Let's analyze the given transformation:

Transformation $T: P_2(\mathbb{R}) \to \mathbb{R}^2$, defined by:

$T(p(x)) = (p(0), p'(0))$

where $P_2(\mathbb{R})$ represents the space of polynomials of degree at most 2, and $\mathbb{R}^2$ is the two-dimensional Euclidean space.

Step 1: Understanding the mapping

• Domain: $P_2(\mathbb{R})$, the space of polynomials of degree at most 2, can be written as:

  $p(x) = a_0 + a_1x + a_2x^2$ where $a_0, a_1, a_2 \in \mathbb{R}$.

• Action of T: The transformation $T(p(x))$ takes a polynomial $p(x)$ and maps it to a vector in $\mathbb{R}^2$, consisting of:

  1. $p(0)$, which is the constant term of the polynomial $a_0$.
  2. $p'(0)$, which is the value of the derivative of the polynomial at $x = 0$, and equals $a_1$, the coefficient of $x$ in the polynomial.

Thus, we can express the output of $T(p(x))$ as: $T(p(x)) = (a_0, a_1)$

Step 2: Injectivity

A transformation is injective if different inputs map to different outputs, meaning that if $T(p(x)) = T(q(x))$, then $p(x) = q(x)$.

• Let $p(x) = a_0 + a_1x + a_2x^2$ and $q(x) = b_0 + b_1x + b_2x^2$.
• $T(p(x)) = T(q(x))$ implies: $(a_0, a_1) = (b_0, b_1)$ Therefore, $a_0 = b_0$ and $a_1 = b_1$, but this doesn't necessarily imply that $a_2 = b_2$. The value of $a_2$ and $b_2$ does not affect the output of $T$, because the second derivative of $p(x)$ and $q(x)$ is not involved in the mapping.

Thus, T is not injective, because two different polynomials (differing in their $a_2$-terms) could map to the same vector in $\mathbb{R}^2$.

Step 3: Surjectivity

A transformation is surjective if every element in the target space $\mathbb{R}^2$ has a preimage in $P_2(\mathbb{R})$.

• Given any point $(b_0, b_1) \in \mathbb{R}^2$, we need to find a polynomial $p(x) \in P_2(\mathbb{R})$ such that $T(p(x)) = (b_0, b_1)$.
• We can choose a polynomial $p(x) = b_0 + b_1x$, which clearly satisfies: $p(0) = b_0, \quad p'(0) = b_1$ Hence, for every pair $(b_0, b_1) \in \mathbb{R}^2$, we can find a polynomial in $P_2(\mathbb{R})$ that maps to it.

Thus, T is surjective, because every element of $\mathbb{R}^2$ is the image of some polynomial in $P_2(\mathbb{R})$.

Step 4: Bijectivity

A transformation is bijective if it is both injective and surjective. Since we've established that $T$ is surjective but not injective, T is not bijective.

Final Answer:

• T is surjective.
• T is not injective.
• T is not bijective.

Would you like more details or have any further questions?

Related questions:

1. How can we determine if a linear transformation is injective?
2. What would happen if we modified the transformation to include higher derivatives?
3. Can you describe other examples of surjective transformations in linear algebra?
4. How do the concepts of rank and nullity relate to injectivity and surjectivity?
5. What is the null space of this transformation?

Tip: To check if a transformation is injective, see if the kernel (null space) of the transformation contains only the zero vector.