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Chapter 6: Problem 1

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{array}{l} 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{array} \quad \text { and so } \quad A=\left[\begin{array}{rr} 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}{rr} 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) \(^{\prime}\) 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{aligned} x+2 y &=a \\ 2 x+3 y &=b \end{aligned} \\] 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{array}{l} 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{array} \quad \text { and so } \quad B=\left[\begin{array}{rr} -49 & -76 \\ 30 & 47 \end{array}\right] \\]

Short Answer

Expert verified
The matrix representation of the given linear mapping F relative to basis E is: \[A=\left[\begin{array}{rr} 3 & 4 \\ 2 & -5 \end{array}\right]\] and relative to basis S is: \[B=\left[\begin{array}{rr} -49 & -76 \\ 30 & 47 \end{array}\right]\]

Step by step solution

01

Understand the linear mapping

We have a linear mapping F from R虏 to R虏 defined by F(x, y) = (3x + 4y, 2x - 5y). It maps a vector (x, y) in R虏 to another vector (3x + 4y, 2x - 5y) in R虏.
02

Express F in terms of basis E

Apply the linear mapping F on the basis vectors e鈧� and e鈧�, and write the result as a linear combination of the basis vectors. This creates the matrix A representing F relative to E: \[ A=\left[\begin{array}{rr} 3 & 4 \\ 2 & -5 \end{array}\right] \]
03

Express F in terms of basis S

Apply the linear mapping F on the basis vectors u鈧� and u鈧�, and write the result as a linear combination of the basis vectors. To do so, one needs to solve for x and y in two systems of linear equations for vectors F(u鈧�) and F(u鈧�). After solving, we form the matrix B representing F relative to S: \[ B=\left[\begin{array}{rr} -49 & -76 \\ 30 & 47 \end{array}\right] \] Putting it all together, we have found both the matrix representations A and B for the given linear mapping F relative to the bases E and S.

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Key Concepts

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

Linear Algebra
Linear algebra is a fundamental field within mathematics that deals with vectors, vector spaces (also known as linear spaces), and linear mappings between such spaces. In this discipline, concepts like lines and planes are represented by linear equations that are explored in higher dimensions using vectors and matrices.

For instance, in our exercise, we look at a linear mapping which is a function that preserves vector addition and scalar multiplication. The coordinates of a vector within a linear space can be represented differently depending on the chosen basis, and the same linear mapping can have different matrix representations relative to different bases. Understanding these matrix representations is crucial for solving equations, transforming geometrical figures, and more sophisticated applications like computer graphics and machine learning.
Basis Vectors
Basis vectors form the 'building blocks' of vector spaces. Every vector in the space can be written uniquely as a combination of these basis vectors, usually via a linear combination. A set of basis vectors must be linearly independent and span the entire vector space.

In our exercise, we encounter two different sets of basis vectors for the same space, \( \mathbf{R}^{2} \): the standard basis E and another basis S. This illustrates that there are many ways to describe the same vector space, and calculations that involve changing the basis, like in the provided example, are common in linear algebra.
Linear Transformations
Linear transformations are mappings between vector spaces that preserve the operations of vector addition and scalar multiplication. They play a key role in linear algebra because they can be depicted as matrices with respect to basis vectors, making computations more manageable.

In the original exercise, the linear transformation F takes a vector from \( \mathbf{R}^{2} \) and maps it to another vector in the same space. By expressing this transformation with respect to different bases, we get different matrices鈥擜 relative to the standard basis E, and B relative to basis S. These matrix representations make it easier to visualize and perform the transformation on vectors within the space.

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Most popular questions from this chapter

Let \(A=\left[a_{i j}\right]\) and \(B=\left[b_{i j}\right]\) be row equivalent \(m \times n\) matrices over a field \(K,\) and let \(v_{1}, v_{2}, \ldots, v_{n}\) be any vectors in a vector space \(V\) over \(K\). For \(i=1,2, \ldots, m\), let \(u_{i}\) and \(w_{i}\) be defined by \\[ u_{i}=a_{i 1} v_{1}+a_{i 2} v_{2}+\cdots+a_{i n} v_{n} \quad \text { and } \quad w_{i}=b_{i 1} v_{1}+b_{i 2} v_{2}+\cdots+b_{i n} v_{n} \\]

Prove Theorem 6.2: Let \(S=\left\\{u_{1}, u_{2}, \ldots, u_{n}\right\\}\) be a basis for \(V\) over \(K,\) and let \(\mathbf{M}\) be the algebra of \(n\) -square matrices over \(K .\) Then the mapping \(m: A(V) \rightarrow \mathbf{M}\) defined by \(m(T)=[T]_{S}\) is a vector space isomorphism. That is, for any \(F, G \in A(V)\) and any \(k \in K,\) we have (i) \([F+G]=[F]+[G]\) (ii) \([k F]=k[F]\) (iii) \(\quad m\) is one-to-one and onto. (i) Suppose, for \(i=1, \ldots, n\) \\[ F\left(u_{i}\right)=\sum_{j=1}^{n} a_{i j} u_{j} \quad \text { and } \quad G\left(u_{i}\right)=\sum_{j=1}^{n} b_{i j} u_{j} \\] Consider the matrices \(A=\left[a_{i j}\right]\) and \(B=\left[b_{i j}\right] .\) Then \([F]=A^{T}\) and \([G]=B^{T} .\) We have, for \(i=1, \ldots, n\) \\[ (F+G)\left(u_{i}\right)=F\left(u_{i}\right)+G\left(u_{i}\right)=\sum_{j=1}^{n}\left(a_{i j}+b_{i j}\right) u_{j} \\] Because \(A+B\) is the matrix \(\left(a_{i j}+b_{i j}\right),\) we have \\[ [F+G]=(A+B)^{T}=A^{T}+B^{T}=[F]+[G] \\] (ii) Also, for \(i=1, \ldots, n\) \\[ (k F)\left(u_{i}\right)=k F\left(u_{i}\right)=k \sum_{j=1}^{n} a_{i j} u_{j}=\sum_{j=1}^{n}\left(k a_{i j}\right) u_{j} \\] Because \(k A\) is the matrix \(\left(k a_{i j}\right),\) we have \\[ [k F]=(k A)^{T}=k A^{T}=k[F] \\] (iii) Finally, \(m\) is one-to-one, because a linear mapping is completely determined by its values on a basis. Also, \(m\) is onto, because matrix \(A=\left[a_{y}\right]\) in \(\mathbf{M}\) is the image of the linear operator, \\[ F\left(u_{i}\right)=\sum_{j=1}^{n} a_{i j} u_{j}, \quad i=1, \ldots, n \\] Thus, the theorem is proved.

Prove Theorem 6.1: Let \(T: V \rightarrow V\) be a linear operator, and let \(S\) be a (finite) basis of \(V\). Then, for any vector \(v\) in \(V,[T]_{S}[v]_{S}=[T(v)]_{S}\) Suppose \(S=\left\\{u_{1}, u_{2}, \ldots, u_{n}\right\\},\) and suppose, for \(i=1, \ldots, n\) \\[ T\left(u_{i}\right)=a_{i 1} u_{1}+a_{i 2} u_{2}+\cdots+a_{i n} u_{n}=\sum_{j=1}^{n} a_{i j} u_{j} \\] Then \([T]_{S}\) is the \(n\) -square matrix whose \(j\) th row is \\[ \left(a_{1 j}, a_{2 j}, \ldots, a_{n j}\right) \\] Now suppose \\[ v=k_{1} u_{1}+k_{2} u_{2}+\cdots+k_{n} u_{n}=\sum_{i=1}^{n} k_{i} u_{i} \\] Writing a column vector as the transpose of a row vector, we have \\[ [v]_{S}=\left[k_{1}, k_{2}, \ldots, k_{n}\right]^{T} \\] Furthermore, using the linearity of \(T\) \\[ \begin{aligned} T(v) &=T\left(\sum_{i=1}^{n} k_{i} u_{i}\right)=\sum_{i=1}^{n} k_{i} T\left(u_{i}\right)=\sum_{i=1}^{n} k_{i}\left(\sum_{j=1}^{n} a_{i j} u_{j}\right) \\ &=\sum_{j=1}^{n}\left(\sum_{i=1}^{n} a_{i j} k_{i}\right) u_{j}=\sum_{j=1}^{n}\left(a_{1 j} k_{1}+a_{2 j} k_{2}+\cdots+a_{n j} k_{n}\right) u_{j} \end{aligned} \\] Thus, \([T(v)]_{S}\) is the column vector whose \(j\) th entry is \\[ a_{1 j} k_{1}+a_{2 j} k_{2}+\dots+a_{n j} k_{n} \\] On the other hand, the \(j\) th entry of \([T]_{S}[v]_{S}\) is obtained by multiplying the \(j\) th row of \([T]_{S}\) by \([v]_{S}-\) that is (1) by \((2) .\) But the product of (1) and (2) is \((3) .\) Hence, \([T]_{S}[v]_{S}\) and \([T(v)]_{S}\) have the same entries. Thus, \([T]_{S}[v]_{S}=[T(v)]_{S}\).

Consider the linear transformation \(F\) on \(\mathbf{R}^{2}\) defined by \(F(x, y)=(5 x-y, 2 x+y)\) 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\\}=\\{(1,4),(2,7)\\} \\]

The vectors \(u_{1}=(1,2,0), u_{2}=(1,3,2), u_{3}=(0,1,3)\) form a basis \(S\) of \(\mathbf{R}^{3}\). Find (a) The change-of-basis matrix \(P\) from the usual basis \(E=\left\\{e_{1}, e_{2}, e_{3}\right\\}\) to \(S\) (b) The change-of-basis matrix \(Q\) from \(S\) back to \(E\). (a) Because \(E\) is the usual basis, simply write the basis vectors of \(S\) as columns: \(P=\left[\begin{array}{lll}1 & 1 & 0 \\ 2 & 3 & 1 \\ 0 & 2 & 3\end{array}\right]\) (b) Method 1. Express each basis vector of \(E\) as a linear combination of the basis vectors of \(S\) by first finding the coordinates of an arbitrary vector \(v=(a, b, c)\) relative to the basis \(S .\) We have \\[ \left[\begin{array}{l} a \\ b \\ c \end{array}\right]=x\left[\begin{array}{l} 1 \\ 2 \\ 0 \end{array}\right]+y\left[\begin{array}{l} 1 \\ 3 \\ 2 \end{array}\right]+z\left[\begin{array}{l} 0 \\ 1 \\ 3 \end{array}\right] \quad \text { or } \quad \begin{array}{r} x+y=a \\ 2 x+3 y+z=b \\ 2 y+3 z=c \end{array} \\] Solve for \(x, y, z\) to get \(x=7 a-3 b+c, y=-6 a+3 b-c, z=4 a-2 b+c .\) Thus, \\[ v=(a, b, c)=(7 a-3 b+c) u_{1}+(-6 a+3 b-c) u_{2}+(4 a-2 b+c) u_{3} \\] or $$[v]_{S}=[(a, b, c)]_{S}=[7 a-3 b+c,-6 a+3 b-c, 4 a-2 b+c]^{T}$$ Using the above formula for \([v]_{S}\) and then writing the coordinates of the \(e_{i}\) as columns yields \\[ \begin{array}{l} e_{1}=(1,0,0)=7 u_{1}-6 u_{2}+4 u_{3} \\ e_{2}=(0,1,0)=-3 u_{1}+3 u_{2}-2 u_{3} \quad \text { and } \quad Q=\left[\begin{array}{rrr} 7 & -3 & 1 \\ -6 & 3 & -1 \\ 4 & -2 & 1 \end{array}\right] \end{array} \\] Method 2. Find \(P^{-1}\) by row reducing \(M=[P, I]\) to the form \(\left[I, P^{-1}\right]\) \\[ \begin{array}{l} M=\left[\begin{array}{llllll} 1 & 1 & 0 & 1 & 0 & 0 \\ 2 & 3 & 1 & 0 & 1 & 0 \\ 0 & 2 & 3 & 0 & 0 & 1 \end{array}\right] \sim\left[\begin{array}{lllllll} 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 1 & -2 & 1 & 0 \\ 0 & 2 & 3 & 1 & 0 & 0 & 1 \end{array}\right] \\ \sim & {\left[\begin{array}{lllrrl} 1 & 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 1 & -2 & 1 & 0 \\ 0 & 0 & 1 & 1 & 4 & -2 & 1 \end{array}\right] \sim\left[\begin{array}{lllll} 1 & 0 & 0 & 1 & 7 & -3 & 1 \\ 0 & 1 & 0 & 1 & -6 & 3 & -1 \\ 0 & 0 & 1 & 1 & 4 & -2 & 1 \end{array}\right]=\left[I, P^{-1}\right]} \end{array} \\] Thus, \(Q=P^{-1}=\left[\begin{array}{rrr}7 & -3 & 1 \\ -6 & 3 & -1 \\ 4 & -2 & 1\end{array}\right]\).
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