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Chapter 5: Problem 19

Find a linear map \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{4}\) whose image is spanned by (1,2,0,-4) and (2,0,-1,-3) Form a \(4 \times 3\) matrix whose columns consist only of the given vectors, say \\[ A=\left[\begin{array}{rrr} 1 & 2 & 2 \\ 2 & 0 & 0 \\ 0 & -1 & -1 \\ -4 & -3 & -3 \end{array}\right] \\]

Short Answer

Expert verified
The linear map \(F: \mathbb{R}^3 \rightarrow \mathbb{R}^4\) whose image is spanned by (1,2,0,-4) and (2,0,-1,-3) is represented by the matrix: \( F=\left[\begin{array}{rrr} 1 & 2 & 2 \\ 2 & 0 & 5 \\ 0 & -1 & -4 \\ -4 & -3 & 0 \end{array}\right] \)

Step by step solution

01

Check Linear Independence of Given Vectors

To form a basis for the image, we need linearly independent vectors. Check if the given vectors (1,2,0,-4) and (2,0,-1,-3) are linearly independent. Take the determinant of the \(2\times2\) matrix formed by the upper two elements of each vector: \( \text{det}\left[\begin{array}{rr} 1 & 2 \\ 2 & 0 \end{array}\right] = (1)(0) - (2)(2) = -4 \) Since the determinant is not zero, the given vectors are linearly independent.
02

Find a Linearly Independent Third Vector

Now, we need to find a third vector linearly independent of the given vectors (1,2,0,-4) and (2,0,-1,-3). One way to do this is to find a vector which is orthogonal to the two given vectors, by taking the cross product of the given vectors. Let \( \mathbf{a}=(1,2,0,-4) \) and \( \mathbf{b}=(2,0,-1,-3) \). The cross product of the two vectors, which we denote as \( \mathbf{c} \), is given by the following formula: \( \mathbf{c} = \left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{array}\right| \) After evaluating the cross product, we get: \( \mathbf{c} = (2-0)\mathbf{i} - (-(1+(-4)))\mathbf{j} + ((1)(0)-2(2))\mathbf{k} = (2, 5, -4) \)
03

Form the 4脳3 Matrix with the Basis Vectors

Now that we have three linearly independent vectors, (1,2,0,-4), (2,0,-1,-3), and (2,5,-4), we can form a \(4\times3\) matrix using these vectors as columns: \( F=\left[\begin{array}{rrr} 1 & 2 & 2 \\ 2 & 0 & 5 \\ 0 & -1 & -4 \\ -4 & -3 & 0 \end{array}\right] \) The linear map F: 鈩漗3 鈫� 鈩漗4 whose image is spanned by (1,2,0,-4) and (2,0,-1,-3) is represented by the matrix F.

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Key Concepts

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

Linear Independence
Linear independence is a fundamental concept in linear algebra. When we talk about whether vectors are linearly independent, we're basically asking if one vector can be written as a combination of the others. If no vector can be expressed this way, they are independent.
For example, consider vectors \((1, 2, 0, -4)\) and \((2, 0, -1, -3)\). To check their linear independence, you can find the determinant of a matrix using their components:
  • Set up a matrix using parts of both vectors (for simplicity, the first two components can often suffice).
  • Calculate the determinant.
  • If the determinant is non-zero, as in our case, the vectors are independent.
So, the vectors \((1, 2, 0, -4)\) and \((2, 0, -1, -3)\) are independent. This independence means they can form part of a basis, an important idea that leads to defining spaces in linear algebra.
Basis Vectors
Basis vectors are crucial in understanding how spaces are formed and structured. A set of vectors is a basis for a vector space if they are linearly independent and can span the vector space. In simpler terms, you can create any vector in the space from a combination of these basis vectors.
Let's explore further:
  • Each basis vector contributes uniquely to the space configuration, ensuring complete coverage without redundancy.
  • For the image space in the exercise, we built a matrix of three vector columns: the two given vectors (1, 2, 0, -4) and (2, 0, -1, -3), plus an additional vector found through the cross product, (2, 5, -4).
  • This new vector is crucial because it provides an orthogonal addition that supports spanning the whole space we're examining.
Understanding basis vectors helps in knowing why we need them for the linear map \(F: \mathbf{R}^{3} \to \mathbf{R}^{4}\), since they ensure we properly define the mapping and coverage in \(\mathbf{R}^{4}\).
Cross Product
The cross product is a mathematical operation essential for finding orthogonal vectors, helping to ensure a set of linearly independent vectors. Although commonly used in three-dimensional space, it adapts easily to find additional vectors needed for higher-dimensional matrices.
In our context:
  • We use the cross product to seek a third vector independent of our given set.
  • Begin by setting up a determinant with the components of the original vectors acting as parts of the rows or columns.
  • Calculate it properly, and the result gives you a new vector, in our case \((2, 5, -4)\), which doesn't align with the existing ones.
The cross product result confirms you have a vector that neither copies nor depends on the previous ones, providing a perfect addition to your set of basis vectors. With this understanding, using a cross product becomes vital when ensuring your matrix spans all dimensions of the space when working with transformations and mapping like in this exercise.

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Most popular questions from this chapter

Let \(V\) be of finite dimension and let \(T\) be a linear operator on \(V\) for which \(T R=I\), for some operator \(R\) on \(V .\) (We call \(R\) a right inverse of \(T\).) (a) Show that \(T\) is invertible. (b) Show that \(R=T^{-1}\). (c) Give an example showing that the above need not hold if \(V\) is of infinite dimension. (a) Let \(\operatorname{dim} V=n .\) By Theorem \(5.14, T\) is invertible if and only if \(T\) is onto; hence, \(T\) is invertible if and only if \(\operatorname{rank}(T)=n .\) We have \(n=\operatorname{rank}(I)=\operatorname{rank}(T R) \leq \operatorname{rank}(T) \leq n .\) Hence, \(\operatorname{rank}(T)=n\) and \(T\) is invertible. (b) \(T T^{-1}=T^{-1} T=I .\) Then \(R=I R=\left(T^{-1} T\right) R=T^{-1}(T R)=T^{-1} I=T^{-1}\). (c) Let \(V\) be the space of polynomials in \(t\) over \(K\); say, \(p(t)=a_{0}+a_{1} t+a_{2} t^{2}+\cdots+a_{s} t^{5}\). Let \(T\) and \(R\) be the operators on \(V\) defined by $$ T(p(t))=0+a_{1}+a_{2} t+\cdots+a_{s} t^{s-1} \quad \text { and } \quad R(p(t))=a_{0} t+a_{1} t^{2}+\cdots+a_{s} t^{s+1} $$ We have $$ (T R)(p(t))=T(R(p(t)))=T\left(a_{0} t+a_{1} t^{2}+\cdots+a_{s} t^{s+1}\right)=a_{0}+a_{1} t+\cdots+a_{5} t^{s}=p(t) $$ and so \(T R=I\), the identity mapping. On the other hand, if \(k \in K\) and \(k \neq 0\), then $$ (R T)(k)=R(T(k))=R(0)=0 \neq k $$ Accordingly, \(R T \neq I\)

Show that the following mappings are linear: (a) \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y, z)=(x+2 y-3 z, 4 x-5 y+6 z)\). (b) \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y)=(a x+b y, c x+d y)\), where \(a, b, c, d\) belong to \(\mathbf{R}\).

Consider the matrix mapping \(A: \mathbf{R}^{4} \rightarrow \mathbf{R}^{3}\), where \(A=\left[\begin{array}{rrrr}1 & 2 & 3 & 1 \\ 1 & 3 & 5 & -2 \\ 3 & 8 & 13 & -3\end{array}\right]\). Find a basis and the dimension of (a) the image of \(A,(b)\) the kernel of \(A .\) (a) The column space of \(A\) is equal to \(\operatorname{Im} A\). Now reduce \(A^{T}\) to echelon form: $$ A^{T}=\left[\begin{array}{rrr} 1 & 1 & 3 \\ 2 & 3 & 8 \\ 3 & 5 & 13 \\ 1 & -2 & -3 \end{array}\right] \sim\left[\begin{array}{rrr} 1 & 1 & 3 \\ 0 & 1 & 2 \\ 0 & 2 & 4 \\ 0 & -3 & -6 \end{array}\right] \sim\left[\begin{array}{lll} 1 & 1 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] $$ Thus, \(\\{(1,1,3),(0,1,2)\\}\) is a basis of \(\operatorname{Im} A\), and \(\operatorname{dim}(\operatorname{Im} A)=2\). (b) Here \(\operatorname{Ker} A\) is the solution space of the homogeneous system \(A X=0\), where \(X=\left\\{x, y, z_{1} t\right)^{T}\). Thus, reduce the matrix \(A\) of coefficients to echelon form: $$ \left[\begin{array}{rrrr} 1 & 2 & 3 & 1 \\ 0 & 1 & 2 & -3 \\ 0 & 2 & 4 & -6 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & 2 & 3 & 1 \\ 0 & 1 & 2 & -3 \\ 0 & 0 & 0 & 0 \end{array}\right] \quad \text { or } \quad \begin{array}{r} x+2 y+3 z+t=0 \\ y+2 z-3 t=0 \end{array} $$ The free variables are \(z\) and \(t\). Thus, \(\operatorname{dim}(\) Ker \(A)=2\). (i) Set \(z=1, t=0\) to get the solution \((1,-2,1,0)\). (ii) Set \(z=0, t=1\) to get the solution \((-7,3,0,1)\). Thus, \((1,-2,1,0)\) and \((-7,3,0,1)\) form a basis for Ker \(A\).

Give an example of a nonlinear map \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) such that \(F^{-1}(0)=\\{0\\}\) but \(F\) is not one-to-one.

Show that the following mappings are not linear: (a) \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y)=(x y, x)\) (b) \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{3}\) defined by \(F(x, y)=(x+3,2 y, x+y)\) (c) \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y, z)=(|x|, y+z)\)
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