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Let \(A=\left[\begin{array}{cc}{2} & {0} \\ {0} & {2}\end{array}\right],\) and define \(T : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) by \(T(\mathbf{x})=A \mathbf{x}\) Find the images under \(T\) of \(\mathbf{u}=\left[\begin{array}{r}{1} \\\ {-3}\end{array}\right]\) and \(\mathbf{v}=\left[\begin{array}{l}{a} \\\ {b}\end{array}\right]\)

Short Answer

Expert verified
The image of \( \mathbf{u} \) is \( \begin{bmatrix} 2 \\ -6 \end{bmatrix} \); the image of \( \mathbf{v} \) is \( \begin{bmatrix} 2a \\ 2b \end{bmatrix} \).

Step by step solution

01

Define the Linear Transformation

Given the matrix \( A \) and the transformation \( T \), where \( T(\mathbf{x}) = A \mathbf{x} \), we know the transformation applied to any vector \( \mathbf{x} \) in \( \mathbb{R}^2 \) is accomplished by multiplying the matrix \( A \) with the vector \( \mathbf{x} \). The matrix \( A \) is defined as \[ A = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix} \].
02

Compute T(u)

To find the image of \( \mathbf{u} = \begin{bmatrix} 1 \ -3 \end{bmatrix} \) under \( T \), calculate \( T(\mathbf{u}) = A \mathbf{u} \):\[A \begin{bmatrix} 1 \ -3 \end{bmatrix} = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix} \begin{bmatrix} 1 \ -3 \end{bmatrix} = \begin{bmatrix} 2(1) + 0(-3) \ 0(1) + 2(-3) \end{bmatrix} = \begin{bmatrix} 2 \ -6 \end{bmatrix}\].Thus, \( T(\mathbf{u}) = \begin{bmatrix} 2 \ -6 \end{bmatrix} \).
03

Compute T(v)

Now find the image of \( \mathbf{v} = \begin{bmatrix} a \ b \end{bmatrix} \) under \( T \), calculate \( T(\mathbf{v}) = A \mathbf{v} \):\[A \begin{bmatrix} a \ b \end{bmatrix} = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix} \begin{bmatrix} a \ b \end{bmatrix} = \begin{bmatrix} 2a \ 2b \end{bmatrix}\].Thus, \( T(\mathbf{v}) = \begin{bmatrix} 2a \ 2b \end{bmatrix} \).

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Key Concepts

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

Matrix Multiplication
Matrix multiplication is a crucial concept in linear algebra that allows us to perform linear transformations. When multiplying a matrix by a vector, we apply a function to transform the vector. This process involves each component of the vector being combined with elements from the matrix through specific arithmetic operations.

To multiply a matrix with a vector in a 2D vector space, we follow a straightforward procedure:
  • Take each row of the matrix.
  • Multiply the elements of the row by the corresponding elements of the vector.
  • Add the results together to get a single value, which becomes an element in the resulting vector.
For example, with matrix \( A = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix} \) and vector \( \mathbf{u} = \begin{bmatrix} 1 \ -3 \end{bmatrix} \), multiplying gives:\[A \mathbf{u} = \begin{bmatrix} 2(1) + 0(-3) \ 0(1) + 2(-3) \end{bmatrix} = \begin{bmatrix} 2 \ -6 \end{bmatrix}.\]This result is the vector's image under the transformation defined by matrix \( A \).
Image of a Vector
The image of a vector in the context of linear transformations is the result of applying a transformation to the vector. When we have a function \( T \) defined by a linear transformation, such as \( T(\mathbf{x}) = A \mathbf{x} \), applying this to any vector \( \mathbf{x} \) in \( \mathbb{R}^2 \) transforms it.

The matrix \( A = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix} \) stretches every component of a vector by a factor of 2:
  • The first component \( x_1 \) becomes \( 2x_1 \).
  • The second component \( x_2 \) becomes \( 2x_2 \).
For vector \( \mathbf{u} = \begin{bmatrix} 1 \ -3 \end{bmatrix} \), the image under \( T \) is \( T(\mathbf{u}) = \begin{bmatrix} 2 \ -6 \end{bmatrix} \). Similarly, for vector \( \mathbf{v} = \begin{bmatrix} a \ b \end{bmatrix} \), the image is \( \begin{bmatrix} 2a \ 2b \end{bmatrix} \). Each coordinate is scaled directly by the corresponding elements in the matrix, illustrating the transformation's effect.
2D Vector Space
A 2D vector space is a collection of vectors that can be added together and multiplied by scalars to produce another vector within the same space. It is often visualized as the flat plane on which vectors like \( \mathbf{u} \) and \( \mathbf{v} \) lie.

In this particular exercise, the vectors \( \mathbf{u} = \begin{bmatrix} 1 \ -3 \end{bmatrix} \) and \( \mathbf{v} = \begin{bmatrix} a \ b \end{bmatrix} \) are part of the 2D vector space \( \mathbb{R}^2 \). This space includes all possible pairs of real numbers \( (x_1, x_2) \). The transformation \( T \) maps every vector in this space to another vector, making it a function that reshapes or repositions vectors.

Understanding 2D vector spaces is fundamental in linear algebra as it provides the backdrop where transformations like \( T(\mathbf{x}) = A \mathbf{x} \) occur. These vector spaces allow us to model and solve real-world problems involving direction and magnitude, often represented as vectors.

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Most popular questions from this chapter

a. Fill in the blank in the following statement: "If \(A\) is an \(m \times n\) matrix, then the columns of \(A\) are linearly independent if and only if \(A\) has ______ pivot columns.鈥 b. Explain why the statement in (a) is true.

Let \(\mathbf{u}=\left[\begin{array}{l}{7} \\ {2} \\ {5}\end{array}\right], \mathbf{v}=\left[\begin{array}{l}{3} \\ {1} \\ {3}\end{array}\right],\) and \(\mathbf{w}=\left[\begin{array}{l}{6} \\ {1} \\ {0}\end{array}\right]\) It can be shown that \(3 \mathbf{u}-5 \mathbf{v}-\mathbf{w}=\mathbf{0} .\) Use this fact (and no row operations) to find \(x_{1}\) and \(x_{2}\) that satisfy the equation $$ \left[\begin{array}{ll}{7} & {3} \\ {2} & {1} \\ {5} & {3}\end{array}\right]\left[\begin{array}{l}{x_{1}} \\\ {x_{2}}\end{array}\right]=\left[\begin{array}{l}{6} \\ {1} \\\ {0}\end{array}\right] $$

Let \(\mathbf{v}_{1}=\left[\begin{array}{r}{1} \\ {0} \\ {-1} \\\ {0}\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{r}{0} \\ {-1} \\\ {0} \\ {1}\end{array}\right], \mathbf{v}_{3}=\left[\begin{array}{r}{1} \\\ {0} \\ {0} \\ {-1}\end{array}\right]\) Does \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\) span \(\mathbb{R}^{4} ?\) Why or why not?

Find the value(s) of \(h\) for which the vectors are linearly dependent. Justify each answer. \(\left[\begin{array}{r}{1} \\ {-1} \\\ {4}\end{array}\right],\left[\begin{array}{r}{3} \\ {-5} \\\ {7}\end{array}\right],\left[\begin{array}{r}{-1} \\ {5} \\\ {h}\end{array}\right]\)

1\( produces ore that contains 20 metric tons of copper and 550 kilograms of silver, while one day's opera鈥 # A mining company has two mines. One day's operation at mine \)\\# 1\( produces ore that contains 20 metric tons of copper and 550 kilograms of silver, while one day's operation at mine \)\\# 2\( produces ore that contains 30 metric tons of copper and 500 kilograms of silver. Let \)\mathbf{v}_{1}=\left[\begin{array}{c}{20} \\ {550}\end{array}\right]\( and \)\mathbf{v}_{2}=\left[\begin{array}{c}{30} \\ {500}\end{array}\right] .\( Then \)\mathbf{v}_{1}\( and \)\mathbf{v}_{2}\( represent the "output per day" of mine \)\\# 1\( and mine \)\\# 2\( , respectively. a. What physical interpretation can be given to the vector 5 \)\mathbf{v}_{1} ?\( b. Suppose the company operates mine \)\\# 1\( for \)x_{1}\( days and mine \)\\# 2\( for \)x_{2}\( days. Write a vector equation whose solution gives the number of days each mine should operate in order to produce 150 tons of copper and 2825 kilograms of silver. Do not solve the equation. c. \)[\mathbf{M}]\( Solve the equation in \)(b)$

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