Questions: Define T: P2 → R^2 by T(p)=[p(0) p(1)]. For instance, if p(t)=3+5t+7t^2, then T(p)=[3 15]. a. Show that T is a linear transformation. [Hint: For arbitrary polynomials p, q in P2, compute T(p+q) and T(p).] b. Find a polynomial p in P2 that spans the kernel of T, and describe the range of T.

Solution Steps

a. To show that \( T \) is a linear transformation, we need to verify two properties: additivity and homogeneity. For additivity, compute \( T(\mathbf{p} + \mathbf{q}) \) and show it equals \( T(\mathbf{p}) + T(\mathbf{q}) \). For homogeneity, compute \( T(c\mathbf{p}) \) and show it equals \( cT(\mathbf{p}) \) for any scalar \( c \).

b. To find a polynomial that spans the kernel of \( T \), solve for \( \mathbf{p}(t) \) such that \( T(\mathbf{p}) = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \). The range of \( T \) is the set of all possible outputs, which can be described by evaluating \( T \) on a basis of \( \mathbb{P}_2 \).

To solve the given problem, we will follow the steps outlined in the question. Let's begin with the first part of the problem.

A transformation \( T: V \rightarrow W \) is linear if it satisfies the following two properties for all vectors \( \mathbf{u}, \mathbf{v} \in V \) and scalars \( c \in \mathbb{R} \):

1. Additivity: \( T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \)
2. Homogeneity: \( T(c\mathbf{u}) = cT(\mathbf{u}) \)

Let's verify these properties for the transformation \( T: \mathbb{P}_{2} \rightarrow \mathbb{R}^{2} \) defined by \( T(\mathbf{p}) = \begin{bmatrix} \mathbf{p}(0) \\ \mathbf{p}(1) \end{bmatrix} \).

Let \( \mathbf{p}(t) = a_0 + a_1 t + a_2 t^2 \) and \( \mathbf{q}(t) = b_0 + b_1 t + b_2 t^2 \) be arbitrary polynomials in \( \mathbb{P}_{2} \).

Then, \( \mathbf{p}(t) + \mathbf{q}(t) = (a_0 + b_0) + (a_1 + b_1)t + (a_2 + b_2)t^2 \).

Now, compute \( T(\mathbf{p} + \mathbf{q}) \):

\[ T(\mathbf{p} + \mathbf{q}) = \begin{bmatrix} (\mathbf{p} + \mathbf{q})(0) \\ (\mathbf{p} + \mathbf{q})(1) \end{bmatrix} = \begin{bmatrix} a_0 + b_0 \\ a_0 + a_1 + a_2 + b_0 + b_1 + b_2 \end{bmatrix} \]

Compute \( T(\mathbf{p}) + T(\mathbf{q}) \):

\[ T(\mathbf{p}) = \begin{bmatrix} a_0 \\ a_0 + a_1 + a_2 \end{bmatrix}, \quad T(\mathbf{q}) = \begin{bmatrix} b_0 \\ b_0 + b_1 + b_2 \end{bmatrix} \]

\[ T(\mathbf{p}) + T(\mathbf{q}) = \begin{bmatrix} a_0 + b_0 \\ (a_0 + a_1 + a_2) + (b_0 + b_1 + b_2) \end{bmatrix} = \begin{bmatrix} a_0 + b_0 \\ a_0 + a_1 + a_2 + b_0 + b_1 + b_2 \end{bmatrix} \]

Since \( T(\mathbf{p} + \mathbf{q}) = T(\mathbf{p}) + T(\mathbf{q}) \), the additivity property holds.

For a scalar \( c \) and polynomial \( \mathbf{p}(t) = a_0 + a_1 t + a_2 t^2 \), compute \( T(c\mathbf{p}) \):

\[ c\mathbf{p}(t) = ca_0 + ca_1 t + ca_2 t^2 \]

\[ T(c\mathbf{p}) = \begin{bmatrix} (c\mathbf{p})(0) \\ (c\mathbf{p})(1) \end{bmatrix} = \begin{bmatrix} ca_0 \\ ca_0 + ca_1 + ca_2 \end{bmatrix} \]

Compute \( cT(\mathbf{p}) \):

\[ cT(\mathbf{p}) = c \begin{bmatrix} a_0 \\ a_0 + a_1 + a_2 \end{bmatrix} = \begin{bmatrix} ca_0 \\ c(a_0 + a_1 + a_2) \end{bmatrix} = \begin{bmatrix} ca_0 \\ ca_0 + ca_1 + ca_2 \end{bmatrix} \]

Since \( T(c\mathbf{p}) = cT(\mathbf{p}) \), the homogeneity property holds.

Thus, \( T \) is a linear transformation.

The kernel of \( T \), denoted as \( \ker(T) \), is the set of all polynomials \( \mathbf{p} \in \mathbb{P}_{2} \) such that \( T(\mathbf{p}) = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \).

Let \( \mathbf{p}(t) = a_0 + a_1 t + a_2 t^2 \). Then:

\[ T(\mathbf{p}) = \begin{bmatrix} a_0 \\ a_0 + a_1 + a_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]

This gives the system of equations:

1. \( a_0 = 0 \)
2. \( a_0 + a_1 + a_2 = 0 \)

From the first equation, \( a_0 = 0 \). Substituting into the second equation gives:

\[ 0 + a_1 + a_2 = 0 \quad \Rightarrow \quad a_1 = -a_2 \]

Thus, a polynomial in the kernel is of the form \( \mathbf{p}(t) = 0 + a_1 t + (-a_1) t^2 = a_1(t - t^2) \).

A polynomial that spans the kernel is \( \mathbf{p}(t) = t - t^2 \).

The range of \( T \), denoted as \( \text{Range}(T) \), is the set of all possible outputs of \( T \).

For \( \mathbf{p}(t) = a_0 + a_1 t + a_2 t^2 \), we have:

\[ T(\mathbf{p}) = \begin{bmatrix} a_0 \\ a_0 + a_1 + a_2 \end{bmatrix} \]

The range of \( T \) is all vectors of the form \( \begin{bmatrix} a_0 \\ a_0 + a_1 + a_2 \end{bmatrix} \), which is a subspace of \( \mathbb{R}^2 \).

To find a basis for the range, consider the vectors:

• \( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \) from \( \mathbf{p}(t) = 1 \)
• \( \begin{bmatrix} 0 \\ 1 \end{bmatrix} \) from \( \mathbf{p}(t) = t + t^2 \)

These vectors are linearly independent and span the range of \( T \).

Final Answer