--- Suppose 𝑏, 𝑐 ∈ 𝐑 . Define 𝑇 ∶ 𝒫(𝐑) → 𝐑^2 by 𝑇𝑝 = (3𝑝(4) + 5𝑝'(6) + 𝑏𝑝(1)𝑝(2) , ∫𝑥^3 𝑝(𝑥) 𝑑𝑥 + 𝑐 sin 𝑝(0)). Show that 𝑇 is linear if and only if 𝑏 = 𝑐 = 0 . SOLUTION: First suppose 𝑏 = 𝑐 = 0 . Then 𝑇𝑝 = (3𝑝(4) + 5𝑝'(6), ∫𝑥^3 𝑝(𝑥) 𝑑𝑥) , which easily implies that 𝑇 is linear. >[!explain] >since all elements are linear: diff, int >but need more explanation about integral of x^3 Conversely, now suppose that 𝑇 is linear. Note that 𝑇1 = (3 + 𝑏, $\frac{15}{4}$ + 𝑐 sin 1) and 𝑇2 = (6 + 4𝑏 , $\frac{15}{2}$ + 𝑐 sin 2) >[!why?] EF picks two polynomials >T1 means p is the polynomial 1, the answer is correct, substituting 1. >T2 means the polynomial = 2. then 6+4b is correct, 15/2 is correct, c sin 2 is correct >So EF shows that linearity requires that T2 = 2T1; both these equations to be true must have b and c = 0 The equation 𝑇2 = 2𝑇1 implies that 6 + 4𝑏 = 6 + 2𝑏 and 15/2 + 𝑐 sin 2 = 15/2 + 2𝑐 sin 1 , which implies that 𝑏 = 0 and 𝑐 = 0 . Edward Frenkel --- What if the p is x, or x^2? More complicated. Seems a lot to use 1 and 2 Tp is (3p(4) + 5p'(6)) means 3 times value of p when x = 4 and p' at x=6? This is just bypassed when the polynomial value is just 1 or 2, no matter where it is evaluated. --- Commentary: $\textbf{Exercise 2.}$ Suppose $b, c \in \mathbb{R}$. Define $T : \mathcal{P}(\mathbb{R}) \to \mathbb{R}^2$ by $Tp = \left(3p(4) + 5p'(6) + bp(1)p(2), \int_{-1}^2 x^3p(x) dx + c \sin p(0)\right).$ Show that $T$ is linear if and only if $b = c = 0$. $\textbf{Solution 2.}$ First suppose $b = c = 0$. Then $Tp = \left(3p(4) + 5p'(6), \int_{-1}^2 x^3p(x) dx\right),$ which easily implies that $T$ is linear. Conversely, now suppose that $T$ is linear. Note that $T1 = \left(3 + b, \frac{15}{4} + c \sin 1\right) \text{ and } T2 = \left(6 + 4b, \frac{15}{2} + c \sin 2\right).$ The equation $T2 = 2T1$ implies that $6 + 4b = 6 + 2b$ and $\frac{15}{2} + c \sin 2 = \frac{15}{2} + 2c \sin 1$, which implies that $b = 0$ and $c = 0$. $\textit{Commentary:}$ This exercise is similar to Exercise 1, but involves a linear map $T$ from the space of polynomials $\mathcal{P}(\mathbb{R})$ to $\mathbb{R}^2$. The map $T$ is defined in terms of an evaluation, a derivative evaluation, an integral, and a sine function, along with two parameters $b$ and $c$. The exercise asks to prove that $T$ is linear if and only if both $b$ and $c$ are zero. The proof of the forward direction is again straightforward: if $b = c = 0$, then $Tp$ involves only the linear operations of evaluation, differentiation, and integration, so $T$ is linear. The proof of the reverse direction follows a similar strategy to Exercise 1. It considers the specific polynomials 1 and 2 and uses the fact that $T(2p) = 2T(p)$ for linear maps to conclude that $b$ and $c$ must be 0. This exercise reinforces the practice of using the defining properties of linear maps, and also illustrates that these properties hold for linear maps defined on infinite-dimensional spaces like $\mathcal{P}(\mathbb{R})$. $\textit{Examples:}$ 1. Define $T : \mathcal{P}(\mathbb{R}) \to \mathbb{R}$ by $Tp = p(1) + p'(0)p(0)$. Then $T$ is not linear, because $T(p + q) \neq Tp + Tq$ in general (consider $p(x) = 1$ and $q(x) = x$). 2. Define $T : \mathcal{P}(\mathbb{C}) \to \mathbb{C}$ by $Tp = \int_0^1 p(z) dz + ip(i)$. Then $T$ is linear, because integration and evaluation are linear operations. 3. Define $T : \mathcal{P}(\mathbb{F}_2) \to \mathbb{F}_2$ by $Tp = p(0) + p(1)$. Then $T$ is linear, because evaluation is a linear operation, and the sum of two evaluations is linear over $\mathbb{F}_2$ (because $1 + 1 = 0$ in $\mathbb{F}_2$). These examples showcase a variety of linear and nonlinear maps defined on polynomial spaces over different fields. They highlight that the linearity of a map depends on the linearity of the operations used to define it, and that the same operation (like addition) can be linear or nonlinear depending on the underlying field. --- Here's the Claude proof: Suppose b, c ∈ ℝ. Define T: 𝒫(ℝ) → ℝ² by T(p) = (3p(4) + 5p'(6) + bp(1)p(2), ∫₋₁²x³p(x) dx + c sin p(0)). Show that T is linear if and only if b = c = 0. Proof: Assume b = c = 0. 1. To show that T is linear, we need to prove that for any polynomials p, q ∈ 𝒫(ℝ) and any scalar α ∈ ℝ: 2. a. T(p + q) = T(p) + T(q) 3. b. T(αp) = αT(p)  4. Let's prove each property:  - a. T(p + q) = (3(p + q)(4) + 5(p + q)'(6), ∫₋₁²x³(p + q)(x) dx) 1. = (3p(4) + 3q(4) + 5p'(6) + 5q'(6), ∫₋₁²x³p(x) dx + ∫₋₁²x³q(x) dx) 2. = (3p(4) + 5p'(6), ∫₋₁²x³p(x) dx) + (3q(4) + 5q'(6), ∫₋₁²x³q(x) dx) = T(p) + T(q)  - b. T(αp) = (3(αp)(4) + 5(αp)'(6), ∫₋₁²x³(αp)(x) dx) - = (3α·p(4) + 5α·p'(6), α·∫₋₁²x³p(x) dx) - = α(3p(4) + 5p'(6), ∫₋₁²x³p(x) dx) = αT(p)  - Therefore, when b = c = 0, T is linear. - Assume T is linear. We will prove that b = c = 0 by contradiction. Suppose at least one of b or c is non-zero.  **Case 1: b ≠ 0** - Consider the polynomials p(x) = x and q(x) = 1. - T(p + q) = (3(p + q)(4) + 5(p + q)'(6) + b(p + q)(1)(p + q)(2), ∫₋₁²x³(p + q)(x) dx + c sin(p + q)(0)) - Claude errors? or my errors? - by hand: 3((x+1)(4)) + 5(x+1)'(6) should be - 3(5)+ 5 (1) = 15 + 5=20; how could it get 23? - b(x+1)(1) (x+1)(2) = b(2)(3)= 6b, not 5b; - integral is 15/4 if p is 1; if it's x, then it's integral from -1 to 2 of x^4, or x^5/5 from -1 to 2, or 32/5 - -1/5 = 33/5 - = (23 + 5b, 27/4 + c)  - T(p) + T(q) = (3p(4) + 5p'(6) + bp(1)p(2), ∫₋₁²x³p(x) dx + c sin p(0)) + (3q(4) + 5q'(6) + bq(1)q(2), ∫₋₁²x³q(x) dx + c sin q(0)) - = (23, 27/4 + c) + (3 + b, c) - = (26 + b, 27/4 + 2c)  - Since b ≠ 0, T(p + q) ≠ T(p) + T(q), contradicting the linearity of T.  **Case 2: c ≠ 0** - Consider the polynomial p(x) = 1 and the scalar α = 2. - T(αp) = (3(αp)(4) + 5(αp)'(6) + b(αp)(1)(αp)(2), ∫₋₁²x³(αp)(x) dx + c sin(αp)(0)) - = (6 + 4b, c) αT(p) = α(3p(4) + 5p'(6) + bp(1)p(2), ∫₋₁²x³p(x) dx + c sin p(0)) = (6 + 4b, 2c)  - Since c ≠ 0, T(αp) ≠ αT(p), contradicting the linearity of T. Therefore, if T is linear, both b and c must be equal to 0. In conclusion, T is linear if and only if b = c = 0. ## Claude gets right answer, but wrong details?