Question

In each case, show that \(M_{D B}(T)\) is invertible and use the fact that \(M_{B D}\left(T^{-1}\right)=\left[M_{B D}(T)\right]^{-1}\) to determine the action of \(T^{-1}\). a. \(T: \mathbf{P}_{2} \rightarrow \mathbb{R}^{3}, T\left(a+b x+c x^{2}\right)=(a+c, c, b-c) ;\) \(B=\left\\{1, x, x^{2}\right\\}, D=\) standard b. \(T: \mathbf{M}_{22} \rightarrow \mathbb{R}^{4}\) \(T\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=(a+b+c, b+c, c, d)\) \(B=\left\\{\begin{array}{l}\left.\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]\right\\} \\ D=\text { standard }\end{array}\right.\)

Short answer

Both transformation matrices are invertible, and their respective inverse actions are represented by their inverse matrices.

Step-by-step solution

Determine Basis Transformation Matrix for Part a

For the linear transformation \(T: \mathbf{P}_2 \rightarrow \mathbb{R}^3\) defined by \(T(a + bx + cx^2) = (a+c, c, b-c)\), we need to find the matrix \(M_{DB}(T)\) in the bases \(B = \{1, x, x^2\}\) and \(D = \{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}\), which is the standard basis for \(\mathbb{R}^3\). To do this, apply \(T\) to each element of \(B\) and express the result as a vector in \(D\).1. \(T(1) = (1, 0, 0)\) - The vector representation is \([1, 0, 0]^T\).2. \(T(x) = (0, 0, 1)\) - The vector representation is \([0, 0, 1]^T\).3. \(T(x^2) = (1, 1, -1)\) - The vector representation is \([1, 1, -1]^T\).The matrix \(M_{DB}(T)\) is formed by these columns: \[M_{DB}(T) = \begin{bmatrix} 1 & 0 & 1 \ 0 & 0 & 1 \ 0 & 1 & -1 \end{bmatrix}\]

Check Invertibility of Transformation Matrix for Part a

To determine if \(M_{DB}(T)\) is invertible, calculate its determinant.\[det(M_{DB}(T)) = \begin{vmatrix} 1 & 0 & 1 \ 0 & 0 & 1 \ 0 & 1 & -1 \end{vmatrix} = 1(0 \times (-1) - 1 \times 1) - 0 + 1(0 \times 0 - 0 \times 1) = -1 eq 0\]Since the determinant is non-zero, \(M_{DB}(T)\) is invertible.

Determine the Inverse Transformation for Part a

Use the relation \(M_{BD}(T^{-1}) = [M_{DB}(T)]^{-1}\) to find the inverse.First, find the inverse matrix \([M_{DB}(T)]^{-1}\). Utilizing the formula for the inverse of a 3x3 matrix or using computational tools:\[[M_{DB}(T)]^{-1} = \begin{bmatrix} 1 & 0 & 1 \ 0 & 0 & 1 \ 0 & 1 & -1 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & 1 & 1 \ 0 & -1 & 0 \ 0 & 1 & 0 \end{bmatrix}\]Thus, the action of \(T^{-1}\) is represented by this inverse matrix.

Determine Basis Transformation Matrix for Part b

For \(T: \mathbf{M}_{22} \rightarrow \mathbb{R}^4\) with \(T\left(\begin{bmatrix} a & b \ c & d \end{bmatrix}\right) = (a+b+c, b+c, c, d)\), determine \(M_{DB}(T)\) using the bases:- \(B\) is the set of matrices corresponding to the standard basis matrices for \(\mathbf{M}_{22}\).- \(D\) is the standard basis for \(\mathbb{R}^4\).1. \(T\left(\begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}\right) = (1, 0, 0, 0)\)2. \(T\left(\begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}\right) = (1, 1, 0, 0)\)3. \(T\left(\begin{bmatrix} 0 & 0 \ 1 & 0 \end{bmatrix}\right) = (1, 1, 1, 0)\)4. \(T\left(\begin{bmatrix} 0 & 0 \ 0 & 1 \end{bmatrix}\right) = (0, 0, 0, 1)\)Thus, the matrix is:\[M_{DB}(T) = \begin{bmatrix} 1 & 1 & 1 & 0 \ 0 & 1 & 1 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \end{bmatrix}\]

Check Invertibility of Transformation Matrix for Part b

To check if \(M_{DB}(T)\) is invertible, calculate its determinant. Given its upper triangular structure with determinant equal to the product of diagonal elements:\[det(M_{DB}(T)) = 1 \times 1 \times 1 \times 1 = 1 eq 0\]The matrix \(M_{DB}(T)\) is invertible since its determinant is non-zero.

Determine the Inverse Transformation for Part b

Find \(M_{BD}(T^{-1})\) which is \([M_{DB}(T)]^{-1}\):Compute \([M_{DB}(T)]^{-1}\):Using computational tools or formulas:\[[M_{DB}(T)]^{-1} = \begin{bmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & -1 & 1 & 0 \ 0 & 1 & 0 & 1 \end{bmatrix}\]This matrix represents the action of \(T^{-1}\).

Key concepts

Invertibility

In the realm of linear transformations, invertibility is an essential concept. When we say a transformation is invertible, we mean that there exists another transformation that can reverse its effects. In simpler terms, for a linear transformation \(T\) to be invertible, there must exist a transformation \(T^{-1}\) such that applying \(T\) followed by \(T^{-1}\) (or vice versa) ends up with no change at all.

To determine if a matrix represents an invertible transformation, we typically check its determinant. If the determinant is non-zero, the matrix has an inverse, therefore the transformation is invertible. This comes from the property that the existence of a matrix inverse hinges on a non-zero determinant. In both examples from the exercise, once we calculated the determinants of the matrices \(M_{DB}(T)\), they were non-zero, confirming that \(T\) is invertible.

Basis Transformation

Linear transformations between vector spaces often involve changing the basis of a vector. This is where the concept of a basis transformation comes in. A basis transformation matrix allows you to move from one basis to another, which can simplify problems and calculations.

Each vector space might have several possible bases, from which we can express any vector uniquely. In the exercise, you see two bases: \(B\) and \(D\). The basis transformation matrix holds the vectors of one basis expressed in terms of another. For example, \(M_{DB}(T)\) holds the transformation of each element of basis \(B\) expressed in basis \(D\).

• The role of a basis transformation matrix is essential in effectively translating the action of a transformation into a clear mathematical form.
• It becomes a bridge for understanding how different bases relate within the linear transformation.

By understanding this matrix, you can switch between different vector space descriptions seamlessly.

Matrix Inversion

Once identifying a matrix as invertible, such as \(M_{DB}(T)\), the next task is often to find its inverse. Matrix inversion involves returning to an identity state when the matrix is multiplied by its inverse.

For a 3x3 or larger matrix, computing the inverse can be challenging. The inverse of a matrix, \(A\), is typically found using computational tools or formula approaches like the adjugate matrix method. In the exercise, after identifying the invertibility of \(M_{DB}(T)\), computing \([M_{DB}(T)]^{-1}\) is necessary to guide us on \(T^{-1}\).

• The computation of the inverse is central to understanding inverse transformations, as it reflects reversing the linear transformation's action.
• Good knowledge of matrix properties expedites finding inverses, aiding in exploring the nature and effects of transformations.

Determinant Calculation

Determinant calculation is a powerful tool in linear algebra. It reveals a wealth of information about matrices and the transformations they represent. Generally, the determinant gives us insights into the area or volume scaling factor imposed by the transformation and helps in understanding the invertibility of the matrix.

To find the determinant, particularly in 2x2 and 3x3 matrices, simple arithmetic operations can be applied. For instance, with a 3x3 matrix, it involves calculating a mixture of products and sums using its elements. In the exercise, the determinant of either \(M_{DB}(T)\) matrices was calculated to be non-zero, confirming the invertibility.

• A non-zero determinant confirms that the linear transformation does not squish the entire space onto a lower dimension.
• Determinant value sign can indicate if the transformation reverses orientation.

Knowing how to calculate and interpret determinants is hugely beneficial in solving linear transformation problems.