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In Exercises 5-12, assume that \(T\) is a linear transformation. Refer to Example 7 for Exercises \(9-12\), if necessary. If \(T([-1,2])=[1,0,0]\) and \(T([2,1])=\) \([0,1,2]\), find \(T([0,10])\).

Short Answer

Expert verified
\(T([0,10]) = [4, 2, 4]\)."

Step by step solution

01

Understanding the Problem

We are given a linear transformation \(T\) with known outputs for two vectors: \(T([-1,2]) = [1, 0, 0]\) and \(T([2,1]) = [0, 1, 2]\). Our task is to find \(T([0,10])\).
02

Using Properties of Linear Transformations

Recall that for a linear transformation \(T\), if we know \(T(\mathbf{u})\) and \(T(\mathbf{v})\) for any vectors \(\mathbf{u}\) and \(\mathbf{v}\), we can express and find \(T(a\mathbf{u} + b\mathbf{v})\) using \(T(a\mathbf{u}) + T(b\mathbf{v}) = aT(\mathbf{u}) + bT(\mathbf{v})\).
03

Expressing the Target Vector

We want to express the vector \([0,10]\) as a linear combination of \([-1, 2]\) and \([2, 1]\). We set up the equation: \(a[-1, 2] + b[2, 1] = [0, 10]\).
04

Setting up the Equations

Solving for \(a\) and \(b\), we have two equations: \(-a + 2b = 0\) (for the first component) and \(2a + b = 10\) (for the second component).
05

Solving the Linear System

From \(-a + 2b = 0\), we have \(a = 2b\). Substitute \(a = 2b\) into the second equation \(2(2b) + b = 10\), which simplifies to \(5b = 10\). Hence, \(b = 2\). Substitute back to find \(a\): \(a = 2(2) = 4\).
06

Apply the Linear Combination

We found \([0, 10] = 4[-1, 2] + 2[2, 1]\). Using linearity, \(T([0,10]) = 4T([-1,2]) + 2T([2,1])\).
07

Calculate the Final Result

Substitute the given \(T([-1,2]) = [1, 0, 0]\) and \(T([2,1]) = [0, 1, 2]\) into the equation: \(T([0, 10]) = 4[1, 0, 0] + 2[0, 1, 2] = [4, 0, 0] + [0, 2, 4]\). Perform the addition to get \([4, 2, 4]\).

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Key Concepts

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

Vector Spaces
Understanding vector spaces is fundamental for grasping concepts involving linear transformations. A vector space is a collection of vectors where vector addition and scalar multiplication are defined. These spaces are equipped with certain rules, such as the existence of a zero vector and the ability to perform linear operations.
A vector space can be formed by a set of vectors, like \([-1, 2]\) and \[2, 1]\], which can be combined linearly to produce any vector in that space, such as \[0, 10]\].
This idea becomes crucial when dealing with linear transformations, as it allows us to express new vectors in terms of known ones. The set of all linear combinations of a given set of vectors forms a vector subspace, illustrating the powerful concept of span in vector spaces.
Linearity
Linearity is a fundamental property of linear transformations and is key to solving the problem given in the exercise. Linear transformations preserve vector addition and scalar multiplication, which mathematically translates to: \[ T(a \, \mathbf{u} + b \, \mathbf{v}) = a \, T(\mathbf{u}) + b \, T(\mathbf{v}) \].
This property allows us to decompose complex vectors into simpler components whose transformations are known.
The linear transformation doesn't distort the vector space's structure, ensuring that parallelism and origin point remain consistent across transformations. For example, in our problem, knowing the transformations of \([-1, 2]\) and \[2, 1]\], we could confidently deduce the transformation for \[0, 10]\] using their linear combination.
Linear Combination
A linear combination involves expressing a vector as the sum of scalar multiples of other vectors. In the exercise, the methodology revolved around expressing the vector \[0, 10]\] as a linear combination of \([-1, 2]\) and \[2, 1]\].
Mathematically, this expression took the form: \[ a[-1, 2] + b[2, 1] = [0, 10] \]. By solving this equation, we found the values of \("a"\) and \("b"\) to be 4 and 2, respectively.
Linear combinations allow for the simplification and breakdown of complex vectors into known vectors within the vector space, facilitating easier transformations and calculations. This process is central to vector space theory and problem-solving in linear algebra.
Matrix Representation
To understand transformations better, we can use matrices. A linear transformation can be represented by a matrix, which acts on a vector to produce another vector. This representation is very powerful as it simplifies computations and provides a clear framework for understanding complex transformations.
In matrix form, transformations utilize rows and columns to define how exactly the vectors are altered. If we had \[T(\mathbf{x}) = A\mathbf{x}\], the matrix \(A\) would fully encode the transformation \(T\).
Though not explicitly used in this exercise, knowing the matrix representation of a transformation allows us to capture all aspects of its linearity and quickly determine how different vectors are mapped. This is especially helpful for visualizing and computing transformations on a larger scale.

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Most popular questions from this chapter

Let \(\mathbf{v}\) and \(\mathbf{w}\) be independent column vectors in \(\mathbb{R}^{3}\), and let \(A\) be an invertible \(3 \times 3\) matrix. Prove that the vectors \(A v\) and \(A w\) are independent.

Mark each of the following True or False. a. The number of independent row vectors in a matrix is the same as the number of independent column vectors. b. If \(H\) is a row-echelon form of a matrix \(A\), then the nonzero column vectors in \(H\) form a basis for the column space of \(A\). c. If \(H\) is a row-echelon form of a matrix \(A\), then the nonzero row vectors in \(H\) are a basis for the row space of \(A\). d. If an \(n \times n\) matrix \(A\) is invertible, then rank \((A)=n\). e. For every matrix \(A\), we have rank \((A)>0\). f. For all positive integers \(m\) and \(n\), the rank of an \(m \times n\) matrix might be any number from 0 to the maximum of \(m\) and \(n\). g. For all positive integers \(m\) and \(n\), the rank of an \(m \times n\) matrix might be any number from 0 to the minimum of \(m\) and \(n\). h. For all positive integers \(m\) and \(n\), the nullity of an \(m \times n\) matrix might be any number from 0 to \(n\). i. For all positive integers \(m\) and \(n\), the nullity of an \(m \times n\) matrix might be any number from 0 to \(m\). j. For all positive integers \(m\) and \(n\), with m \(\geq n\), the nullity of an \(m \times n\) matrix might be any number from 0 to \(n\).

For each of the giveas pairs of lines in \(R^{3}\), determine whether the lines intersect. If they do intersect, find the point of intersection, and determine whether the lines are orthogonal. a. \(\begin{aligned} x_{1} &=4+t_{2} \\ x_{3} &=-3+5 t \end{aligned} \quad x_{2}=2-3 t\), and $$ \begin{aligned} &x_{1}=11+3 s, \quad \ddot{r}_{2}=-9-4 s, \\ &x_{1}=-4-3 s \end{aligned} $$ b. \(x_{1}=11+3 t, \quad x_{2}=-3-t\), \(x_{3}=4+3 t\) and \(x_{1}=6-2 s, \quad x_{2}=-2+s\), \(x_{3}=-15+7 s\)

Find all scalars \(s\), if any exist, such that \([1,0,1],[2, s, 3],[2,3,1]\) are independent.

Let \(A\) be an \(m \times n\) matrix with row-echelon form \(H\), and let \(V\) be the row space of \(A\) (and thus of \(H\) ). Let \(W_{k}=\operatorname{sp}\left(\mathrm{e}_{1}, \mathrm{e}_{2}, \ldots, \mathbf{e}_{k} \bar{\jmath}\right.\) be the subspace of \(\mathbb{R}^{n}\) generated by the first \(k\) rows of the \(n \times n\) identity matrix. Consider \(T_{k}: V \rightarrow W_{k}\) defined by $$ \begin{aligned} T_{k}\left(\left[x_{1}, x_{2}, \ldots,\right.\right.&\left.\left.x_{n}\right]\right) \\ &=\left[x_{1}, x_{2}, \ldots, x_{k}, 0, \ldots, 0\right] . \end{aligned} $$ a. Show that \(T_{k}\) is a linear transformation of \(V\) into \(W_{\text {h }}\) and that \(T_{k}[V]=\) \(\left\\{T_{k}(\mathbf{v}) \mid \mathbf{v}\right.\) in \(\left.V\right\\}\) is a subspace of \(W_{k}\). b. If \(T_{k}[V]\) has dimension \(d_{k}\), show that, for each \(j

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