91Ó°ÊÓ

In linear algebra, the concept of basis change is fundamental when dealing with transformations. It involves expressing vectors from one basis in terms of another. This is essential for understanding how vectors change when represented in different coordinate systems.

For the exercise, we use two bases: \( \mathcal{B} = \{1+x, 1-x\} \) and \( C = \{1, x\} \). To facilitate the transformation, each vector in \( \mathcal{B} \) needs to be rewritten in terms of \( C \).

Here's how the transformation works for the basis vectors:

• \( 1+x \) becomes \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \) in basis \( C \).
• \( 1-x \) becomes \( \begin{bmatrix} 1 \ -1 \end{bmatrix} \) in basis \( C \).

This process lays the groundwork for applying the linear transformation, as we need a common basis to switch between different forms of polynomial expressions effectively.

Polynomial transformations are functions that map polynomials to other polynomials, under a specific rule. In our problem, the transformation \( T: \mathscr{P}_{1} \rightarrow \mathscr{P}_{1} \) modifies a linear polynomial \( a + bx \) to \( b - ax \).

This simple linear transformation flips coefficients with certain modifications. Applying it to the basis vectors:

• For \( 1+x \), \( T(1+x) = -x \)
• For \( 1-x \), \( T(1-x) = -1-x \)

These changes define the action of \( T \) on the polynomial space, allowing us to understand how the entire space is transformed.

The matrix representation of a transformation is crucial for computation and understanding of linear transformations. The goal is to express the transformation in matrix form relative to chosen bases.

The process involves finding the transformation matrix \([T]_{C \leftarrow B}\) which maps vectors from basis \( \mathcal{B} \) to basis \( C \). For the given transformation:

• The vector \( -x \) in terms of \( C \) is \( \begin{bmatrix} 0 \ -1 \end{bmatrix} \).
• The vector \( -1-x \) in terms of \( C \) is \( \begin{bmatrix} -1 \ -1 \end{bmatrix} \).

These column vectors create the transformation matrix: \[ \begin{bmatrix} 0 & -1 \ -1 & -1 \end{bmatrix} \].
This matrix is essential for efficiently mapping any vector from \( \mathcal{B} \) to \( C \) and subsequently applying the transformation \( T \).

Verifying theorems in linear algebra often involves checking the consistency of different methods of achieving the same result. In our case, we're confirming that the matrix operation aligns with a direct transformation computation.

Let's verify Theorem 6.26 using the vector \( \mathbf{v} = 4 + 2x \):
First, compute \( T(4+2x) \) directly:

• Transformation leads to \( 2 - 4x \).

Then, apply the matrix: express \( \mathbf{v} \) in \( \mathcal{B} \):

• \( \mathbf{v}_{\mathcal{B}} = \begin{bmatrix} 3 \ -1 \end{bmatrix} \)

Multiply with \([T]_{C \leftarrow B}\): \[ \begin{bmatrix} 0 & -1 \ -1 & -1 \end{bmatrix} \begin{bmatrix} 3 \ -1 \end{bmatrix} \] results in \( \begin{bmatrix} 1 \ -7 \end{bmatrix} \).
This vector in basis \( C \) corresponds to the polynomial \( 2 - 4x \), confirming that the matrix operation matches the direct transformation.