat(n,f) is an F-algebra for x, y, z

Ben Owens

Beginner2022-01-06Added 27 answers

That is not full answer, here is full:

Note that,

and

We can check

So (Not associative)

Comm? Note that

Unitary? Mean it sholud have identity element operation *

Let T be identity element

Then

and

So there are no identity elements for all x

Not Unitary

The permutations and combinations of $abcd$ taken $3$ at a time are respectively.

Show that the image of ${\mathbb{P}}^{n}\times {\mathbb{P}}^{m}$ under the Segre embedding $\psi $ is actually irreducible.

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If $B$ and $C$ are $A$-algebras with ring morphisms $f:A\to B$ and $g:A\to C$, and $D=B{\otimes}_{A}C$ is an $A$-algebra with morphism $a\mapsto f(a)\otimes g(a)$, then $uf=vg$, where $u:B\to D$ is $u(b)=b\otimes 1$.The map $v:C\to D$ is not defined in the text, but my guess is it's $v(c)=1\otimes c$.I don't understand why the diagram is commutative though. That would imply $f(a)\otimes 1=1\otimes g(a)$ for all $a\in A$. Is that true, or is v something else?Added: On second thought, does this follow since $f(a)\otimes 1=a\cdot (1\otimes 1)$ and $1\otimes g(a)=a\cdot (1\otimes 1)$ where ⋅ is the $A$-module structure on $D$?

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$a\vee b=b\vee a$

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In any case, whether it concerns only commutative Banach unitary algebras or commutative algebras as defined above, I think we must intend contained as properly contained. Am I right?Let $H$ be a Hilbert space and $\mathcal{A}$ a commutative norm-closed unital $\ast $-subalgebra of $\mathcal{B}(H)$. Let $\mathcal{M}$ be the weak operator closure of $\mathcal{A}$.

Question: For given a projection $P\in \mathcal{M}$, is the following true?

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and

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