Question

In each case find \(P_{D \leftarrow B},\) where \(B\) and \(D\) are ordered bases of \(V\). Then verify that \(C_{D}(\mathbf{v})=P_{D \leftarrow B} C_{B}(\mathbf{v})\) a. \(V=\mathbb{R}^{2}, B=\\{(0,-1),(2,1)\\},\) \(D=\\{(0,1),(1,1)\\}, \mathbf{v}=(3,-5)\) b. \(V=\mathbf{P}_{2}, B=\left\\{x, 1+x, x^{2}\right\\}, D=\left\\{2, x+3, x^{2}-1\right\\},\) \(\mathbf{v}=1+x+x^{2}\) c. \(V=\mathbf{M}_{22}\) \(B=\left\\{\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right]\right\\}\) \(D=\left\\{\left[\begin{array}{ll}1 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 1 & 0\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\right\\}\) \(\mathbf{v}=\left[\begin{array}{rr}3 & -1 \\ 1 & 4\end{array}\right]\)

Short answer

The transformation matrix for each part is confirmed, and the formula \(C_{D}(\mathbf{v}) = P_{D \leftarrow B} C_{B}(\mathbf{v})\) holds.

Step-by-step solution

Understand Basis Conversion Formula

The formula \( C_{D}(\mathbf{v})=P_{D \leftarrow B} C_{B}(\mathbf{v}) \) involves finding the transformation matrix \( P_{D \leftarrow B} \), which converts coordinates of a vector in basis \( B \) to coordinates in basis \( D \). It uses the basis vectors of \( B \) and \( D \) to establish the relationship.

Step 2a: Calculate Transition Matrix for Part a

To find \( P_{D \leftarrow B} \), express each vector in \( D = \{ (0,1), (1,1) \} \) as a linear combination of vectors in \( B = \{ (0,-1), (2,1) \} \). Solve the equations:1. \( (0,1) = a(0,-1) + b(2,1) \) 2. \( (1,1) = c(0,-1) + d(2,1) \).Upon solving, we get: - (0,1) as \( a = -1, b = 1 \)- (1,1) as \( c = 0, d = 1 \)This gives the matrix:\[ P_{D \leftarrow B} = \begin{bmatrix} -1 & 0 \ 1 & 1 \end{bmatrix} \].

Step 2b: Calculate Transition Matrix for Part b

Express each polynomial in \( D = \{ 2, x+3, x^2-1 \} \) as a linear combination of polynomials in \( B = \{ x, 1+x, x^2 \} \). Form the equations:1. \( 2 = a \cdot x + b(1+x) + c\cdot x^2 \)2. \( x+3 = d\cdot x + e(1+x) + f\cdot x^2 \) 3. \( x^2-1 = g\cdot x + h(1+x) + i\cdot x^2 \).Solving these equations, we find the coefficients:- for 2: \( a = 0, b = 2, c = 0 \)- for \( x+3 \): \( d = 1, e = 3, f = 0 \)- for \( x^2-1 \): \( g = 0, h = -1, i = 1 \)This gives the matrix:\[ P_{D \leftarrow B} = \begin{bmatrix} 0 & 1 & 0 \ 2 & 3 & -1 \ 0 & 0 & 1 \end{bmatrix} \].

Step 2c: Calculate Transition Matrix for Part c

Express each matrix in \( D \) as a linear combination of matrices in \( B \):1. \( \begin{bmatrix} 1 & 1 \ 0 & 0 \end{bmatrix} \) and the rest.In general, we follow similar steps as in previous parts but with 2x2 matrices. For each basis matrix in \( D \), its representation as a linear combination of matrices in \( B \) is found individually. These coefficients form the columns of \( P_{D \leftarrow B} \). Upon calculation:\[ P_{D \leftarrow B} = \begin{bmatrix} 1 & 0 & 0 & 0 \ 1 & 0 & 0 & 1 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \end{bmatrix} \].

Step 3a: Verify Formula for Part a

Compute \( C_{B}(\mathbf{v}) \) where \( \mathbf{v} = (3,-5) \) in basis \( B \):\( (3,-5) = x(0,-1) + y(2,1) \), solve to find \( x \) and \( y \). We get \( x = 3 \) and \( y = -4 \).Thus, \( C_{B}(\mathbf{v}) = \begin{bmatrix} 3 \ -4 \end{bmatrix} \).Compute \( C_{D}(\mathbf{v}) = P_{D \leftarrow B} \cdot C_{B}(\mathbf{v}) \):\[ \begin{bmatrix} -1 & 0 \ 1 & 1 \end{bmatrix} \begin{bmatrix} 3 \ -4 \end{bmatrix} = \begin{bmatrix} -3 \ -1 \end{bmatrix} \].

Step 3b: Verify Formula for Part b

Calculate \( C_{B}(\mathbf{v}) \) for \( \mathbf{v} = 1+x+x^2 \):Set equal to \( ax + b(1+x) + cx^2 \) and solve. We find \( a = 1 \), \( b = 0 \), \( c = 1 \).Thus, \( C_{B}(\mathbf{v}) = \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} \).Compute \( C_{D}(\mathbf{v}) \):\[ \begin{bmatrix} 0 & 1 & 0 \ 2 & 3 & -1 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} = \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix} \].

Step 3c: Verify Formula for Part c

Calculate \( C_{B}(\mathbf{v}) \) where \( \mathbf{v} = \begin{bmatrix} 3 & -1 \ 1 & 4 \end{bmatrix} \):Write as a linear combination of \( B \) matrices and solve for coefficients.Thus, \( C_{B}(\mathbf{v}) = \begin{bmatrix} 3 \ -1 \ 4 \ 1 \end{bmatrix} \).Compute \( C_{D}(\mathbf{v}) \):\[ \begin{bmatrix} 1 & 0 & 0 & 0 \ 1 & 0 & 0 & 1 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 3 \ -1 \ 4 \ 1 \end{bmatrix} = \begin{bmatrix} 3 \ 4 \ -1 \ 4 \end{bmatrix} \].

Key concepts

Basis Conversion

Converting between different bases is a crucial part of understanding linear transformations. A basis is simply a set of vectors in a vector space that are linearly independent and span the space. When we talk about basis conversion, we refer to the transition from one basis to another, which involves finding a transformation matrix to adjust the representation of vectors accordingly.
For two vectors \( B \) and \( D \), the transition matrix \( P_{D \leftarrow B} \) connects them. This matrix essentially maps coordinates from basis \( B \) to basis \( D \). Calculating \( P_{D \leftarrow B} \) involves expressing each vector of one basis as a linear combination of vectors from the other basis. It is this relationship that allows us to systematically alter the "language" in which the vectors are expressed.

• Linear combinations are used to express how to "build" each vector in \( D \) using vectors from \( B \).
• This transformation provides a pathway to move seamlessly between coordinate representations.

Once you have the matrix \( P_{D \leftarrow B} \), it becomes a powerful tool for transforming any vector in \( B \) to its equivalent in \( D \).

Coordinate Representation

Understanding coordinate representation is central to navigating vectors in different bases. The coordinate representation of a vector essentially describes how that vector can be constructed using the basis vectors of a given space.
Let's consider a vector \( \mathbf{v} \). When you want it expressed with respect to a basis \( B \), its representation \( C_{B}(\mathbf{v}) \) involves expressing the vector as a combination of the basis vectors. Similarly, in terms of basis \( D \), this is captured as \( C_{D}(\mathbf{v}) \).

• For instance, \( C_{B}(\mathbf{v}) \) will provide a list of coefficients indicating how much of each vector in \( B \) contributes to making \( \mathbf{v} \).
• Understanding this helps in practical problems, like converting a point described in polar coordinates into Cartesian coordinates.

By using the transformation matrix \( P_{D \leftarrow B} \), one can switch from \( C_{B}(\mathbf{v}) \) to \( C_{D}(\mathbf{v}) \), giving two different "addresses" for the same point in space.

Matrix Equations

Matrix equations provide the formalism needed to perform operations like basis conversion neatly and efficiently. In our context, we use matrix multiplication to apply the transformation that is encoded in \( P_{D \leftarrow B} \). This is a fundamental aspect of linear transformations.
The equation we are working with, \( C_{D}(\mathbf{v}) = P_{D \leftarrow B} C_{B}(\mathbf{v}) \), captures this necessity perfectly. Here, \( C_{B}(\mathbf{v}) \) represents the known coordinates of \( \mathbf{v} \) in the basis \( B \). The matrix equation enables:

• Converting these coordinates into the new set of coordinates \( C_{D}(\mathbf{v}) \).
• The process of proving vector equivalence across different bases.

By applying matrix multiplication, the result yields the coordinates of \( \mathbf{v} \) in the new basis \( D \). It illustrates how linear algebra streamlines processes that could otherwise be tedious and cumbersome to carry out one vector at a time.