1. **Stating the problem:** We have a linear transformation $$T : \mathbb{P}_2(\mathbb{R}) \to \mathbb{P}_3(\mathbb{R})$$ defined by $$T(p(x)) = p'(x) + \int_0^x p(t) \, dt,$$ where $$\mathbb{P}_n(\mathbb{R})$$ is the space of real polynomials of degree at most $$n$$. We want to verify which statements (a, b, c, d, e) about $$T$$ and its standard matrix representations are true. 2. **Find the matrix representation of $$T$$ relative to the bases $$\{1, x, x^2\}$$ and $$\{1, x, x^2, x^3\}$$:** Let $$p(x) = a + bx + cx^2$$. - Compute the derivative: $$p'(x) = b + 2cx.$$ - Compute the integral from 0 to $$x$$: $$\int_0^x p(t) \, dt = \int_0^x (a + bt + ct^2) \, dt = a x + \frac{b}{2} x^2 + \frac{c}{3} x^3.$$ - So, $$T(p(x)) = p'(x) + \int_0^x p(t) \, dt = (b) + (2c x) + \left(a x + \frac{b}{2} x^2 + \frac{c}{3} x^3\right) = b + (2c + a) x + \frac{b}{2} x^2 + \frac{c}{3} x^3.$$ 3. **Express this as a vector with respect to the $$\{1, x, x^2, x^3\}$$ basis:** $$T(p(x)) = \begin{pmatrix} b \\ a + 2c \\ \frac{b}{2} \\ \frac{c}{3} \end{pmatrix}.$$ 4. **Find the matrix columns by applying $$T$$ to each basis vector of $$\mathbb{P}_2$$:** - $$T(1) = 0 + \int_0^x 1 \, dt = x = 0 \cdot 1 + 1 \cdot x + 0 \cdot x^2 + 0 \cdot x^3$$ gives the first column $$\begin{pmatrix}0 \\ 1 \\ 0 \\ 0\end{pmatrix}$$. - $$T(x) = 1 + \int_0^x t \, dt = 1 + \frac{x^2}{2} = 1 \cdot 1 + 0 \cdot x + \frac{1}{2} x^2 + 0 \cdot x^3$$ gives the second column $$\begin{pmatrix}1 \\ 0 \\ \frac{1}{2} \\ 0\end{pmatrix}$$. - $$T(x^2) = 2x + \int_0^x t^2 \, dt = 2x + \frac{x^3}{3} = 0 \cdot 1 + 2 \cdot x + 0 \cdot x^2 + \frac{1}{3} x^3$$ gives the third column $$\begin{pmatrix}0 \\ 2 \\ 0 \\ \frac{1}{3}\end{pmatrix}$$. 5. **Construct the standard matrix of $$T$$:** $$\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{3} \end{pmatrix}.$$ This matches option (a). 6. **Check injectivity (one-to-one):** The domain $$\mathbb{P}_2$$ has dimension 3, codomain $$\mathbb{P}_3$$ has dimension 4. - Suppose $$T(p) = 0$$. Then $$b = 0,$$ $$a + 2c = 0,$$ $$\frac{b}{2} = 0$$ (already zero), $$\frac{c}{3} = 0 \implies c=0.$$ From $$a + 2c = a = 0$$. So $$a=b=c=0$$ means only the zero polynomial maps to zero; hence, $$T$$ is injective. 7. **Check surjectivity (onto):** Since the codomain dimension (4) is greater than the domain dimension (3), $$T$$ cannot be onto. 8. **Is $$T$$ an isomorphism?** No, because it is not onto. 9. **Option (e) matrix check:** The matrix in (e) has $$\frac{1}{3}$$ replaced by $$\frac{1}{3}$$ with $$\frac{1}{3}$$ changed to $$\frac{1}{4}$$ in the last entry, which does not match our computed matrix, so (e) is false. **Summary:** - (a) True. - (b) True. - (c) False. - (d) False. - (e) False.