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In Exercises \(1-10\) , assume that \(T\) is a linear transformation. Find the standard matrix of \(T\) . \(T : \mathbb{R}^{2} \rightarrow \mathbb{R}^{4}, T\left(\mathbf{e}_{1}\right)=(3,1,3,1)\) and \(T\left(\mathbf{e}_{2}\right)=(-5,2,0,0)\) where \(\mathbf{e}_{1}=(1,0)\) and \(\mathbf{e}_{2}=(0,1)\)

Short Answer

Expert verified
The standard matrix is \[ A = \begin{bmatrix} 3 & -5 \\ 1 & 2 \\ 3 & 0 \\ 1 & 0 \end{bmatrix} \]

Step by step solution

01

Understand the Vectors

The standard basis vectors for \( \mathbb{R}^2 \) are \( \mathbf{e}_1 = (1,0) \) and \( \mathbf{e}_2 = (0,1) \). We are given that \( T(\mathbf{e}_1) = (3,1,3,1) \) and \( T(\mathbf{e}_2) = (-5,2,0,0) \).
02

Assemble the Standard Matrix

The standard matrix of a linear transformation \( T \) is formed by using the images of the standard basis vectors as its columns. Thus, the first column of the matrix is \( T(\mathbf{e}_1) = (3,1,3,1) \) and the second column is \( T(\mathbf{e}_2) = (-5,2,0,0) \).
03

Write the Standard Matrix

The standard matrix of the transformation \( T \) is then given by arranging the column vectors as follows: \[A = \begin{bmatrix}3 & -5 \1 & 2 \3 & 0 \1 & 0\end{bmatrix}\]

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Key Concepts

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

Standard Matrix
The standard matrix of a linear transformation is a vital concept in linear algebra. It is essentially a matrix that represents a linear transformation in terms of standard basis vectors. The process of finding this matrix involves understanding the transformation's effect on some basic vectors and combining these effects into a matrix.In our example, we have a linear transformation from \( \mathbb{R}^2 \) to \( \mathbb{R}^4 \). The vectors \( \mathbf{e}_1 = (1,0) \) and \( \mathbf{e}_2 = (0,1) \) serve as our starting points. These are the standard basis vectors for \( \mathbb{R}^2 \). The transformation \( T \) maps these vectors to new vectors in \( \mathbb{R}^4 \):
  • \( T(\mathbf{e}_1) = (3,1,3,1) \)
  • \( T(\mathbf{e}_2) = (-5,2,0,0) \)
The standard matrix of \( T \) will have these resulting vectors as its columns. This means the first column of the matrix is \( T(\mathbf{e}_1) \) and the second column is \( T(\mathbf{e}_2) \). Thus, the matrix looks like this:\[A = \begin{bmatrix} 3 & -5 \ 1 & 2 \ 3 & 0 \ 1 & 0 \end{bmatrix}. \] This matrix, \( A \), is called the standard matrix representation of the transformation \( T \). It's used to understand and compute the effect of \( T \) on any vector in \( \mathbb{R}^2 \).
Basis Vectors
Basis vectors are fundamental elements in vector spaces that define a coordinate system. For a domain like \( \mathbb{R}^2 \), the standard basis vectors are \( \mathbf{e}_1 = (1, 0) \) and \( \mathbf{e}_2 = (0, 1) \). These vectors are special because they span the entire space and provide a straightforward method to represent any vector within it as a linear combination of \( \mathbf{e}_1 \) and \( \mathbf{e}_2 \).In linear transformations, the basis vectors of the original space (\( \mathbb{R}^2 \) in our case) are used as the starting point. When the transformation \( T \) is applied, these vectors transform to new vectors in the codomain, \( \mathbb{R}^4 \), as follows:
  • \( T(\mathbf{e}_1) = (3, 1, 3, 1) \)
  • \( T(\mathbf{e}_2) = (-5, 2, 0, 0) \)
These new vectors place the transformed version of the basis into the new space, influencing the shape and orientation of our transformation matrix. They are fundamentally important in creating the standard matrix because they serve as the columns of this matrix. This connection highlights the role that basis vectors play in bridging coordinates between different vector spaces through linear transformations.
Matrix Representation
The matrix representation of a linear transformation is a compact and powerful way to describe the operation of the transformation on any vector from the domain vector space. By organizing the transformation of the standard basis vectors into a matrix, we achieve an elegant means to apply the transformation to any vector in \( \mathbb{R}^2 \) through simple matrix multiplication.In the case of our example, the transformation \( T \) is captured completely by the matrix:\[A = \begin{bmatrix} 3 & -5 \ 1 & 2 \ 3 & 0 \ 1 & 0 \end{bmatrix}.\]This matrix becomes a tool for easily transforming any vector \( \mathbf{v} = \begin{bmatrix} a \ b \end{bmatrix} \) in \( \mathbb{R}^2 \) by performing the operation \( A\mathbf{v} = \begin{bmatrix} 3 & -5 \ 1 & 2 \ 3 & 0 \ 1 & 0 \end{bmatrix} \begin{bmatrix} a \ b \end{bmatrix} \). This results in a new vector in \( \mathbb{R}^4 \), representing the transformed outcome that \( T \) describes.The elegance of this method lies in its simplicity: once the matrix is established, transforming vectors is effortless. This exemplifies why understanding the matrix representation is invaluable in studying linear transformations, as it connects theory with computation efficiently and concisely.

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Most popular questions from this chapter

[M] Budget® Rent A Car in Wichita, Kansas, has a fleet of about 500 cars, at three locations. A car rented at one location may be returned to any of the three locations. The various fractions of cars returned to the three locations are shown in the matrix below. Suppose that on Monday there are 295 cars at the airport (or rented from there), 55 cars at the east side office, and 150 cars at the west side office. What will be the approximate distribution of cars on Wednesday? \(\left[\begin{array}{ccc}{.97} & {.05} & {.10} \\ {.00} & {.90} & {.05} \\\ {.03} & {.05} & {.85}\end{array}\right]\)

Let \(A=\left[\begin{array}{rrr}{2} & {0} & {6} \\ {-1} & {8} & {5} \\ {1} & {-2} & {1}\end{array}\right],\) let \(\mathbf{b}=\left[\begin{array}{r}{10} \\\ {3} \\ {3}\end{array}\right],\) and let \(W\) be the set of all linear combinations of the columns of \(A .\) a. Is \(\mathbf{b}\) in \(W ?\) b. Show that the third column of \(A\) is in \(W .\)

Construct \(3 \times 2\) matrices \(A\) and \(B\) such that \(A \mathbf{x}=\mathbf{0}\) has only the trivial solution and \(B \mathbf{x}=\mathbf{0}\) has a nontrivial solution.

Suppose \(A\) is a \(3 \times 3\) matrix and \(y\) is a vector in \(\mathbb{R}^{3}\) such that the equation \(A \mathbf{x}=\mathbf{y}\) does not have a solution. Does there exist a vector \(\mathbf{z}\) in \(\mathbb{R}^{3}\) such that the equation \(A \mathbf{x}=\mathbf{z}\) has a unique solution? Discuss.

[M] In Exercises \(37-40\) , determine if the columns of the matrix span \(\mathbb{R}^{4} .\) $$ \left[\begin{array}{rrrr}{5} & {-7} & {-4} & {9} \\ {6} & {-8} & {-7} & {5} \\\ {4} & {-4} & {-9} & {-9} \\ {-9} & {11} & {16} & {7}\end{array}\right] $$

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