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please help, this linear transformation has me completely confused!

Let $T:P_2 \to R$ be given by $T(p(t))=\int_{-2}^{3}p(t)dt$

a.) prove that $T$ is a linear transformation. - For this i know that i need to show that $T$ preserves vector addition and scalar multiplication, but i do not know how to show this abstractly, i mean i have know idea how to set up my vectors.

b.) find the kernel of $T$. - i know that i need to solve for $T(p(t)) = 0$, but again the integral is confusing me

c.) find the nullity of $T$ and the rank of $T$ - only way i know to find these is with the rre form of the standard matrix representation (SMR), which im not sure how to get

d.) is $T$ an isomorphism? - I know that $T$ must be one-to-one and onto if it is an isomorphism, which can be shown with rank and nullity or if the $T(v) = 0 \implies v = 0 $ right?

I think the integral is really confusing me and the fact that $T$ is a mapping from $P_2 \to R$ its just not making sense to me right now, any help is greatly appreciated! thanks in advance.

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a) The integral is linear, so $T$ is linear too : $$ \int_{-2}^3(ap_1(t)+bp_2(t))\,dt=a\int_{-2}^3p_1(t)dt+b\int_{-2}^3p_2(t)\,dt $$

b) Your polynomials are in $P_2$, so in general $p(t)=at^2+bt+c$ for some coefficients $a,b,c\in\mathbb{R}$. Then $$ \int_{-2}^3(at^2+bt+c)\,dt=0 $$ is equivalent to (verify this) $$ 14a+3b+6c=0\tag{1} $$ So, the kernel is $\{at^2+bt+c\in P_2:14a+3b+6c=0\}$

c) The nullity of $T$ is the dimension of $\ker T$. Note that $(1)$ gives you two degrees of freedom, e.g. $a$ and $b$ can be chosen to be whatever real numbers you want, and $c$ is then $$ c=-\frac{14}{6}a-\frac{1}{2}b $$ So, you can write $\ker T$ as $$ \left\{p(t)\in P_2:p(t)=a\left(t^2-\frac{14}{6}\right)+b\left(t-\frac{1}{2}\right)\text{ for some }a,b\in\mathbb{R}\right\} $$ or $$ \ker T=\text{Span}\left\{t^2-\frac{14}{6},t-\frac{1}{2}\right\} $$ where $t^2-\frac{14}{6}$ and $t-\frac{1}{2}$ are obviously linearly independent. Hence, $\dim\ker T=2$, that is to say, the nullity of $T$ is $2$.

Since $\dim P_2=3$, the Rank-Nullity Theorem tells you that the rank of $T$, that is, the dimension of the image of $T$, is $3-2=1$.

d) Since $\ker T\neq\{0\}$, $T$ is not injective. Hence $T$ is not an isomorphism.

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Let $p(t)$ and $q(t) \in P_2$ and k $\in R$.

$$T\big (p(t) + kq(t)\big ) = \int_{-2}^{3} \big (p(t) + kq(t)\big )dt$$ $$=\int_{-2}^{3}p(t)dt + k\int_{-2}^{3}q(t)dt $$ $$= T\big(p(t)\big) + kT\big(q(t)\big)$$

$\implies T$ is a linear transformation.

$$N\big(T\big) = \big\{p(t) \mid T\big(p(t)\big) = 0\big\} $$ $$\iff \int_{-2}^{3}p(t)dt = 0$$ $$\iff \int_{-2}^{3}(at^2 + bt + c)dt = 0 $$ $$\iff \frac{a}{3}(3^3 - (-2)^3) + \frac{b}{2}(3^2 - (-2)^2) + c(3 -(-2))= 0$$ $$\iff \frac{35a}{3} + \frac{5b}{2} + 5c = 0$$ $$\iff c = -\frac{7}{3}a - \frac{1}{2}b$$

Thus it is all polynomials of the form $$p(t)= at^2 + bt + (-\frac{7}{3}a - \frac{1}{2}b)$$

Thus $$N\big(T\big)= \big\{p(t) \mid p(t) = at^2 + bt + (-\frac{7}{3}a - \frac{1}{2}b) \big\} $$

Notice that $$p(t) = at^2 + bt + (-\frac{7}{3}a - \frac{1}{2}b)$$ $$= a(t^2 - \frac{7}{3}) + b(t - \frac{1}{2})$$

Since these two polynomials are of different degrees then they are linearly independent and thus they constitute a basis for the null space, that is $$Basis = \big \{t^2 - \frac{7}{3}, t - \frac{1}{2} \big \}$$

$$\implies dim(N(T)) = 2$$ $$\implies dim(R(T)) = 3 -2 = 1$$

Since this function is not one-to-one, that is $dim(N(T)) \neq 0$, then it cannot be an isomorphism.

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