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Chapter 4: Problem 4

Let \(L: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) be a linear operator. If \\[ L\left((1,2)^{T}\right)=(-2,3)^{T} \\] and \\[ L\left((1,-1)^{T}\right)=(5,2)^{T} \\] find the value of \(L\left((7,5)^{T}\right)\)

Short Answer

Expert verified
The short answer for the given problem is: \(L\left((7,5)^{T}\right) = (7, 18)^{T}\).

Step by step solution

01

Find the linear combination of given input vectors

First, we need to find scalars \(a\) and \(b\) such that \((7,5)^{T} = a(1,2)^{T} + b(1,-1)^{T}\). This results in a system of two linear equations: \\[ \begin{cases} a + b = 7 \\ 2a - b = 5 \end{cases} \\]
02

Solve the system of equations for a and b

We can solve this system of equations to find the values of \(a\) and \(b\) using substitution or elimination method. In this case, we'll use substitution: Solving the first equation for \(a\), we get: \\[ a = 7 - b \\] Now, we can substitute this expression for \(a\) into the second equation: \\[ 2(7 - b) - b = 5 \\] Simplifying the equation, we get: \\[ 14 - 2b - b = 5 \\] Which gives us: \\[ -3b = -9 \\] Dividing by -3, we obtain: \\[ b = 3 \\] Now, we can substitute this value of \(b\) back into the equation for \(a\): \\[ a = 7 - 3 = 4 \\] So, we found that \(a = 4\) and \(b = 3\). Therefore, \((7,5)^{T} = 4(1,2)^{T} + 3(1,-1)^{T}\).
03

Use the linearity property

Now that we have the linear combination, we can use the linearity property to find the value of \(L\left((7,5)^{T}\right)\): \\[ L\left((7,5)^{T}\right) = L\left(4(1,2)^{T} + 3(1,-1)^{T}\right) = 4L\left((1,2)^{T}\right) + 3L\left((1,-1)^{T}\right) \\] Since \(L\left((1,2)^{T}\right) = (-2,3)^{T}\) and \(L\left((1,-1)^{T}\right) = (5,2)^{T}\), we can substitute these values into the equation: \\[ L\left((7,5)^{T}\right) = 4(-2,3)^{T} + 3(5,2)^{T} \\]
04

Compute the result

We can now calculate the result of the linear transformation: \\[ L\left((7,5)^{T}\right) = (4 \times -2, 4 \times 3)^{T} + (3 \times 5, 3 \times 2)^{T} = (-8, 12)^{T} + (15, 6)^{T} \\] Adding the vectors, we get: \\[ L\left((7,5)^{T}\right) = (-8+15, 12+6)^{T} = (7, 18)^{T} \\] So, the value of \(L\left((7,5)^{T}\right) = (7, 18)^{T}\).

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Most popular questions from this chapter

Suppose that \(L_{1}: V \rightarrow W\) and \(L_{2}: W \rightarrow Z\) are linear transformations and \(E, F,\) and \(G\) are ordered bases for \(V, W,\) and \(Z,\) respectively. Show that, if \(A\) represents \(L_{1}\) relative to \(E\) and \(F\) and \(B\) represents \(L_{2}\) relative to \(F\) and \(G,\) then the matrix \(C=B A\) represents \(L_{2} \circ L_{1}: V \rightarrow Z\) relative to \(E\) and \(G .\left[\text { Hint: Show that } B A[\mathbf{v}]_{E}=\left[\left(L_{2} \circ L_{1}\right)(\mathbf{v})\right]_{G}\right.\) for all \(\mathbf{v} \in V .]\)

Let \(L: V \rightarrow W\) be a linear transformation, and let \(T\) be a subspace of \(W\). The inverse image of \(T\) denoted \(L^{-1}(T),\) is defined by \\[ L^{-1}(T)=\\{\mathbf{v} \in V | L(\mathbf{v}) \in T\\} \\] Show that \(L^{-1}(T)\) is a subspace of \(V\)

For each \(f \in C[0,1]\), define \(L(f)=F,\) where \\[ F(x)=\int_{0}^{x} f(t) d t \quad 0 \leq x \leq 1 \\] Show that \(L\) is a linear operator on \(C[0,1]\) and then find \(L\left(e^{x}\right)\) and \(L\left(x^{2}\right)\)

Let \(L\) be the operator on \(P_{3}\) defined by $$L(p(x))=x p^{\prime}(x)+p^{\prime \prime}(x)$$ (a) Find the matrix \(A\) representing \(L\) with respect to \(\left[1, x, x^{2}\right]\) (b) Find the matrix \(B\) representing \(L\) with respect to \(\left[1, x, 1+x^{2}\right]\) (c) Find the matrix \(S\) such that \(B=S^{-1} A S\). (d) If \(p(x)=a_{0}+a_{1} x+a_{2}\left(1+x^{2}\right),\) calculate \(L^{n}(p(x))\)

Determine whether the following are linear transformations from \(\mathbb{R}^{2}\) into \(\mathbb{R}^{3}:\) (a) \(L(\mathbf{x})=\left(x_{1}, x_{2}, 1\right)^{T}\) (b) \(L(\mathbf{x})=\left(x_{1}, x_{2}, x_{1}+2 x_{2}\right)^{T}\) (c) \(L(\mathbf{x})=\left(x_{1}, 0,0\right)^{T}\) (d) \(L(\mathbf{x})=\left(x_{1}, x_{2}, x_{1}^{2}+x_{2}^{2}\right)^{T}\)
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