# Thread: linear algebra change of basis

1. ## linear algebra change of basis

Let p(x) = 7 -12x - 8x^2 + 12x^3. Find the coordinate vector of p relative to B.
Hint: Determine {c0, c1, c2, c3} such that p(x) = the sum from i = 0 to 3 of c_i * H_i(x)

H_0(x) = 1
H_1(x) = 2x
H_2(x) = -2 + 4x^2
H_3(x) = -12x + 8x^3

x is in (-infinity, infinity)

Can someone please explain to me how to do this problem?

2. isn't the hint enough ? have you tired to follow it? you want to equate and compare coefficients. this will give you some simple linear equations relating you c_i's.

3. Maybe I'm just not thinking about it right but, I don't understand how to do that.

4. Maybe this isn't how you are use to doing change of basis, if you follow the hint the question is just a simple case of basic algebra and nothing more. I'll explain how it works to get you started.

In any bases your polynomial function has to be the same function, changing the bases just changes how you look it at. Normally we represent a polynomial just looking at the coefficient of power of x.

under this basis the polynomial p(x) = 7 -12x - 8x^2 + 12x^3 is represented by the vector ( 7 , -12 , -8 , 12 ). Now for some reason we want to change the basis of our vector space.

Under the new basis the vector (c0 , c1 , c2 ,c3) represents the polynomial p(x) = c0*(1) + c1*(2x) + c2*(-2 + 4x^2) + c3*(-12x + 8x^3). So you want to find a vector under this new basis that gives you the polynomial you want.

understand the hint now ?

5. Yeah, we've never done change of basis like that before. My lecture on this was only about half an hour since my professor has to rush through things so I'm a little slower about this than I should be. It makes a little more sense now. But i don't understand how to find the vectors c0 - c3

6. Originally Posted by SpiffyEh
Let p(x) = 7 -12x - 8x^2 + 12x^3. Find the coordinate vector of p relative to B.
Hint: Determine {c0, c1, c2, c3} such that p(x) = the sum from i = 0 to 3 of c_i * H_i(x)

H_0(x) = 1
H_1(x) = 2x
H_2(x) = -2 + 4x^2
H_3(x) = -12x + 8x^3
Are we to assume that B= {H_0(x), H_1(x), H_2(x), H_3(x)}? It would have been a good idea to say that!

x is in (-infinity, infinity)

Can someone please explain to me how to do this problem?
Exactly what the hint says:
$\displaystyle 7 -12x - 8x^2 + 12x^3= c_0(1)+ c_1(2x)+ c_2(-2+ 4x^2)+ c_3(-12x+ 8x^3)$
That is to be true for all x. Since there are 4 unknown values, $\displaystyle c_0$, $\displaystyle c_1$, $\displaystyle c_2$, and $\displaystyle c_3$, you will need 4 equations.

One way to get 4 equations is to take x to be 4 different values:
If x= 0, $\displaystyle 7= c_0- 2c_2$
If x= 1, $\displaystyle 7- 12- 8+ 12= -1= c_0+ 2c_1+ 2c_2- 4c_3$, etc.

Another way is to combine like powers:
$\displaystyle 7 -12x - 8x^2 + 12x^3= (c_0- 2c_2)+ (2c_1- 12c_3)x+ (4c_2)x^2+ (8c_3)x^3$
In order to be true for all values of x, we must have $\displaystyle c_0- 2c_2= 7$, $\displaystyle 2c_1- 12c_3= -12$, $\displaystyle 4c_2= -8$, $\displaystyle 8c_3= 12$.

Yet a third way is to evaluate the function and its derivatives at x= 0.
function at x= 0: $\displaystyle 7= c_0- 2c_2$
Taking the derivative of both sides, $\displaystyle -12- 16x+ 36x^2= c_1+ 8c_2 x+ (-12+ 24x^2)c_3$ and at x= 0 that gives
$\displaystyle -12= c_1- 12 c_3$, etc.

7. umm there is a B given far above the problem but I'm not sure if its part of this particular problem. It's B = {1/sqrt(2)[1 1]^T, 1/sqrt(2)[-1 1]^T} It doesn't seem like it would be part of this problem. I think the B would be what you mentioned. I'll try doing it one of those ways and post what I get. If someone could check it, that would be great