Question

In each case, find \(P=P_{B_{0} \leftarrow B}\) and verify that \(P^{-1} M_{B_{0}}(T) P=M_{B}(T)\) for the given operator \(T\). a. \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}, T(a, b, c)=(2 a-b, b+c, c-3 a) ;\) \(B_{0}=\\{(1,1,0),(1,0,1),(0,1,0)\\}\) and \(B\) is the standard basis. b. \(T: \mathbf{P}_{2} \rightarrow \mathbf{P}_{2},\) \(\quad T\left(a+b x+c x^{2}\right)=(a+b)+(b+c) x+(c+a) x^{2}\) \(B_{0}=\left\\{1, x, x^{2}\right\\}\) and \(B=\left\\{1-x^{2}, 1+x, 2 x+x^{2}\right\\}\) c. \(T: \mathbf{M}_{22} \rightarrow \mathbf{M}_{22}\) \(T\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{ll}a+d & b+c \\ a+c & b+d\end{array}\right]\) \(B_{0}=\left\\{\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]\right\\}\) and \(B=\left\\{\left[\begin{array}{ll}1 & 1 \\\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 1 & 1\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]\right\\}\)

Short answer

Find transition matrix and verify for each operator.

Step-by-step solution

Understanding the Problem

We need to find the transition matrix \( P_{B_{0} \leftarrow B} \) for each given operator \( T \) and verify the relation \( P^{-1} M_{B_{0}}(T) P = M_{B}(T) \). We'll approach this by first finding the matrix representations for the operator in both bases and then finding the transition matrix.

Finding Matrix Representations

For each part, we will find the matrix representation of \( T \) using the bases \( B_{0} \) and \( B \). This involves expressing the images of basis vectors in the chosen bases.

Example (a): Operator on \( \mathbb{R}^3 \)

Using the transformation \( T(a, b, c) = (2a-b, b+c, c-3a) \) with \( B_{0} = \{(1,1,0), (1,0,1), (0,1,0)\} \) as a non-standard basis and \( B \) as the standard basis:- Evaluate \( T(1,0,0), T(0,1,0), T(0,0,1) \) to represent \( M_{B}(T) \).- Evaluate \( T(1,1,0), T(1,0,1), T(0,1,0) \) in terms of \( B_{0} \) to obtain \( M_{B_{0}}(T) \).

Constructing Transition Matrix \( P \)

The transition matrix \( P_{B_{0} \leftarrow B} \) is determined by expressing the elements of \( B \) in terms of \( B_{0} \). Write each standard basis vector as a linear combination of the basis vectors in \( B_{0} \).

Verify the Matrix Equation

Compute \( P^{-1}M_{B_{0}}(T)P \) and check if it equals \( M_{B}(T) \). Verify through matrix multiplication.

Continue with Other Examples

For case (b), use polynomials and for case (c), use matrices. Repeat Steps 3 to 5 for these cases.

Key concepts

Transition Matrix

The transition matrix is a fundamental concept in linear algebra. It is primarily used to convert vector coordinates from one basis to another within a vector space. In simpler terms, if you have a vector expressed in one basis, the transition matrix helps to express it in terms of another basis.

To find the transition matrix from a basis \( B \) to another basis \( B_0 \), you need to express each vector in \( B \) as a linear combination of vectors in \( B_0 \). The coefficients of these linear combinations form the columns of the transition matrix. For example, in problem (a), when converting from the standard basis to \( B_0 \), you calculate the coefficients that express the standard basis vectors in terms of the non-standard basis vectors provided by \( B_0 \).

This matrix allows seamless transformations between different basis representations, helping simplify complex linear transformations by providing a systematic way to change coordinate representations.

Matrix Representation

Matrix representation is a convenient way to handle linear transformations in vector spaces. For any linear operator \( T \), its action on a vector can be represented as a matrix product of \( T \)'s matrix form and the vector in question.

Essentially, a matrix representation \( M(T) \) of an operator \( T \) depends on the choice of a basis. If you have a basis \( B \), and a set of transformed vectors resulting from \( T \), the columns of \( M(T) \) will represent these vectors as combinations of the basis vectors.

In the given exercises, you create matrices \( M_{B}(T) \) and \( M_{B_0}(T) \) by calculating how \( T \) acts on each basis vector, either returning a standard coordinate form or as a coordinate transformation relative to a specific basis set. This facilitates both computation and analysis of linear transformations.

Basis Change

Basis change in a vector space refers to the process of converting the representation of a vector from one basis to another. This is crucial in many calculations as it can simplify the representation and calculation of linear operators.

When performing a basis change, it involves determining how each vector of the original basis expresses in terms of the new basis. The transition matrix plays a pivotal role here, as it provides the means by which these calculations can be consistently carried out.

In the context of the exercise, verifying that \( P^{-1} M_{B_0}(T) P = M_{B}(T) \) is a means of ensuring that the transformation behaves consistently across different bases. This checks the correctness and consistency of the change of basis and allows for a double-check of matrix computations.

Operator on Vector Space

An operator on a vector space is a function that takes any vector from a vector space and maps it onto another vector within the same or another vector space. In mathematical terms, it's a function \( T: V \to V \), where \( V \) is some vector space.

In linear algebra, these operators can be represented using matrices, allowing for operations like addition and scalar multiplication to maintain certain algebraic properties. In the exercises provided, various operators were given in terms of transformations, such as those acting on polynomials or matrices.

Understanding these operators in terms of their matrix representations provides valuable insight into their behavior and properties. This allows for a more intuitive grasp of complex transformations and leads to more straightforward computational strategies, forming the backbone for more advanced applications in differential equations, quantum mechanics, and beyond.