Dom V. answered • 05/07/21

Cornell Engineering grad specializing in advanced math subjects

The answer above is incorrect; it analyzes the wrong aspect of the stated matrix. Your question is asking if you can take combinations of the p(x):

p1 = 2x^{2} - 2x + 1 = [2, -2, 1]^{T}

p2 = 0x^{2} + 1x + 3 = [0, 1, 3]^{T}

p3 = 4x^{2 }- 1x + 5 = [4, -1, 5]^{T}

p4 = 2x^{2} + 2x -2 = [2, 2, 2]^{T}

to create a general quadratic in the vector space, meaning

A*p1 + B*p2 + C*p3 + D*p4 = ax^{2}+bx+c = [a,b,c]^{T}.

As a matrix equation, this is

[2, 0, 4, 2]

[-2, 1, -1, 2]. *. [A, B, C, D]^{T }=. [a, b, c]^{T}

[1, 3, 5, -2]

Coefficients of each p(x) should form the ** columns** of the matrix you analyze. It will span P2 if it has full row rank. We can row reduce the matrix:

[2, 0, 4, 2]

[-2, 1, -1, 2]

[1, 3, 5, -2]

Row reduce the first column:

[2, 0, 4, 2]

[0, 1, 3, 4]

[0, 3, 3, -3]

Row reduce the second column to arrive at echelon form:

[2, 0, 4, 2]

[0, 1, 3, 4]

[0, 0, -6, -9]

Rescale pivots:

[1, 0, 2, 1]

[0, 1, 3, 4]

[0, 0, 1, 3/2]

Row reduce the third column to get to RREF

[1, 0, 0, -2]

[0, 1, 0, -1/2]

[0, 0, 1, 3/2].

Therefore the first three polynomials can be taken in linear combination to span the P2 space. The fourth polynomial is a linear combination of the first three, but the set of four will still span. For example, using the reduced form we can always take the combination [a, b, c, 0]^{T} to produce ax^{2}+bx+c.

This should make sense--we were given four 3-dimensional vectors, so from the start we expect at least one of them to be dependent. To span the space we had to be sure the remaining three were independent, which we verified. You can still span the space with four vectors, only now the output polynomials will not correspond to a unique input because of the redundant vector.

The other answer places each p(x) across a row of the matrix. In that case, taking linear combinations of the functions is equivalent to multiplying on the left by a row vector:

[A B C D] *[2 -1 1] = [a b c]

_________[0 1 3]

_________[4 -1 5]

_________[2 2 -2]

If the operations are done this way, then the reduced matrix needs to have full column rank to produce any arbitrary values in the columns of the output (row) vector.

The reduced matrix in this case is

1 0 0

0 1 0

0 0 0

0 0 1

So now taking combinations of the reduced p(x) like [a, b, 0, c] will produce ax^{2}+bx+c, equivalent to before.

Ashley P.

Thank you very much, that makes sense now!11/08/20