Chapter 6: Problem 23
Determine whether the linear transformation \(\mathrm{T}\) is invertible by considering its matrix with respect to the standard bases. If \(T\) is invertible, use Theorem 6.28 and the method of Example 6.82 to find \(T^{-1}\). \(T: \mathscr{P}_{2} \rightarrow \mathscr{P}_{2}\) defined by \(T(p(x))=p(x)+p^{\prime}(x)\)
Short Answer
Step by step solution
Determine the Standard Matrix A
Find the Determinant of A
Calculate the Inverse of A
Express T^{-1}
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Standard Matrix
For the transformation \( T: \mathscr{P}_2 \to \mathscr{P}_2 \) defined as \( T(p(x))=p(x)+p'(x) \), we consider polynomials up to the second degree. This leads to forming a standard matrix that represents this transformation.
The general polynomial \( p(x) = a_0 + a_1 x + a_2 x^2 \) transforms under \( T \) to \( a_0 + (2a_1 + 2a_2)x + a_2 x^2 \). By comparing coefficients, we see how each component of the polynomial is transformed.
The corresponding standard matrix is:
\[A = \begin{pmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 1 & 1 \end{pmatrix}.\]
Understanding this matrix allows us to work with polynomials in a structured, linear fashion, providing the groundwork to examine properties like invertibility.
Determinant
In our exercise, for matrix \( A \):
\[\text{det}(A) = \begin{vmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 1 & 1 \end{vmatrix} = 1(2 \times 1 - 0) - 0 + 0 = 2.\]
The determinant here is 2, which indicates that the matrix is invertible. This means the linear transformation \( T \) is also invertible. The non-zero determinant directly leads us to explore the matrix inverse, which is essential for determining the inverse of the transformation.
Polynomial Transformation
Given the polynomial \( p(x) = a_0 + a_1 x + a_2 x^2 \), its derivative is \( p'(x) = a_1 + 2a_2 x \). The transformation \( T(p(x)) = p(x) + p'(x) \) yields:
- The constant term \( a_0 \) remains unchanged.
- The coefficient for \( x^1 \) becomes \( 2a_1 + 2a_2 \).
- The coefficient for \( x^2 \) stays \( a_2 \).
Matrix Inverse
If a matrix is invertible, meaning its determinant is non-zero, it has an inverse. For the specified matrix \( A \):
\[A^{-1} = \begin{pmatrix} 1 & 0 & 0 \ 0 & 0.5 & 0 \ 0 & -0.5 & 1 \end{pmatrix}.\]
This inverse matrix is used to define the inverse of the transformation \( T \), symbolized as \( T^{-1} \). By applying \( A^{-1} \) to the transformed polynomial, we retrieve the original polynomial components.
For polynomial \( q(x) = b_0 + b_1 x + b_2 x^2 \), the inverse transformation gives:
- The constant term remains \( b_0 \).
- The coefficient for \( x^1 \) is adjusted to \( 0.5 b_1 \).
- For \( x^2 \), the coefficient becomes \( b_2 - 0.5b_1 \).