Problem 12
Question
Determine \(T(\mathbf{v})\) for the given linear transformation \(T\) and vector in \(V\) by (a) Computing \([T]_{B}^{C}\) and \([\mathbf{v}]_{B}\) and using Theorem 6.5 .4 (b) Direct calculation. \(T: M_{2}(\mathbb{R}) \rightarrow M_{2}(\mathbb{R})\) via $$ T\left(\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\right)=\left[\begin{array}{cc} 2 a-b+d & -a+3 d \\ 0 & -a-b+3 c \end{array}\right] $$ relative to the standard basis \(B=C\); $$ A=\left[\begin{array}{rr} -7 & 2 \\ 1 & -3 \end{array}\right] $$
Step-by-Step Solution
Verified Answer
The transformation of the given matrix A under the linear transformation T is:
$$
T(\mathbf{A}) = \left[\begin{array}{rr}
-17 & 7 \\
0 & 5
\end{array}\right].
$$
1Step 1: Find \(T(E_1), T(E_2), T(E_3),\) and \(T(E_4)\)
By applying the linear transformation T to the basis matrices, we get:
$$
T(E_1) = \left[\begin{array}{cc}
2 & -1 \\
0 & -1
\end{array}\right],
T(E_2) = \left[\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right],
T(E_3) = \left[\begin{array}{cc}
1 & 0 \\
0 & 3
\end{array}\right],
T(E_4) = \left[\begin{array}{cc}
1 & 3 \\
0 & 0
\end{array}\right].
$$
2Step 2: Compute \([T]_{B}^{C}\)
Now, we can form the matrix of the linear transformation:
$$
[T]_{B}^{C} = \left[\begin{array}{cccc}
2 & -1 & 1 & 1 \\
-1 & 0 & 0 & 3 \\
0 & 0 & 0 & 0 \\
-1 & -1 & 3 & 0
\end{array}\right].
$$
3Step 3: Compute \([\mathbf{A}]_{B}\)
To find the coordinate matrix \([\mathbf{A}]_{B}\), we write the initial matrix A as a linear combination of the basis matrices:
$$
A = -7E_1 + 2E_2 + E_3 - 3E_4.
$$
So, the coordinate matrix is
$$
[\mathbf{A}]_{B} = \left[\begin{array}{c}
-7 \\
2 \\
1 \\
-3
\end{array}\right].
$$
4Step 4: Apply Theorem 6.5.4
By using Theorem 6.5.4, we can compute the matrix of the transformation applied to A as follows:
$$
[T(\mathbf{A})]_{C} = [T]_{B}^{C} [\mathbf{A}]_{B} = \left[\begin{array}{cccc}
2 & -1 & 1 & 1 \\
-1 & 0 & 0 & 3 \\
0 & 0 & 0 & 0 \\
-1 & -1 & 3 & 0
\end{array}\right]
\left[\begin{array}{c}
-7 \\
2 \\
1 \\
-3
\end{array}\right].
$$
5Step 5: Compute T(v)
Now we can find T(v) by multiplying the matrices:
$$
T(\mathbf{A}) = \left[\begin{array}{c}
-17 \\
7 \\
0 \\
5
\end{array}\right].
$$
Re-write T(v) as a 2x2 matrix:
$$
T(\mathbf{A}) = \left[\begin{array}{rr}
-17 & 7 \\
0 & 5
\end{array}\right].
$$
(b) Direct calculation:
6Step 6: Apply T to the Matrix A
We apply the linear transformation T to the matrix A directly:
$$
T\left(\left[\begin{array}{rr}
-7 & 2 \\
1 & -3
\end{array}\right]\right) = \left[\begin{array}{cc}
2(-7) - 2 + (-3) & -(-7) + 3(-3) \\
0 & -(-7) - 2 + 3(1)
\end{array}\right].
$$
7Step 7: Simplify the Result
Simplify and obtain T(A):
$$
T(\mathbf{A}) = \left[\begin{array}{cc}
-17 & 7 \\
0 & 5
\end{array}\right].
$$
In both methods (a) and (b), we have found T(v) to be the same matrix:
$$
T(\mathbf{A}) = \left[\begin{array}{rr}
-17 & 7 \\
0 & 5
\end{array}\right].
$$
Key Concepts
Matrix RepresentationCoordinate MatrixStandard BasisDifferential Equations
Matrix Representation
The concept of matrix representation is central in understanding linear transformations. It encapsulates how a linear transformation can be described using a matrix. This description is dependent on the choice of bases for the vector spaces involved.
For instance, consider a linear transformation \( T: V \rightarrow W \), where \( V \) and \( W \) are vector spaces with bases \( B \) and \( C \), respectively. The matrix representation \( [T]_{B}^{C} \) is a compact way to express the effect of \( T \) on any vector \( \textbf{v} \text{ in } V \). To construct this matrix, one would apply \( T \) to each vector in the basis \( B \), and express the result as a linear combination of the basis vectors in \( C \). Each of these expressions fills a column of the matrix representation.
To use this matrix for actual transformations, we first express the vector \( \textbf{v} \) as a coordinate matrix relative to \( B \), labeled \( [\textbf{v}]_{B} \). Multiplication of \( [T]_{B}^{C} \) with \( [\textbf{v}]_{B} \) yields the coordinates of the transformed vector in the \( C \) basis, \( [T(\textbf{v})]_{C} \). This approach is powerful because it turns abstract linear transformations into concrete matrix computations.
For instance, consider a linear transformation \( T: V \rightarrow W \), where \( V \) and \( W \) are vector spaces with bases \( B \) and \( C \), respectively. The matrix representation \( [T]_{B}^{C} \) is a compact way to express the effect of \( T \) on any vector \( \textbf{v} \text{ in } V \). To construct this matrix, one would apply \( T \) to each vector in the basis \( B \), and express the result as a linear combination of the basis vectors in \( C \). Each of these expressions fills a column of the matrix representation.
To use this matrix for actual transformations, we first express the vector \( \textbf{v} \) as a coordinate matrix relative to \( B \), labeled \( [\textbf{v}]_{B} \). Multiplication of \( [T]_{B}^{C} \) with \( [\textbf{v}]_{B} \) yields the coordinates of the transformed vector in the \( C \) basis, \( [T(\textbf{v})]_{C} \). This approach is powerful because it turns abstract linear transformations into concrete matrix computations.
Coordinate Matrix
A coordinate matrix, denoted as \( [\textbf{v}]_{B} \), is a tool used to represent a vector \( \textbf{v} \) in terms of a specific basis \( B \) of a vector space. Essentially, it translates the vector's components relative to the chosen basis.
Let's simplify this with an example. If we have a basis \( B \) consisting of vectors \( \textbf{b}_1, \textbf{b}_2, \textbf{b}_3, \text{...} \textbf{b}_n \), any vector \( \textbf{v} \) in that space can be expressed as a combination of these basis vectors: \( \textbf{v} = c_1\textbf{b}_1 + c_2\textbf{b}_2 + c_3\textbf{b}_3 + \text{...} + c_n\textbf{b}_n \). The scalars \( c_1, c_2, c_3, \text{...}, c_n \) form the coordinate matrix \( [\textbf{v}]_{B} = [c_1, c_2, c_3, \text{...}, c_n]^T \).
This representation is key in converting between different bases and simplifying calculations relating to linear transformations, as demonstrated in the exercise.
Let's simplify this with an example. If we have a basis \( B \) consisting of vectors \( \textbf{b}_1, \textbf{b}_2, \textbf{b}_3, \text{...} \textbf{b}_n \), any vector \( \textbf{v} \) in that space can be expressed as a combination of these basis vectors: \( \textbf{v} = c_1\textbf{b}_1 + c_2\textbf{b}_2 + c_3\textbf{b}_3 + \text{...} + c_n\textbf{b}_n \). The scalars \( c_1, c_2, c_3, \text{...}, c_n \) form the coordinate matrix \( [\textbf{v}]_{B} = [c_1, c_2, c_3, \text{...}, c_n]^T \).
This representation is key in converting between different bases and simplifying calculations relating to linear transformations, as demonstrated in the exercise.
Standard Basis
The standard basis for a vector space provides the simplest and most intuitive framework for representing vectors and matrices. In \( \reals^n \), the standard basis comprises vectors that have a one in one position and zeros elsewhere. For instance, in \( \reals^2 \), the standard basis vectors are \( \textbf{e}_1 = [1, 0]^T \) and \( \textbf{e}_2 = [0, 1]^T \).
Working with the standard basis often simplifies problems because the coordinate matrix of a vector in the standard basis is simply the vector itself. Similarly, the matrix representation of a linear transformation using the standard basis is often easier to interpret, as seen in the exercise where the standard basis \( B = C \) is used for both domain and codomain of the transformation \( T \).
However, while the standard basis is convenient, it is not always the most efficient choice for every problem, particularly when dealing with more complex structures like differential equations or intricate vector spaces.
Working with the standard basis often simplifies problems because the coordinate matrix of a vector in the standard basis is simply the vector itself. Similarly, the matrix representation of a linear transformation using the standard basis is often easier to interpret, as seen in the exercise where the standard basis \( B = C \) is used for both domain and codomain of the transformation \( T \).
However, while the standard basis is convenient, it is not always the most efficient choice for every problem, particularly when dealing with more complex structures like differential equations or intricate vector spaces.
Differential Equations
Differential equations are mathematical equations that relate some function with its derivatives. They are pervasive in science and engineering as they describe many phenomena, from physics to finance. Solving a differential equation means finding the function that satisfies the relationship specified by the equation.
In the context of linear algebra, linear differential equations can be represented and solved using matrix methods. For example, systems of first order linear differential equations can be written in matrix form as \( \frac{d\textbf{x}}{dt} = A\textbf{x} \), where \( A \) is a matrix that describes the system and \( \textbf{x} \) is a vector of functions we are solving for. By computing the eigenvalues and eigenvectors of \( A \), we can find solutions to the differential equations.
Though differential equations were not directly the focus of our original exercise, understanding how to manipulate matrices and vector spaces is a foundational skill that greatly facilitates finding solutions to these more complex problems.
In the context of linear algebra, linear differential equations can be represented and solved using matrix methods. For example, systems of first order linear differential equations can be written in matrix form as \( \frac{d\textbf{x}}{dt} = A\textbf{x} \), where \( A \) is a matrix that describes the system and \( \textbf{x} \) is a vector of functions we are solving for. By computing the eigenvalues and eigenvectors of \( A \), we can find solutions to the differential equations.
Though differential equations were not directly the focus of our original exercise, understanding how to manipulate matrices and vector spaces is a foundational skill that greatly facilitates finding solutions to these more complex problems.
Other exercises in this chapter
Problem 12
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