**Example 2** The mapping \( T: P_2 \to P_2 \) defined by \[ T(a_0 + a_1t + a_2t^2) = a_1 + 2a_2t \] is a linear transformation. (Calculus students will recognize \( T \) as the differentiation operator.) **a.** Find the \( B \)-matrix for \( T \), when \( B \) is the basis \(\{1, t, t^2\}\). **b.** Verify that \([T(p)]_B = [T]_B [p]_B \) for each \( p \) in \( P_2 \). **Solution** **a.** Compute the images of the basis vectors: - \( T(1) = 0 \) - \( T(t) = 1 \) - \( T(t^2) = 2t \) Then write the \( B \)-coordinate vectors of \( T(1), T(t), \) and \( T(t^2) \) (which are found by inspection in this example) and place them together as the \( B \)-matrix for \( T \): \[ [T(1)]_B = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}, \quad [T(t)]_B = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad [T(t^2)]_B = \begin{bmatrix} 0 \\ 2 \\ 0 \end{bmatrix} \] The zero polynomial: A polynomial whose value is always 0. The polynomial whose value is always 1.

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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I'm looking at page 292 of the Linear Algebra and Its Applications 5th Edition Textbook by David C. Lay. Could someone break down Example 2 some more for me? I don't understand how T(1) = 0, T(t) = 1, and T(t^2) = 2t or which function we should look t for the mapping. The entire part a section of that Example 2 problem I need help with.

**Example 2**

The mapping \( T: P_2 \to P_2 \) defined by 

\[ T(a_0 + a_1t + a_2t^2) = a_1 + 2a_2t \]

is a linear transformation. (Calculus students will recognize \( T \) as the differentiation operator.)

**a.** Find the \( B \)-matrix for \( T \), when \( B \) is the basis \(\{1, t, t^2\}\).

**b.** Verify that \([T(p)]_B = [T]_B [p]_B \) for each \( p \) in \( P_2 \).

**Solution**

**a.** Compute the images of the basis vectors:

- \( T(1) = 0 \)
- \( T(t) = 1 \)
- \( T(t^2) = 2t \)

Then write the \( B \)-coordinate vectors of \( T(1), T(t), \) and \( T(t^2) \) (which are found by inspection in this example) and place them together as the \( B \)-matrix for \( T \):

\[
[T(1)]_B = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}, \quad [T(t)]_B = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad [T(t^2)]_B = \begin{bmatrix} 0 \\ 2 \\ 0 \end{bmatrix}
\]

The zero polynomial: A polynomial whose value is always 0.

The polynomial whose value is always 1.
Transcribed Image Text:**Example 2** The mapping \( T: P_2 \to P_2 \) defined by \[ T(a_0 + a_1t + a_2t^2) = a_1 + 2a_2t \] is a linear transformation. (Calculus students will recognize \( T \) as the differentiation operator.) **a.** Find the \( B \)-matrix for \( T \), when \( B \) is the basis \(\{1, t, t^2\}\). **b.** Verify that \([T(p)]_B = [T]_B [p]_B \) for each \( p \) in \( P_2 \). **Solution** **a.** Compute the images of the basis vectors: - \( T(1) = 0 \) - \( T(t) = 1 \) - \( T(t^2) = 2t \) Then write the \( B \)-coordinate vectors of \( T(1), T(t), \) and \( T(t^2) \) (which are found by inspection in this example) and place them together as the \( B \)-matrix for \( T \): \[ [T(1)]_B = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}, \quad [T(t)]_B = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad [T(t^2)]_B = \begin{bmatrix} 0 \\ 2 \\ 0 \end{bmatrix} \] The zero polynomial: A polynomial whose value is always 0. The polynomial whose value is always 1.
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