Question 1168921: Let vector p=2−x+x^2 and let B be the basis for P2:
{1+x,1+x^2,x+x^2}.
Find [p]B.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Solution:
Let the basis $B = \{v_1, v_2, v_3\}$, where $v_1 = 1+x$, $v_2 = 1+x^2$, and $v_3 = x+x^2$.
We are given the vector $p = 2 - x + x^2$.
We need to find the coordinates of $p$ with respect to the basis $B$, which means finding scalars $c_1, c_2, c_3$ such that:
$p = c_1 v_1 + c_2 v_2 + c_3 v_3$
$2 - x + x^2 = c_1(1+x) + c_2(1+x^2) + c_3(x+x^2)$
$2 - x + x^2 = c_1 + c_1 x + c_2 + c_2 x^2 + c_3 x + c_3 x^2$
Group the terms by powers of $x$:
$2 - x + x^2 = (c_1 + c_2) + (c_1 + c_3)x + (c_2 + c_3)x^2$
For this equation to hold for all $x$, the coefficients of the corresponding powers of $x$ on both sides must be equal. This gives us a system of linear equations:
1. Coefficient of $x^0$ (constant term): $c_1 + c_2 = 2$
2. Coefficient of $x^1$ (term with $x$): $c_1 + c_3 = -1$
3. Coefficient of $x^2$ (term with $x^2$): $c_2 + c_3 = 1$
We need to solve this system for $c_1, c_2, c_3$.
From equation (1), $c_2 = 2 - c_1$.
From equation (2), $c_3 = -1 - c_1$.
Substitute these expressions for $c_2$ and $c_3$ into equation (3):
$(2 - c_1) + (-1 - c_1) = 1$
$1 - 2c_1 = 1$
$-2c_1 = 0$
$c_1 = 0$
Now, substitute the value of $c_1$ back into the expressions for $c_2$ and $c_3$:
$c_2 = 2 - c_1 = 2 - 0 = 2$
$c_3 = -1 - c_1 = -1 - 0 = -1$
So, the coordinates of $p$ with respect to the basis $B$ are $c_1 = 0$, $c_2 = 2$, and $c_3 = -1$.
The coordinate vector $[p]_B$ is given by:
$[p]_B = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 2 \\ -1 \end{bmatrix}$
To verify, we can substitute these values back into the linear combination:
$0(1+x) + 2(1+x^2) + (-1)(x+x^2) = 0 + 2 + 2x^2 - x - x^2 = 2 - x + x^2 = p$.
Final Answer: The final answer is $\boxed{\begin{bmatrix} 0 \\ 2 \\ -1 \end{bmatrix}}$
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