MHB Finding a basis for a Vector Space

[Dethrone]

View attachment 4284

4b). How can I find a basis? I was thinking of the standard basis $\{1,x,x^2\}$, but that doesn't work under the scalar multiplication definition in the vector space.

EDIT: I think it is $\{0,x,x^2\}$ and we take $1$ to be the $0$ vector!
$a(0)+b(x)+c(x^2)=1$ implies $a=b=c=0$.
It is strange, because I've done nearly all the questions in the book and I never encountered a question like this one.

 

[I like Serena]

EDIT: I think it is $\{0,x,x^2\}$ and we take $1$ to be the $0$ vector!
$a(0)+b(x)+c(x^2)=1$ implies $a=b=c=0$.
It is strange, because I've done nearly all the questions in the book and I never encountered a question like this one.

Hey Rido!
You've got it. (Nod)
How is it different?

 

[Dethrone]

Hey ILS :D

Thanks for the reply!

It is different because in my book, vector spaces and linear independence are in different sections. So most of the time, I have only tried $a(1)+b(x)+c(x^2)=0$ where I set it to $0$, not $1$, and I would solve it under regular addition and multiplication.

This time:
$a(0)+b(x)+c(x^2)=1$
$a(0)-a+1+bx-b+1+cx^2-c+1=1$
$-a+bx-b+cx^2-c=0$

$\implies a=0$ (when $x=1$)
$\implies b=0$ (when $x=-1$)
And finally, $c=0$.

 

[I like Serena]

It is different because in my book, vector spaces and linear independence are in different sections. So most of the time, I have only tried $a(1)+b(x)+c(x^2)=0$ where I set it to $0$, not $1$, and I would solve it under regular addition and multiplication.

So it's a good exercise! (Mmm)

It alerts us to the fact that the addition and scalar multiplication defined for the vector space have to be used whenever checking anything.
Btw, to avoid confusion, a better notation for linear independence would be:
$$a \odot (0) \quad\oplus\quad b \odot (x) \quad\oplus\quad c \odot (x^2) \quad=\quad 0_{\mathbb P_2} \quad=\quad 1$$
(Nerd)