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December 9th, 2016, 11:50 AM   #1
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Unhappy Linear combination

Let S = {p1, p2, p3} be the set formed by the polynomials p1 (x) = 1 + x - x ^ 2 + x ^ 3, p2 (x) = 1 + x ^ 2 + x ^ 3 and p3 (x ) = 1 - x + x ^ 2. Find p (x) = 1 + 3x - 17x ^ 2 - 5x ^ 3 as a linear combination of S. The coordinate vector of p (x) written on base S is?
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December 9th, 2016, 04:35 PM   #2
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let $bx = (1,~x,~x^2,~x^3)$

$S = bx \begin{pmatrix}
1 &1 &1 \\
1 &0 &-1 \\
-1 &1 &1 \\
1 &1 &0
\end{pmatrix}$

$p = bx\cdot (1, 3, -17, -5)$

we are looking for a coefficient matrix $c$ such that

$Sc = p$

$ c = bx \cdot (1, 3, -17, -5)$

$bx \begin{pmatrix}
1 &1 &1 \\
1 &0 &-1 \\
-1 &1 &1 \\
1 &1 &0
\end{pmatrix} c = bx \cdot (1, 3, -17, -5)$

$\begin{pmatrix}
1 &1 &1 \\
1 &0 &-1 \\
-1 &1 &1 \\
1 &1 &0
\end{pmatrix} c = (1, 3, -17, -5)$

you can solve this matrix equation in the usual fashion. I leave it to you so show that

$c = (9, -14, 6)$

and thus

$p =1+3x-17x^2-5x^3= 9(1 + x - x^2 + x^3) - 14(1 + x^2 + x^3) + 6(1 - x + x^2)$

the coordinate vector of p with respect to the basis S should be pretty obvious
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December 10th, 2016, 07:07 AM   #3
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Equivalently: You want numbers a, b, and c such that p= ap1+ bp2+ cp3
That is, you want
1+ 3x- 17x^2- 5x^3= a(1+ x- x^2+ x^3)+ b(1+ x^2+ x^3)+ c(1- x+ x^2)
= a+ ax- ax^2+ ax^3+ b+ bx^2+ bx^3+ c- cx+ cx^2
= (a+ b+ c)+ (a- c)x+ (-a+ b+ c)x^2+ (a+ b)x^3

In order that two polynomials be the same for all x, the "corresponding coefficient" must be the same so you must have
a+ b+ c= 1
a- c= 3
-a+ b+ c= -17
a+ b= -5

That is four equations for only three unknowns- it might be that this vector, p, is not in the three dimensional subspace of the four dimensional space of polynomials of order three or less spanned by these three vectors.

(Notice that the second equation can be written c= a- 3 and the fourth as b= -a- 5. Put those into the first and third equations.)



p1 (x) = 1 + x - x ^ 2 + x ^ 3, p2 (x) = 1 + x ^ 2 + x ^ 3 and p3 (x ) = 1 - x + x ^ 2. Find p (x) = 1 + 3x - 17x ^ 2 - 5x ^ 3
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