Question

Let's start over again. How do I show this?

Answer 1

Here's the question

Answer 2

SOMEBODY? Everything before this got me confused!

Answer 3

i'm reading
i need to review this stuff before i can help

Answer 4

First, why does the standard basis span?

An arbitrary polynomial in the vector space is of the form
\[v = a_1x^3 + b_1x^2 + c_1x + d_1\]

Can it be expressed as a linear combination of the standard basis?
\[\mathbb{B}=\{x^3, x^2, x, 1\}\]
An arbitrary linear combination is as follows
\[a_2(x^3) + b_2(x^2) + c_2(x) + d_2(1)\]

Clearly
\[v = a_1x^3 + b_1x^2 + c_1x + d_1 = a_2(x^3) + b_2(x^2) + c_2(x) + d_2(1)\]
If a_1 = a_2, b_1 = b_2, c_1 = c_2, d_1 = d_2

So the standard basis spans.

You can also get each element of the standard basis using the second set of vectors.

Answer 5

I can tell that I can show that since it has 1,x,x^2,x^3 but how do I actually write it out? I mean I can't just tell my professor that when I'm handing in the assignment..

Answer 6

My previous post proves that the standard basis spans. Furthermore you can show that you can express each standard basis element as a linear combination of the second set. This will show that the second set also spans.

Answer 7

Yes, i mean it's a standard basis, of course it spans, but what about B'? There are two conditions:
B is independent we proved that already and:
Span(B') = V <<<<---- How do I prove this?

Answer 8

Can you get x
x = 1*(x)
Can you get x^2
\[x^2 = \frac{1}{3}*(1) + \frac{1}{3}*(3x^2 - 1)\]

etc.

Answer 9

Show that you can get each element of the standard basis.
Since we know the standard basis spans it will show that B spans.

Answer 10

okay ill try that..

Answer 11

wait i made a typo

Answer 12

The basic idea is to show that an arbitrary vector (3rd degree polynomial) can be expressed as a linear combination of the set of vectors.

Answer 13

okay working on this, ill show u what i did..

Answer 14

\[a_3*(5x^3-3x)+b_3*(3x^2-1)+c_3*(x)+d_3*(1)\]
\[=x^3*(5a_3)+x^2*(3b_3)+x(-3a_3+1)+1*(-b_3+d_3)\]

Answer 15

oops i left off the c in the third paranthesis lol i cant type

Answer 16

You can get each element of the standard basis:
\[1 = 1*(1) + 0*(x) + 0*(3x^2 - 1) + 0*(5x^3 - 3x)\]\[x = 0*(1) + 1*(x) + 0*(3x^2 - 1) + 0*(5x^3 - 3x)\]\[x^2= \frac{1}{3}*(1) + 0*(x) + 1*(3x^2 - 1) + 0*(5x^3 - 3x)\]\[x^3= 0*(1) - \frac{3}{5}*(x) + 0*(3x^2 - 1) + \frac{1}{5}*(5x^3 - 3x)\]

Answer 17

\[x^3*(5a_3)+x^2*(3b_3)+x*(-3a_3+c_3)+1*(d_3-b_3)\]

Answer 18

okay this is what i did:

Answer 19

Wrong way around look at my previous post.

Answer 20

i totally don't remember this stuff lol

Answer 21

ohhh I see..

Answer 22

So basically u need to be able to rewrite B as a lin combination of B'..

Answer 23

We know the standard basis spans.
We can get all the elements of the standard basis using \[B'\]

Answer 24

If B spans and we can get B from B' then B' spans.

Answer 25

Also with respect to linear independence I think you should be using column operations instead of row operations to show linear independence. Although I think both approaches are probably valid. I have to think about it.

Answer 26

Oh so all i need to do to determine if something is a basis is be able to rewrite the standard basis as a linear combination of the basis that I have..

Answer 27

Yeah

Answer 28

Also refer to my proof that the standard basis spans.

Answer 29

Its at the top.

Answer 30

Okay thanks a lot!!!!!!!!

Answer 31

Ima login from my other account and give u another medal guys... THANKS A LOT FINALLY!!!

Answer 32

By the way, Alchemista you made a minor mistake..

Answer 33

x^3 = 0(1) + (3/5)x + 0(3x^2-1) + (1/5)(5x^3-3x)

Answer 34

2nd term is not -3/5 it's +3/5, but thanks a lot again!!! U just saved my @$$ for this topic..