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Past Exam Question Help

(a) Let $P_2(R)$ denote the vector space of real polynomial functions of degree less than or equal to two and let $B:= [p_0,p_1,p_2]$ denote the natural ordered basis for $P_2(R) $(so $p_i(x) =x^i).$Define $g$ element of $P_2(R)$ by $g(x) = 3x^2-2x+ 2$. Write$g$ as a linear combination of the elements of $B$. Compute the coordinate vector $g_B$ of $g$ with respect to $B$.

Define $h_1,h_2,h_3$elements of $ P_2(R)$ by $h_1(x) = 2,h_2(x) = 6x+ 4$ and$h_3(x) = 2x^2-7x+ 6$. Define $C:= [h_1,h_2,h_3].$ Write each element of B as a linear combination of the elements of C. Explain why the calculations you have performed prove that C is a basis for $P_2(R).$ Compute the coordinate vector $g_C$.

(b) Let$P_n(R)$ denote the vector space of real polynomial functions of degree less than or equal to $n$, let $p_i$ denote the polynomial determined by $p_i(x) =x^i$, and let $F:P_2(R) ->P_3(R)$. be the linear transformation determined by $$F(f)(x) =\int^{x+1}_{2-x}(1-t)f(t) dt:$$ Determine a basis for the kernel of F. Determine a basis for the image of F. Define $A :=[p_0,p_1,p_2,p_3]$ and compute $M^A_B(F)$, the matrix of F with respect to the ordered bases B and A. Compute $M^A_C(F)$ and give the rank of $M^A_C(F)$.

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  • $\begingroup$ There might be a $\ ^2$ missing in the definition of $g$. What's your question? $\endgroup$ – Roland Apr 23 '14 at 23:19
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(a): $g = 3p_2 - 2p_1 + 2 p_0$ is a linear combination of elements of $B$ and the coordinate vector is $(2,-2,3)$ (note how $B$ is ordered so that $p_0$ is the first basis vector).

Writing $B$ in terms of $C$:

$p_0 = 1 = {1 \over 2}h_1$.

$p_1 = x = {1 \over 6}(h_2 - 2 h_1) $.

$p_2 = x^2 = {1\over 2}(h_3 + {7 \over 6}h_2 - {32 \over 6}h_1) $

This proves that $h_i$ are linearly independent because the $p_i$ are and one can note that because $p_0$ and $p_1$ are linearly independent $h_1$ and $h_2$ are too because otherwise there would exist $c_1$ and $c_2$ not both zero with $c_1 h_1 + c_2 h_2 =0$ and this would imply that there are also $c'_1, c'_2$ not both $0$ with $c'_1 p_0 + c'_2 p_1 = 0$ contradicting that $p_0$ and $p_1$ are linearly independent. By the same argument $h_1,h_2,h_3$ are linearly independent.

The vector $g = 3p_2 - 2p_1 + 2 p_0$ in $C$ is (substituting the identitites above) $$ g = 3p_2 - 2p_1 + 2 {1 \over 2}h_1 = {3\over 2}(h_3 + {7 \over 6}h_2 - {32 \over 6}h_1) - {2 \over 6}(h_2 - 2 h_1) + h_1 = {3 \over 2}h_3 + {17 \over 12}h_2 + {100 \over 12}h_1$$

therefore the coordinate vector is $g_C = ({100\over 12}, {17 \over 12}, {3 \over 2})$.

(b): Let $F: P_2 \to P_3$ be defined by $$F(f)(x) =\int^{x+1}_{2-x}(1-t)f(t) dt$$

We compute $F(p_0)$ and $F(p_1)$(the basis elements of $P_2$):

$$F(p_0)(x) =\int^{x+1}_{2-x}(1-t)1 dt = x+1-(2-x) -{1 \over 2}((x+1)^2 - (2-x)^2) = {1\over 2}-x $$

$$ F(p_1) (x)= \int^{x+1}_{2-x}(1-t)t dt = \int^{x+1}_{2-x}t dt- \int^{x+1}_{2-x}t^2 dt = {1 \over 2}((x+1)^2 - (2-x)^2) - {1\over 3}((x+1)^3 - (2-x)^3) = -{2 \over 3}x^3 + x^2 -2x + {5 \over 6}$$

Note that since $p_0,p_1$ span $P_2$ the images $F(p_0)$ and $F(p_1)$ span the image of $F$. Note that they are linearly independent. Applying the rank nullity theorem yields that the kernel of $F$ is $\{0\}$. The basis for this set is $B= \varnothing$.

I will add some computations for $M_B^A$ and $M_C^A$ later.

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  • $\begingroup$ Hi thanks so much for this, if possible could you do the computations for $M_B^A$ and $M_C^A$, also could you elaborate on how you managed to get the kernel of F to be $0$ and explain the basis of the image of F $\endgroup$ – simon May 8 '14 at 22:27
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I'm fairly sure this answer is partly wrong. part (a) is correct but the person has failed to compute $F(p_2)$ in part b. they have also claimed $p_0,p_1$ span $P_2$ which is incorrect they only span $P_1$. so the person has answered part b for $F:P_1 → P_3$

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