.

[wooden frigate]

What is the first criterion for subspaces?

[hazy dagger]

The additive identity: 0 ∈ U

[wooden frigate]

yes

can that be the case in V2

[hazy dagger]

I don't think so

[wooden frigate]

yes

because?

[hazy dagger]

I don't know how I would have to prove this

[wooden frigate]

aT * x is basically another way to describe the dot product

if x is the nullvector

can there ever come out something that is not 0?

[hazy dagger]

No, it can't

[wooden frigate]

yes that is it

so you see

the nullvector doesnt satisfy the condition

a^T * x = 1

[hazy dagger]

Okay, I get it, but I'd like to show it also for closed under addition and closed under scalar multiplication

[wooden frigate]

that is enough

[wooden frigate]

that is probably for V1 which is probably a subspace

[hazy dagger]

This might be true, but I'm supposed to show it

[wooden frigate]

it doesnt say that

also that is very weird

[hazy dagger]

Okay, but how would I normally show that? I want to learn and expand my knowledge after all

[wooden frigate]

like

you could

it could be that V2 satisifes closed under addition and scalar multiplication

then what??

like do you see

that it doesnt matter

and it doesnt make sense to continue when the first criterion already failed

[hazy dagger]

It does not matter to verfiy that U is a subspace of V, but I want to know the way to show it though. Imagine for another exercise!

[wooden frigate]

Well you define two elements of the subspace

Say a,b is from the subspace V

then you wanna check

if a+b satisfies the condition of the subspace of V

that's it

[hazy dagger]

How would you do it for V2?

[hazy dagger]

a^T(u + v) = a^T * u + a^T * v = ?

[wooden frigate]

We assume x and y of V2

a^Tx = 1
and
a^Ty = 1

what about

a^T(x+y)?

[hazy dagger]

This would equal 2, right?

[wooden frigate]

yea

so that also fails

see

but even if it wouldnt fail

it still wouldnt be a subspace

[hazy dagger]

What if a^Tx = 0.5 and a^Ty = 0.5?

[wooden frigate]

try it

still no subspace as it fails the 1st criterion

[hazy dagger]

Now for scalar multiplication, I'd go with:

[wooden frigate]

,, \lambda (\vec{a}^T\vec{x}) = 1

[sour vault]

𝔸dωn𝓲²s

𝔸dωn𝓲²s

[wooden frigate]

So the 3rd criterion also fails

i believe it was lambda from R

[hazy dagger]

But you just said that lambda = 1, which is right?

[wooden frigate]

yes

so it doesnt hold

we want to use any scalar and be able to stay in our subspace

[hazy dagger]

What you mean is that for lambda = 1, it is right. However, lambda does not equal 1 in this case?

[wooden frigate]

for the equation to be satisfied lamba must be 1

however

lambda can be any real value besides 1

like for lambda 2 it doesnt hold anymore

so scalar multiplication is not closed

[hazy dagger]

So there is nothing that guarantees that lambda could be something else besides 1?

[wooden frigate]

It's the 3rd criterion that we assume lambda from R

close scalar multiplication means to use any scalar lambda

and still remain within that subspace

[slender fox]

If you are done with this channel, please mark your problem as solved by typing .close

[hazy dagger]

.

.close