Use coordinate vectors to determine whether the set of...

Use coordinate vectors to determine whether the set of polynomials B = {p1, p2, p3, p4} forms a basis for the space of all polynomials of degree at most 3, P3, where p1(t) = 1, p2(t) = t, p3(t) = 1/2(3t^2 - 1), p4(t) = 1/2(5t^3 - 3t). Find coordinates of q(t) = 1 + t + t^2 + t^3 relative to B.

Transcript

00:01 So in this problem we have the vector space p3 of all polynomials of degree at most 3.

00:15 Now this vector space is dimension 4 and the standard basis equals 1 t t squared t cube.

00:26 What this means is that any polynomial in the variable t of degree at most 3 can can be written as a linear combination of these polynomials, 1 t squared t cube.

00:41 That is immediately obvious and these are of course linearly independent and so we have p3 has a dimension 4.

00:54 Now consider the set of polynomials b which contains p1, p2, p3 and p4 where p1 equals 1 the constant polynomial p2 is t p 3 is 3 over 2 t squared minus half and p 4 is 5 over 2 t cubed minus 3 over 2 times t now the coordinate vectors of these polynomials written with respect to the standard basis e that is 1 t squared t cubed is as follows for p 1 p 1 1 1 2 2 2 2 2 2 2 2 2 is as follows for p 1 p 1 we have the we get the column matrix one in the first entry and zero is everywhere else so what this denotes is that the p1 is the constant polynomial 1 so when p1 is written as a linear combination of 1 t2 t squared and t cube we have that the coefficient of 1 is 1 and the coefficient of t 2 t squared and t cubed are all zeros so the first entry 1 corresponds to the coefficient of 1, the second corresponds to the coefficient of t, the third to the coefficient corresponding to t squared and the fourth is the coefficient of t cubed in the expression of p1.

02:30 So p1 is just 1.

02:32 Similarly we have that p2 is the column matrix 0 -1 -0 because here we have p2 is t that means only t has a non -zero coefficient which happens to be one and all the others are zeros and p3 is minus half zero three over two zero why because we have the coefficient of one is minus half in case of p3 there's no t no t squared so both the coefficients of t and t squared or is uh t is zero but we do have t squared that is the coefficient of t squared is 3 over 2 and no t cubed so that's zero again so we have minus half zero three over two zero and similarly p4 is 0 minus 3 over 2 0 and 5 over 2 so these are the coordinate matrices of the polynomials p 1 p2 p 3 p 4 with respect to the basis is e.

03:44 Now consider the matrix capital p whose columns are the column matrices p1, p2, p3, and p4.

04:02 So that is the matrix p is 1 -0 minus 0, 0 ,0 minus 3 over 2, 0, 0, 0, 0, 0, 0, 05 over 2.

04:13 So it's just the matrix whose columns are the column matrices representing p1, p2, and p4.

04:24 So then expanding along the first column, we get the determinant.

04:35 Expanding along this first column of the matrix p, we get that the determinant of p is that p equals 1 times determinant of the sub matrix 1 .0 minus 3 over 2, 03 over 2, and 0 .05 over 2.

04:52 And the determinant of this sub matrix, is one times.

05:00 So we have one here, expanding along the first column again.

05:06 We have one times the determinant of this 2x2 matrix, which is 15 over 4 minus 0.

05:13 And the other terms in the expansion are 0 times something plus 0 times something.

05:20 So that's 0.

05:21 So we get 1 times 15 over 4 minus 0.

05:27 Over 4, which is not equal to 0...

Question

Use coordinate vectors to determine whether the set of polynomials B = {p1, p2, p3, p4} forms a basis for the space of all polynomials of degree at most 3, P3, where p1(t) = 1, p2(t) = t, p3(t) = 1/2(3t^2 - 1), p4(t) = 1/2(5t^3 - 3t). Find coordinates of q(t) = 1 + t + t^2 + t^3 relative to B.

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