Question

@ganeshie8 @dan815 @Kainui

Answer 1

I don't understand how to use coordinates if it's asking for linear dependence, because I need a basis and that would require it to be linearly independent. Huh

Answer 2

\[c_1p(x)+c_2q(x)+c_3r(x) = 0 \]

Answer 3

okay so u got 1,-1-1 as 1 solution

Answer 4

so uve shown lin dependance

Answer 5

Not yet I was doing that right now, but I can doing what I am above, one second, \[c_1(2-x+x^2)+c_2(1+3x+3x^2)+c_3(1-4x-2x^2)=0\]

Answer 6

\[2c_1-xc_1+c_1x^2+c_2+3xc_2+3x^2c_2+c_3-4xc_3-2x^2c_3=0\]
\[(2c_1+c_2+c_3)+(-c_1+3c_2-4c_3)x+(c_1+3c_2-2c_3)x^2=0\]
After equating the sides we get
\(2c_1+c_2+c_3=0\)
\(-c_1+3c_2-4c_3=0\)
\(c_1+3c_2-2c_3=0\)

Answer 7

mhm

Answer 8

Did I make a mistake

Answer 9

no its fine

Answer 10

ull get a dependance saying
c2=c3
c1=c2+c3

Answer 11

Right

Answer 12

okey

Answer 13

go onnnn

Answer 14

I thought i had to use coordinates on every part of the question xD so i was like wut

Answer 15

just ignore the coordinate word for now

Answer 16

solve it like u normally wud

Answer 17

Ok, so for a dependence relation I'd plug in the values for c1 c2 c3, so for now I'll just use what you said, of course I'll do all the technical work after, so c1=1, c2=-1, c3=-1
\[1 \times p(x)-1 \times q(x)-1 \times r(x) = 0\]

Answer 18

ya

Answer 19

because u can see right like if u add q(x) + r(x) u get p(x)

Answer 20

Yes

Answer 21

p=q+r
1p=1q+1r
1p-1q-1r=0
so 1,-1,-1 is a solution

Answer 22

Omg >_>

Answer 23

similarily
k*<1,-1,-1> , k E R is a solution

Answer 24

this solves a,b

Answer 25

now what is the definition of span

Answer 26

All the linear combinations possible in the set

Answer 27

the set of vectors in V

Answer 28

okay so just take q and p

Answer 29

that is one basis

Answer 30

So I don't have to show they span S right as that's a given, I must show it's linearly indep however?

Answer 31

Because a basis must span and be lin. indp.

Answer 32

right

Answer 33

Ok

Answer 34

u can take any 2 of {q,p,r}

Answer 35

So I can show its lin indep, is there a technique like the one before haha or do I have to do it the long way as I was before

Answer 36

I'd use p and q to be safe

Answer 37

okay well since its just 2 vectors in this case

Answer 38

since one is not a scalar multiple u can already say its lin independant

Answer 39

Eh I'd rather show it though, like \[c_1p(x)+c_2q(x)=0\] would that work?

Answer 40

ya u can do that too

Answer 41

Ok cool, that will be easy to since it's only 2 vectors \[c_1(2-x+x^2)+c_2(1+3x+3x^2)=0\]
\[2c_1-xc_1+x^2c_1+c_2+3xc_2+3x^2c_2=0\]
\[(2c_1+c_2)+(-c_1+3c_2)x+(c_1+3c_2)=0\]
\(2c_1+c_2=0\)
\(-c_1+3c_2=0\)
\(c_1+3c_2=0\)

Yeah it'll be linearly independent

Answer 42

So is our basis just \[S = \left\{ p(x),q(x) \right\}\] as well

Answer 43

ya

Answer 44

u have an finite number of basis thats just one

Answer 45

infinite*

Answer 46

Ok I see, these terms are annoying at times lol

Answer 47

So then c is done

Answer 48

ya now try solving
c1p+c2q=f

Answer 49

For \[f(x)=x+3x^2\] now I must check if it spans and is lin indep

Answer 50

in S

Answer 51

f(x) doesnt span anything by itself its just 1 "vector"

Answer 52

nono u gotta solve

Answer 53

see if ur basis can hit that vector

Answer 54

if a linear combination exists then f(x) is in S

Answer 55

i think u cant do it

Answer 56

Right, so \[f(x) = c_1p(x)+c_2q(x)\]

Answer 57

Let me try

Answer 58

okay

Answer 59

i found a trick to doing these without writing equations

Answer 60

\[x+3x^2=c_1(2-x+x^2)+c_2(3x+3x^2)\]
Really, because I'm tired of equating sides lol

Answer 61

What are you talking about

Answer 62

Did you discover a right hand rule for this

Answer 63

xD

Answer 64

Ok

Answer 65

I'm going to write it on draw, cause equation takes so long

Answer 66

brb getting food