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Given $p_1 = 40+ 7x + 51x^3$

$p_2=8+x+24x^2+7x^3$

$p_3 = 8x + 4x^3$

$p_4 = 24+3x+21x^3$

" If {p1,p2,p3,p4}is a basis of P3, find the coordinate of p(x)=3+0x+1x2+0x3."

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You lost your LaTeX at the end there. I'm assuming the question is:

"...find the coordinate vector of $p(x) = 3 + 0x + 1x^2 + 0x^3 = 3 + x^2$"

Remember, any vector in a space can be defined according to the basis of that vector space.

So given basis vectors $\{p_1,p_2,p_3,p_4\}$, you can write any vector $v \in P3$ in the form:

$$v = a_1p_1 + a_2p_2 + a_3p_3 + a_4p_4$$

The coordinate vector of $v$ will then be $[v] = (a_1, a_2, a_3, a_4)$.

So plug in what you know:

$$p(x)=a_1p_1 + a_2p_2 + a_3p_3 + a_4p_4$$

$$3 + x^2 = a_1(40+7x+51x^3) + a_2(8+x+24x^2+7x^3) + a_3(8x + 4x^3)+ a_4(24+3x+21x^3)$$

After you multiply everything out you'll get a polynomial in P3. Then group the coefficients for each power of x on the right side together. You'll end up with some polynomial of the form: $$3 + x^2 = b_3x^3 + b_2x^2 + b_1x+b_0$$

Then you need the coefficients of corresponding terms on the left and right to be equal. You would set $b_2 = 1$ (which is the coefficient of $x^2$ on the left side), $b_0 = 3$, and all the other coefficients = $0$.

So for instance, one of the terms on the right side will end up being $(51a_1 + 7a_2 +4a_3 + 21a_4)x^3$. ie. $b_3 = (51a_1 + 7a_2 +4a_3 + 21a_4$). So you would set $51a_1 + 7a_2 +4a_3 + 21a_4 = 0$

After setting all of the grouped coefficients equal to something, you'll end up with a linear system of 4 equations ($b_0 = 3, b_1 = 0, ...$) with 4 unkowns ($a_1, a_2, a_3, a_4$).

Solve this system using matrix algebra to find your coordinate vector ($a_1, a_2, a_3, a_4$).

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  • $\begingroup$ I try solving $\begin{pmatrix}40 & 8& 0& 24 \\ 7 & 1 & 8 & 3 \\ 0& 24 & 0 & 0 \\ 51 & 7 & 4 & 21\end{pmatrix} = \begin{pmatrix} 3\\0 & 1 \\ 0\end{pmatrix}$ but get it wrong. What did you get? $\endgroup$ – user113027 Dec 11 '13 at 3:05
  • $\begingroup$ (Sorry this is late, I've been studying). I'm going to guess that the column matrix was supposed to be $\begin{pmatrix} 3\\0 \\ 1 \\ 0\end{pmatrix}$. So you did write your system correctly. I used Cramer's Rule and some back substitution to solve and I got $(a_1,a_2,a_3,a_4) = (\frac{-13}{80},\frac{1}{24}, \frac{-1}{160},\frac{55}{144})$. (You can verify this is the correct answer by writing out $a_1p_1 + a_2p_2 + a_3p_3 + a_4p_4$ and replacing the known values. It will end up equalling $3+x^2$). What method are you using to solve the matrix? $\endgroup$ – andraiamatrix Dec 12 '13 at 2:53

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