linear algebra

[shut fog]

How to solve this question?

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[warm dome]

There's any constant on this type of equation, it can't describe any vector space.

The only approach I imagine is to prove you can't distribute scalar multiplication to a vector addition.

In other words, show that $\forall u,v \in E, \forall a \in \mathbb{R}, a(u+v) \neq au + av$ (if k is a constant $\neq 0$)

[lilac yarrow]

Yojda

[shut fog]

I didn't understand it

[warm dome]

Okay, if E is a vector space, so $\forall u,v \in E$, I can multiply those vectors $u$ and $v$ by a scalar $a$ such that $a*(u+v) = au + av$

[lilac yarrow]

Yojda

[shut fog]

Yes. Of course we can

[warm dome]

But if $k \neq 0$, you can't

[lilac yarrow]

Yojda

[shut fog]

I guess we can

When k isn't equal to 0

[warm dome]

Try it, and tell me if the result is again in the vector space

Even if you calculate $a*v, a \in \mathbb{R}$ and $v \in E$

[lilac yarrow]

Yojda

[warm dome]

You'll that a*v is no longer in E, so E isn't a vector space

It's a linearity problem

[lavish gust]

the 0 vector has to be in the space

so plugging in (0,0,0) must satisfy the equation, hence k=0

[shut fog]

Yes it will satisfy but I am confused what should we do with the equation?

2x1-x2+3x3=k

[lavish gust]

the vector space is all vectors (x1,x2,x3) that satisfy that equation

[warm dome]

Find what the vectors from E are

[lavish gust]

for that set to be a vector space, 0 must be a solution to that equation, hence k=0

[shut fog]

I guess i have to need more readings

[lavish gust]

or read the question

[shut fog]

So here we are finding any value which satisfy the equation?

[lavish gust]

read the question...

"Let W be (this set). If W is to be made into a vector space, what must k be?"

[shut fog]

(0,1,1)

[lavish gust]

17

I can say random stuff too

[lavish gust]

what does this have to do with anything?

[shut fog]

X1,x2,x3

[lavish gust]

Ill ask again

what does (0,1,1) have to do with anything the question is asking?

look up the axioms of being a vector space.

[shut fog]

It satisfies the equation

[lavish gust]

it doesnt unless k is 2

If $W$ is a vector space, then $0\in W$, so $(0,0,0)$ must, by the definition of $W$, satisfy $2(0)-0+3(0)=k$

[lilac yarrow]

Omegabet_

[lavish gust]

hence k must be 0.

[shut fog]

Yes

You are right

It must satisfy other properties

[lavish gust]

yes, like being closed under + as was initially pointed out

which, if you do the algebra, tells you 2k=k so k=0.

[shut fog]

How you got 2k?

@lavish gust

[lavish gust]

if $x,y\in W$, then $x+y\in W$, so $2(x_1+y_1)-(x_2+y_2)+3(x_3+y_3)=k\to (2x_1-x_2+3x_3)+(2y_1-y_2+3y_3)=k\to k+k=k$

[lilac yarrow]

Omegabet_

[shut fog]

Wow. Nice explanation tq so much