91Ó°ÊÓ

Consider the linear mapping \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y)=(3 x+4 y, \quad 2 x-5 y)\) and the following bases of \(\mathbf{R}^{2}\) : $$ E=\left\\{e_{1}, e_{2}\right\\}=\\{(1,0),(0,1)\\} \quad \text { and } \quad S=\left\\{u_{1}, u_{2}\right\\}=\\{(1,2),(2,3)\\} $$ (a) Find the matrix \(A\) representing \(F\) relative to the basis \(E\). (b) Find the matrix \(B\) representing \(F\) relative to the basis \(S\). (a) Because \(E\) is the usual basis, the rows of \(A\) are simply the coefficients in the components of \(F(x, y)\); that is, using \((a, b)=a e_{1}+b e_{2}\), we have $$ \begin{aligned} &F\left(e_{1}\right)=F(1,0)=(3,2)=3 e_{1}+2 e_{2} \\ &F\left(e_{2}\right)=F(0,1)=(4,-5)=4 e_{1}-5 e_{2} \end{aligned} \quad \text { and so } \quad A=\left[\begin{array}{lr} 3 & 4 \\ 2 & -5 \end{array}\right] $$ Note that the coefficients of the basis vectors are written as columns in the matrix representation. (b) First find \(F\left(u_{1}\right)\) and write it as a linear combination of the basis vectors \(u_{1}\) and \(u_{2} .\) We have $$ F\left(u_{1}\right)=F(1,2)=(11,-8)=x(1,2)+y(2,3), \quad \text { and so } \quad \begin{array}{r} x+2 y=11 \\ 2 x+3 y=-8 \end{array} $$ Solve the system to obtain \(x=-49, y=30 .\) Therefore, $$ F\left(u_{1}\right)=-49 u_{1}+30 u_{2} $$ Next find \(F\left(u_{2}\right)\) and write it as a linear combination of the basis vectors \(u_{1}\) and \(u_{2} .\) We have $$ F\left(u_{2}\right)=F(2,3)=(18,-11)=x(1,2)+y(2,3), \quad \text { and so } \quad \begin{array}{r} x+2 y=18 \\ 2 x+3 y=-11 \end{array} $$ Solve for \(x\) and \(y\) to obtain \(x=-76, y=47 .\) Hence, $$ F\left(u_{2}\right)=-76 u_{1}+47 u_{2} $$ Write the coefficients of \(u_{1}\) and \(u_{2}\) as columns to obtain \(B=\left[\begin{array}{rr}-49 & -76 \\ 30 & 47\end{array}\right]\) (b') Alternatively, one can first find the coordinates of an arbitrary vector \((a, b)\) in \(\mathbf{R}^{2}\) relative to the basis \(S\). We have $$ (a, b)=x(1,2)+y(2,3)=(x+2 y, 2 x+3 y), \quad \text { and so } \quad \begin{array}{r} x+2 y=a \\ 2 x+3 y=b \end{array} $$ Solve for \(x\) and \(y\) in terms of \(a\) and \(b\) to get \(x=-3 a+2 b, y=2 a-b\). Thus, $$ (a, b)=(-3 a+2 b) u_{1}+(2 a-b) u_{2} $$ Then use the formula for \((a, b)\) to find the coordinates of \(F\left(u_{1}\right)\) and \(F\left(u_{2}\right)\) relative to \(S\) : $$ \begin{aligned} &F\left(u_{1}\right)=F(1,2)=(11,-8)=-49 u_{1}+30 u_{2} \\ &F\left(u_{2}\right)=F(2,3)=(18,-11)=-76 u_{1}+47 u_{2} \end{aligned} \quad \text { and so } \quad B=\left[\begin{array}{rr} -49 & -76 \\ 30 & 47 \end{array}\right] $$

Step by step solution

01

(a) Find the matrix A representing F relative to the basis E

We are given the linear mapping F defined by: \(F(x, y)=(3x+4y, 2x-5y)\) First, let's find the images of the standard basis vectors under F: \(F(e_1) = F(1, 0) = (3, 2)\) \(F(e_2) = F(0, 1) = (4, -5)\) Therefore, the matrix A is given by: \[A = \begin{bmatrix} 3 & 4 \\ 2 & -5 \end{bmatrix}\]

02

(b) Find the matrix B representing F relative to the basis S

To find the matrix B, we first need to express the linear mapping F in terms of the basis S vectors. First, let's find the images of the basis S vectors under F: \(F(u_1) = F(1, 2) = (11, -8)\) \(F(u_2) = F(2, 3) = (18, -11)\) Now, we need to express each of these images as a linear combination of the basis S vectors: \(11u_1 + 18u_2 = (11,22) + (36, 54) = (47, 76)\) To find the coefficients, we can write: \(\alpha u_1 + \beta u_2 = F(u_1)\) \(\gamma u_1 + \delta u_2 = F(u_2)\) Setting up a system of equations, we have: \(\alpha + 2\beta = 11\) \(2\alpha + 3\beta = 22\) \(\alpha + 2\gamma = 18\) \(2\alpha + 3\gamma = -11\) Solving the above system of equations, we get \(\alpha = 7, \beta = 2, \gamma = 4, \delta = -5\). Therefore, the matrix B is given by: \[B = \begin{bmatrix} 7 & 4 \\ 2 & -5 \end{bmatrix}\]

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Key Concepts

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

Matrix Representation

Understanding matrix representation is central to linear algebra, especially when dealing with linear transformations. Matrix representation serves as a bridge between abstract vector spaces and concrete computations. In the context of a linear mapping such as \(F: \textbf{R}^{2} \rightarrow \textbf{R}^{2}\), which is defined by \(F(x, y)=(3x+4y, 2x-5y)\), matrix representation simplifies the transformation process by using a matrix to describe the function.

In step 1 of our exercise, matrix \(A\) describes the transformation \(F\) with respect to the standard basis \(E\). This matrix is obtained by applying \(F\) to each vector in the basis \(E\text{, }\{e_1, e_2\}\) and recording the output components as columns in \(A\text{. }\) The process is intuitive since the standard basis vectors like \(e_1 = (1,0)\) and \(e_2 = (0,1)\) align with the axes of the Cartesian plane.

Matrix representation is essential when transitioning between different bases or for representing linear systems in a compact and manageable way. It shows how each basis vector is transformed, and thus how any vector in the space can be transformed by using linear combinations of these outputs.

Basis Transformation

Basis transformation is about changing the 'lens' through which we view our vector space. In linear algebra, we can choose different sets of basis vectors to describe the same vector space. Each choice provides a different perspective and can simplify problems depending on the context.

Diving into step 2 of the exercise, we see how the matrix \(B\) represents the mapping \(F\) relative to a non-standard basis \(S\). This process involves expressing the transformed vectors, which are initially in terms of the standard basis, as linear combinations of the new basis vectors in \(S\): \(u_1\) and \(u_2\).

This process necessitates solving systems of linear equations to find the coefficients that express the images \(F(u_1)\) and \(F(u_2)\) as linear combinations of \(u_1\) and \(u_2\). The resulting coefficients are then arranged as columns to form the matrix \(B\text{, }\) which is a powerful tool as it enables performing linear transformations directly in the new basis without reverting back to the standard basis. Basis transformation is crucial in many applications, ranging from simplifying calculations to optimizing algorithms for computational efficiency.

Linear Combination

Finding Linear Combinations

Linear combinations are the building blocks of linear algebra. They are sums of vectors scaled by coefficients, and they are fundamental in expressing any vector in a given space with respect to a basis. To express a vector as a linear combination of basis vectors means to write it in a way that highlights how much of each basis vector is needed to form the original vector.

In the context of our exercise, finding matrix \(B\text{ }\) involves expressing the images of the basis vectors \(u_1\) and \(u_2\) under \(F\) as linear combinations of the basis itself. This requires us to solve a system of linear equations to determine the coefficients that make up the combinations. These coefficients will directly translate into the elements of matrix \(B\text{. }\)

Understanding how to find these linear combinations is vital when converting the matrix representation of a linear transformation from one basis to another. It also reinforces the concept that vectors can be manipulated as objects that are made up of certain amounts of each piece from the basis, allowing for flexibility and adaptability in various mathematical and practical applications.