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Let \(\mathcal{D}=\left\\{\mathbf{d}_{1}, \mathbf{d}_{2}\right\\}\) and \(\mathcal{B}=\left\\{\mathbf{b}_{1}, \mathbf{b}_{2}\right\\}\) be bases for vector spaces \(V\) and \(W,\) respectively. Let \(T : V \rightarrow W\) be a linear transformation with the property that $$T\left(\mathbf{d}_{1}\right)=2 \mathbf{b}_{1}-3 \mathbf{b}_{2}, \quad T\left(\mathbf{d}_{2}\right)=-4 \mathbf{b}_{1}+5 \mathbf{b}_{2}$$ Find the matrix for \(T\) relative to \(\mathcal{D}\) and \(\mathcal{B}\)

Short Answer

Expert verified
The matrix for \( T \) is \( \begin{bmatrix} 2 & -4 \\ -3 & 5 \end{bmatrix} \).

Step by step solution

01

Understand the Given Data

The bases \( \mathcal{D} = \{ \mathbf{d}_1, \mathbf{d}_2 \} \) for vector space \( V \) and \( \mathcal{B} = \{ \mathbf{b}_1, \mathbf{b}_2 \} \) for vector space \( W \) are given. The transformation \( T: V \rightarrow W \) is described by \( T(\mathbf{d}_1) = 2 \mathbf{b}_1 - 3 \mathbf{b}_2 \) and \( T(\mathbf{d}_2) = -4 \mathbf{b}_1 + 5 \mathbf{b}_2 \). You need to find the matrix representation of \( T \) with respect to the bases \( \mathcal{D} \) and \( \mathcal{B} \).
02

Identify Matrix Columns

Since \( T(\mathbf{d}_1) = 2 \mathbf{b}_1 - 3 \mathbf{b}_2 \), the first column of the matrix \( [T]_{\mathcal{D}\mathcal{B}} \) is \( \begin{bmatrix} 2 \ -3 \end{bmatrix} \). Similarly, since \( T(\mathbf{d}_2) = -4 \mathbf{b}_1 + 5 \mathbf{b}_2 \), the second column of the matrix is \( \begin{bmatrix} -4 \ 5 \end{bmatrix} \).
03

Construct the Transformation Matrix

The transformation matrix \( [T]_{\mathcal{D}\mathcal{B}} \) is constructed by using the identified columns in the previous step. Therefore, the matrix is \( \begin{bmatrix} 2 & -4 \ -3 & 5 \end{bmatrix} \).
04

Confirm Matrix Properties

Verify the linear transformation properties by applying the matrix to \( \mathbf{d}_1 \) and \( \mathbf{d}_2 \) to ensure that results match the original \( T(\mathbf{d}_1) \) and \( T(\mathbf{d}_2) \), confirming the solution is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Bases
A vector base is a fundamental concept in linear algebra. It provides a way to represent any vector within a vector space uniquely as a linear combination of basis vectors. This concept allows us to understand the structure of vector spaces and perform operations on vectors.
A base for a vector space is a set of vectors that is both linearly independent and spans the entire space. "Linearly independent" means that no vector in the base can be expressed as a linear combination of the others. "Spans the space" implies that any vector in the vector space can be written as a combination of these basis vectors.
  • For example, in two-dimensional space, the standard basis vectors are typically denoted as \( \mathbf{i} \) and \( \mathbf{j} \). In the three-dimensional case, it's \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \).
  • In our exercise, the sets \( \mathcal{D} = \{ \mathbf{d}_1, \mathbf{d}_2 \} \) and \( \mathcal{B} = \{ \mathbf{b}_1, \mathbf{b}_2 \} \) serve as bases for their respective vector spaces \( V \) and \( W \).
By studying and manipulating bases, we can simplify complex vector transformations, such as finding the matrix representation of a linear transformation.
Matrix Representation
Matrix representation is a powerful tool in linear algebra. It allows us to express linear transformations in a compact form that's easy to work with mathematically. A matrix essentially encodes the action of a linear transformation on certain bases of vector spaces.
To find a matrix representation for a linear transformation \( T \) between vector spaces, we determine how \( T \) acts on the basis vectors of the domain space and express those images in terms of the basis vectors of the codomain space. This results in a matrix where each column represents the image of a basis vector under the transformation:
  • The transformation matrix \( [T]_{\mathcal{D}\mathcal{B}} \) is formed by stacking these column vectors.
  • Given \( T(\mathbf{d}_1) = 2 \mathbf{b}_1 - 3 \mathbf{b}_2 \), the first column becomes \( \begin{bmatrix} 2 \ -3 \end{bmatrix} \).
  • For \( T(\mathbf{d}_2) = -4 \mathbf{b}_1 + 5 \mathbf{b}_2 \), the second column is \( \begin{bmatrix} -4 \ 5 \end{bmatrix} \).
  • Thus, the matrix \( [T]_{\mathcal{D}\mathcal{B}} \) is \( \begin{bmatrix} 2 & -4 \ -3 & 5 \end{bmatrix} \).
This matrix representation is crucial for simplifying calculations and visualizing linear transformations.
Linear Algebra Concepts
Linear algebra involves the study of vectors, vector spaces, and linear transformations, and is a foundational subject in many areas of mathematics and applied sciences. Here are some core concepts:
**Linear Transformation**: A function \( T: V \to W \) that maps between vector spaces and preserves vector addition and scalar multiplication. If \( T \) is linear, then for any vectors \( \mathbf{u}, \mathbf{v} \) and a scalar \( c \):
  • \( T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \)
  • \( T(c\mathbf{u}) = cT(\mathbf{u}) \)

**Basis and Dimension**: The basis is a minimal set of vectors that spans a vector space, while dimension counts the number of vectors in that basis. The concepts help us understand the "size" and nature of vector spaces.
Linear transformations can often be captured and manipulated using matrices, which make computation easier.
  • In the exercise, understanding how the transformation \( T \) acts on the basis helps construct its matrix representation.
  • This highlights the efficiency and power of matrices in capturing systematic linear mappings.
Mastery of these concepts allows clearer insights into many mathematical and practical problems.

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Most popular questions from this chapter

In Exercises \(3-6,\) solve the initial value problem \(\mathbf{x}^{\prime}(t)=A \mathbf{x}(t)\) for \(t \geq 0,\) with \(\mathbf{x}(0)=(3,2) .\) Classify the nature of the origin as an attractor, repeller, or saddle point of the dynamical system described by \(\mathbf{x}^{\prime}=A \mathbf{x}\) . Find the directions of greatest attraction and/or repulsion. When the origin is a saddle point, sketch typical traiectories. $$ A=\left[\begin{array}{rr}{-2} & {-5} \\ {1} & {4}\end{array}\right] $$

Suppose the eigenvalues of a \(3 \times 3\) matrix \(A\) are \(3,4 / 5,\) and \(3 / 5,\) with corresponding eigenvectors \(\left[\begin{array}{r}{1} \\ {0} \\\ {-3}\end{array}\right],\left[\begin{array}{r}{2} \\ {1} \\\ {-5}\end{array}\right],\) and \(\left[\begin{array}{r}{-3} \\ {-3} \\\ {7}\end{array}\right] .\) Let \(\mathbf{x}_{0}=\left[\begin{array}{r}{-2} \\\ {-5} \\ {3}\end{array}\right] .\) Find the solution of the equation \(\mathbf{x}_{k+1}=A \mathbf{x}_{k}\) for the specified \(\mathbf{x}_{0},\) and describe what happens as \(k \rightarrow \infty\)

Let \(A\) be an \(n \times n\) matrix, and suppose \(A\) has \(n\) real eigenvalues, \(\lambda_{1}, \ldots, \lambda_{n},\) repeated according to multiplicities, so that \(\operatorname{det}(A-\lambda I)=\left(\lambda_{1}-\lambda\right)\left(\lambda_{2}-\lambda\right) \cdots\left(\lambda_{n}-\lambda\right)\) Explain why det \(A\) is the product of the \(n\) eigenvalues of \(A .\) (This result is true for any square matrix when complex eigenvalues are considered.)

In Exercises \(3-6,\) solve the initial value problem \(\mathbf{x}^{\prime}(t)=A \mathbf{x}(t)\) for \(t \geq 0,\) with \(\mathbf{x}(0)=(3,2) .\) Classify the nature of the origin as an attractor, repeller, or saddle point of the dynamical system described by \(\mathbf{x}^{\prime}=A \mathbf{x}\) . Find the directions of greatest attraction and/or repulsion. When the origin is a saddle point, sketch typical traiectories. $$ A=\left[\begin{array}{rr}{7} & {-1} \\ {3} & {3}\end{array}\right] $$

Let \(A=\left[\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right] .\) Use formula \((1)\) for a determinant (given before Example 2\()\) to show that det \(A=a d-b c\) Consider two cases: \(a \neq 0\) and \(a=0\)

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