/sci/ - Science & Math » Thread #11857551

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Anonymous Wed 01 Jul 18:51:13 2020 No.11857551 View Reply Original Report

Quoted By: >>11857655

have you studied linear algebra well, anon?
you can prove this, right?

Anonymous

Anonymous Wed 01 Jul 2020 19:12:49 No.11857624 Report

Quoted By:

write v = v + A(v) - A(v)
(A - I)(v + A(v)) = 0
(A + I)(v - A(v)) = 0
So v belongs to ker(A+I)+k(A-I)
if
(A - I)(v) = 0
(A + I)(v) = 0
then A(v) = v = -v so v = 0

this isnt true in characteristic 2 btw
next time post in /sqt/

Anonymous

Anonymous Wed 01 Jul 2020 19:24:28 No.11857655 Report

Quoted By:

>>11857551
Let $x \in V$ then
$x = \frac{1}{2} (A - I)Ax + \frac{1}{2} (A + I) Ax$

$\frac{1}{2} (A - I)Ax \in \ker(A + I)$ and $\frac{1}{2} (A + I) Ax \in \ker(A - I)$

Anonymous

Anonymous Wed 01 Jul 2020 19:29:48 No.11857675 Report

Quoted By:

ker(A+I) are those vectors u such that Au = -u, similarly ker(A-I) are vectors such that Au = +u

For any vector w in V you can write it as
$w = \frac{1}{2}(w+Aw)+\frac{1}{2}(w-Aw)$
The two terms belong to the two kernels so this is exactly the decomposition asked for

Anonymous

Anonymous Wed 01 Jul 2020 19:38:51 No.11857703 Report

Quoted By: >>11857720

Can it be proven in vector space over a field of characteristic 2?

Anonymous

Anonymous Wed 01 Jul 2020 19:40:06 No.11857712 Report

Quoted By:

Anonymous

Anonymous Wed 01 Jul 2020 19:41:24 No.11857720 Report

Quoted By:

>>11857703
No, because there A+I = A-I

Anonymous

Anonymous Wed 01 Jul 2020 19:42:30 No.11857726 Report

Quoted By: >>11857733

what is the cross with a circle? is it a tensor product?

Anonymous

Anonymous Wed 01 Jul 2020 19:43:42 No.11857733 Report

Quoted By: >>11857739

>>11857726
it's a direct sum

Anonymous

Anonymous Wed 01 Jul 2020 19:44:41 No.11857739 Report

Quoted By:

>>11857733

i c , no wonder i failed abstract algebra lol

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