linear algebra

[rare forum]

Can anyone explain to me the difference between bases, kernel and generator? And how do we calculate them?

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[terse wraith]

a basis is a linearly independent spanning set for the subspace, the kernel is the subspace of vectors that map to 0, and generators form bases (ie a basis is a set of generators for the subspace)

[tawdry surge]

to calculate a base you simply prove that the vectors that generate that base are linear independent so that : c1v1+c2v2 = 0 ===> c1,c2== 0

here's an example of how you calculate the ker , just find the condition for the transformation to be 0 , in this case (a,2a,a5)

[terse wraith]

not sure why you pinged me

[rare forum]

Whats a spanning set ?

[terse wraith]

Given a subspace U, a spanning set of U is a set that spans to U

[rare forum]

The kernel isn't just the number of the pivots of a matrix in scale?

[terse wraith]

ie any set of vectors $S$ such that $\text{span}(S)=U$

[solid hamlet]

Omegabet_

[terse wraith]

it is not

given a linear map $T\colon U\to V$, $\ker(T):={u\in U|T(u)=0_V}$

[solid hamlet]

Omegabet_

[rare forum]

So kernel is the given 0 vector

[terse wraith]

no

kernel is the set of vectors in the domain space, that map to 0

[rare forum]

The solution?

[terse wraith]

what?

use full sentences

[rare forum]

Like given a system a kernel is the amount solution to set the system equal to 0

Or to have one only univoke solution

[terse wraith]

the nullspace/kernel is the solutions to the homogenous system Ax=0

[rare forum]

As if the kernel is inferior to the number of coordinates/incognites we would have infinite^n-r solution

Ok then

That's understandable then

[terse wraith]

no clue what that's suppose to mean

"infinite^(n-r)" is complete and utter nonsense

the kernel is everything that maps to 0

nothing more, nothing less.

[rare forum]

Ye I misunderstood the meaning of it

[terse wraith]

yes you did

[rare forum]

We have infinite solution in the power of n-r in which n is the number of Xs to be found and r is the range

Can you explain me the other ones

[terse wraith]

you're just saying words

with no meaning

[rare forum]

I'm not English first language

But it has meaning...

[terse wraith]

that's abundantly clear

it doesnt

you said r is the range, which is a set

n-r therefore doesnt make sense

[rare forum]

Not range

r(A)

Rango

[terse wraith]

rank

[rare forum]

Ye

I meant rank

[terse wraith]

Type what you mean then.

but anyway, "infinite^(n-r)" is still complete and utter nonsense

n also isnt defined

"infinite solution in the power of n-r" is also nonsense

[rare forum]

We have infinite^(n-r) solution to a system in dépendance of how many X are free coordinates

Like under determined systems

[terse wraith]

stop typing

infinite^(n-r)

it is

meaningless

assuming an infinite field like R or C is the scalar field, any solution set of Ax=b is either empty, or infinite

[rare forum]

Theorem of rouche-capelli

Basically

[terse wraith]

sure

That just says the dimension of the affine subspace is n-rk(A)

and characterizes a condition on Ax=b having solutions

[rare forum]

Ye... I was mentioning that one when I quoted infinite^(n-r)

[terse wraith]

ok

and infinite^(n-r) is still nothing relevant to anything

since it quite literally

isnt a thing

[rare forum]

Ye it's not a number but a clarification of how many free coordinates we have

[terse wraith]

it isnt

it is meaningless

[rare forum]

I'm confused...

[terse wraith]

stop using it/trying to justify it actually means something

[rare forum]

P

[terse wraith]

Rouche-Capelli just says Ax=b has a solution iff rk(A)=rk([A|b]), and if there are solutions, the affine subspace of solutions has dimension n-rk(A)

where n is the number of variables

[rare forum]

That's literally from my professors notes

[terse wraith]

note how I never wrote inf^(n-r)

yeah, your prof is braindead if a professional mathematician is writing inf^(n-r)

since Rouche-Capelli doesnt even assume an infinite scalar field

so taking a finite scalar field gives a finite number of solutions trivially

[rare forum]

Ye

But infinite combinations of solution due to under determined variables

[terse wraith]

Assuming $\tilde{x}$ is a solution to $Ax=b$, then ${x|Ax=b}=\tilde{x}+\ker(A):={\tilde{x}+z|z\in\ker(A)}$

[solid hamlet]

Omegabet_

[rare forum]

That's why he used the infinite, even if he also claimed it not to be completely appropriate

[terse wraith]

hence why the affine subspace is dimension n-rk(A), it's size nullity(A)