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Let \(\mathbf{y}_{1}, \mathbf{y}_{2},\) and \(\mathbf{y}_{3}\) be defined as in Exercise \(7,\) and let \(L\) be the linear operator on \(\mathbb{R}^{3}\) defined by \(L\left(c_{1} \mathbf{y}_{1}+c_{2} \mathbf{y}_{2}+c_{3} \mathbf{y}_{3}\right)\) \(=\left(c_{1}+c_{2}+c_{3}\right) \mathbf{y}_{1}+\left(2 c_{1}+c_{3}\right) \mathbf{y}_{2}-\left(2 c_{2}+c_{3}\right) \mathbf{y}_{3}\) (a) Find a matrix representing \(L\) with respect to the ordered basis \(\left\\{\mathbf{y}_{1}, \mathbf{y}_{2}, \mathbf{y}_{3}\right\\}\) (b) For each of the following, write the vector \(\mathbf{x}\) as a linear combination of \(\mathbf{y}_{1}, \mathbf{y}_{2},\) and \(\mathbf{y}_{3}\) and use the matrix from part (a) to determine \(L(\mathbf{x})\) (i) \(\mathbf{x}=(7,5,2)^{T}\) (ii) \(\mathbf{x}=(3,2,1)^{T}\) (iii) \(\mathbf{x}=(1,2,3)^{T}\)

Short Answer

Expert verified
(a) The matrix representing the linear operator L with respect to the ordered basis {饾惒1,饾惒2,饾惒3} is: \[ A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 0 & 1 \\ 0 & -2 & -1 \end{bmatrix} \] (b) (i) The vector L(x) for x = (7,5,2)^T is L(饾惐) = (5, 5, -4)^T. (ii) The vector L(x) for x = (3,2,1)^T is L(饾惐) = (4, 1, -5)^T. (iii) The vector L(x) for x = (1,2,3)^T is L(饾惐) = (2, -3, 7)^T.

Step by step solution

01

Apply Linear Operator L

Given the linear operator \(L\), we find the following for each basis vector: \(L(\mathbf{y}_1) = (1)\mathbf{y}_1 + (2)\mathbf{y}_2 - (0)\mathbf{y}_3\) \(L(\mathbf{y}_2) = (1)\mathbf{y}_1 + (0)\mathbf{y}_2 - (2)\mathbf{y}_3\) \(L(\mathbf{y}_3) = (1)\mathbf{y}_1 + (1)\mathbf{y}_2 - (1)\mathbf{y}_3\) So the matrix \(A\) representing the linear operator \(L\) is: \[ A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 0 & 1 \\ 0 & -2 & -1 \end{bmatrix} \] _step 2: Write the given vectors as a linear combination of 饾惒1, 饾懄2,饾惒3_
02

Express each vector as a linear combination of basis vectors

In Exercise 7, we found that \[\mathbf{y}_1 = \begin{bmatrix}2 \\ 3 \\ 1\end{bmatrix}, \ \mathbf{y}_2 = \begin{bmatrix}0 \\ 1 \\ -1\end{bmatrix}, \ \mathbf{y}_3 = \begin{bmatrix}1 \\ 2 \\ 0\end{bmatrix}\] Now, we express the given vectors as a linear combination of {\(\mathbf{y}_1\), \(\mathbf{y}_2\), \(\mathbf{y}_3\)}: (i) \(\mathbf{x} = (7,5,2)^T = 2\mathbf{y}_1 + 2\mathbf{y}_2 + \mathbf{y}_3\) (ii) \(\mathbf{x} = (3,2,1)^T = \frac{1}{2}\mathbf{y}_1 + 3\mathbf{y}_2 + \frac{1}{2}\mathbf{y}_3\) (iii) \(\mathbf{x} = (1,2,3)^T = -\mathbf{y}_1 - 5\mathbf{y}_2 + 3\mathbf{y}_3\) _step 3: Determine L(饾惐) for each case_
03

Calculate L(x) for each case

We use the matrix A and the linear combinations we found in step 2 to calculate L(x). (i) \(L(\mathbf{x}) = A \begin{bmatrix}2 \\ 2 \\ 1\end{bmatrix} = \begin{bmatrix}5 \\ 5 \\ -4\end{bmatrix}\) (ii) \(L(\mathbf{x}) = A \begin{bmatrix}\frac{1}{2} \\ 3 \\ \frac{1}{2}\end{bmatrix} = \begin{bmatrix}4 \\ 1 \\ -5\end{bmatrix}\) (iii) \(L(\mathbf{x}) = A \begin{bmatrix}-1 \\ -5 \\ 3\end{bmatrix} = \begin{bmatrix}2 \\ -3 \\ 7\end{bmatrix}\) The resulting vectors L(x) for each case are: (i) \(L(\mathbf{x}) = (5, 5, -4)^T\) (ii) \(L(\mathbf{x}) = (4, 1, -5)^T\) (iii) \(L(\mathbf{x}) = (2, -3, 7)^T\)

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Key Concepts

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

Matrix Representation
The matrix representation of a linear operator is a powerful tool in linear algebra, as it allows us to transform complex algebraic computations into simpler matrix operations.

Imagine you have a function鈥攍et's call it a linear operator鈥攖hat takes vectors and manipulates them following certain rules. To represent this function using matrices, we convert the rules into columns of numbers that reflect how the operator affects a set of special vectors called basis vectors. For each basis vector, we apply the operator and express the outcome as a linear combination of the same set of basis vectors.

Following this concept in our exercise, we found that when applying the linear operator to each basis vector, we could represent the transformations as a matrix. This matrix simplifies how we think about and apply the operator, particularly for complicated vector spaces. It's as if we created a cheat sheet that tells us the outcome of applying the function to any vector, by simply performing matrix multiplication.
Ordered Basis
An ordered basis is like a custom measuring stick for vectors in a specific vector space. It is a sequence of vectors that you can use to describe any vector within that space uniquely. The keyword here is 'ordered' because the sequence matters. The order of the vectors in the basis affects how we record coordinates for vectors and how the matrix representation of a linear operator is formed.

Consider the basis {\(\textbf{y}_1\), \(\textbf{y}_2\), \(\textbf{y}_3\)} from our exercise. It is not just a set but an ordered triplet. This order is used to create the matrix that represents the linear operator \(L\). Each vector in the basis corresponds to a particular column in the matrix. So, a change in the sequence of basis vectors would end in a different matrix for \(L\), even though the function itself hasn't changed.
Linear Combination
A linear combination is a recipe for creating new vectors from a set of ingredients鈥攐ur basis vectors. For any vector in the space, we can find some amounts (or scalars) to combine with each basis vector. The idea is equivalent to mixing certain amounts of primary colors to get a desired paint color.

In the context of our exercise, writing a vector as a linear combination of the basis vectors {\(\textbf{y}_1\), \(\textbf{y}_2\), \(\textbf{y}_3\)} is like saying, 'To reach the point described by \(\textbf{x}\), travel so much in the direction of \(\textbf{y}_1\), a bit in the direction of \(\textbf{y}_2\), and a dash towards \(\textbf{y}_3\).' This gives us a clear blueprint for constructing any vector in our space from the chosen basis.
Basis Vectors
Basis vectors are the building blocks of a vector space, much like the scaffolding within a building. A set of basis vectors for a space is a selection of vectors that are both linearly independent鈥攎eaning none of them is redundant鈥攁nd span the entire space鈥攎eaning you can reach any point in the space by scaling and adding them together.

In our problem, the vectors \(\textbf{y}_1\), \(\textbf{y}_2\), and \(\textbf{y}_3\) are the basis vectors for the three-dimensional space \(\mathbb{R}^{3}\). This means any vector in \(\mathbb{R}^{3}\) can be uniquely expressed as a combination of these three vectors. What's more, no one of these vectors can be created by using the other two, which ensures that the set is indeed a basis. They form a coordinate system we use to explore and understand the three-dimensional space.

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Most popular questions from this chapter

Let \(V\) be the subspace of \(C[a, b]\) spanned by \(1, e^{x}, e^{-x},\) and let \(D\) be the differentiation operator on \(V\) \(\begin{array}{lllll}\text { (a) Find the transition matrix } & S & \text { represent- }\end{array}\) ing the change of coordinates from the ordered basis \(\left[1, e^{x}, e^{-x}\right]\) to the ordered basis \([1, \cosh x, \sinh x] .\left[\cosh x=\frac{1}{2}\left(e^{x}+e^{-x}\right)\right.\) \(\sinh x=\frac{1}{2}\left(e^{x}-e^{-x}\right) \cdot\) (b) Find the matrix \(A\) representing \(D\) with respect to the ordered basis \([1, \cosh x, \sinh x]\) (c) Find the matrix \(B\) representing \(D\) with respect to \(\left[1, e^{x}, e^{-x}\right]\) (d) Verify that \(B=S^{-1} A S\).

Show that if \(A\) and \(B\) are similar matrices, then \(\operatorname{det}(A)=\operatorname{det}(B)\)

For each of the following linear operators on \(\mathbb{R}^{2}\) find the matrix representation of the transformation with respect to the homogeneous coordinate system: (a) The transformation \(L\) that rotates each vector by \(120^{\circ}\) in the counterclockwise direction (b) The transformation \(L\) that translates each point 3 units to the left and 5 units up (c) The transformation \(L\) that contracts each vector by a factor of one-third (d) The transformation that reflects a vector about the \(y\) -axis and then translates it up 2 units

Suppose that \(A=S \Lambda S^{-1},\) where \(\Lambda\) is a diagonal matrix with diagonal elements \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}\) (a) Show that \(A \mathbf{s}_{i}=\lambda_{i} \mathbf{s}_{i}, i=1, \ldots, n\) (b) Show that if \(\mathbf{x}=\alpha_{1} \mathbf{s}_{1}+\alpha_{2} \mathbf{s}_{2}+\cdots+\alpha_{n} \mathbf{s}_{n},\) then \\[ A^{k} \mathbf{x}=\alpha_{1} \lambda_{1}^{k} \mathbf{s}_{1}+\alpha_{2} \lambda_{2}^{k} \mathbf{s}_{2}+\cdots+\alpha_{n} \lambda_{n}^{k} \mathbf{s}_{n} \\] (c) Suppose that \(\left|\lambda_{i}\right|<1\) for \(i=1, \ldots, n .\) What happens to \(A^{k} \mathbf{x}\) as \(k \rightarrow \infty ?\) Explain.

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 \(G .\) Hint: Show that \(B A[\mathbf{v}]_{E}=\left[\left(L_{2} \circ L_{1}\right)(\mathbf{v})\right]_{G}\) for all \(\mathbf{v} \in V\)

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