Chapter 1: Problem 7
The vectors \(u_{1}=(2,-3,1), u_{2}=(1,4,-2), u_{3}=(-8,12,-4), u_{4}=\) \((1,37,-17)\), and \(u_{5}=(-3,-5,8)\) generate \(\mathrm{R}^{3} .\) Find a subset of the set \(\left\\{u_{1}, u_{2}, u_{3}, u_{4}, u_{5}\right\\}\) that is a basis for \(\mathrm{R}^{3}\).
Short Answer
Expert verified
A subset that forms a basis for \(\mathrm{R}^{3}\) from the set \(\left\\{u_{1}, u_{2}, u_{3}, u_{4}, u_{5}\right\\}\) is given by \(\left\\{u_{1}, u_{2}, u_{5}\right\\}\), as these vectors are linearly independent.
Step by step solution
01
Check linear independence of \(u_1\) and \(u_2\)
To check if vectors \(u_1\) and \(u_2\) are linearly independent, we will try to express one of the vectors as a linear combination of the other. If not possible, the vectors are linearly independent. For simplification, we can represent \(u_1\) and \(u_2\) as follows:
\( u_1 = 2i-3j+k \) and \( u_2 = i+4j-2k \)
Now, let \(a\) and \(b\) be constants such that:
\(au_1 + bu_2 = 0\)
Using the simplified notation of \(u_1\) and \(u_2\), the equation becomes:
\(a(2i-3j+k) + b(i+4j-2k) = 0\)
Expanding the equation into its components:
i-component: \(2a + b = 0\)
j-component: \(-3a + 4b = 0\)
k-component: \(a - 2b = 0\)
02
Solve the system of linear equations
Solve the system of linear equations:
i-component: \(b = -2a\)
Substitute the value of \(b\) in the j-component equation:
\(-3a + 4(-2a) = 0\)
\(-3a - 8a = 0\)
\(-11a = 0\)
\(a = 0\)
Substitute the value of \(a\) in the equation for \(b\):
\(b=(-2)(0)\)
\(b=0\)
Since the only solution for this system of linear equations is \(a=0\) and \(b=0\), \(u_1\) and \(u_2\) are linearly independent.
03
Check linear independence of \(u_3\) with \(u_1\) and \(u_2\)
Next, we will check if \(u_3\) is linearly independent with \(u_1\) and \(u_2\). To do that, we will try to write \(u_3\) as a linear combination of \(u_1\) and \(u_2\). If no combination exists, the trio of vectors are linearly independent.
However, on close observation, we notice that \(u_3\) can be represented as a linear combination of \(u_1\) and \(u_2\):
\(u_3 = -4u_1 + u_2\)
Therefore, \(u_3\) is linearly dependent on \(u_1\) and \(u_2\). Hence we do not include \(u_3\) in the basis.
04
Check linear independence of \(u_4\) with \(u_1\) and \(u_2\)
Similarly, let's analyze \(u_4\). We will try to write \(u_4\) as a linear combination of \(u_1\) and \(u_2\). If no combination exists, the trio of vectors is linearly independent.
However, we again notice that \(u_4\) can be represented as a linear combination of \(u_1\) and \(u_2\):
\(u_4 = 7u_1 + 9u_2\)
Thus, \(u_4\) is linearly dependent on \(u_1\) and \(u_2\). Hence we do not include \(u_4\) in the basis.
05
Check linear independence of \(u_5\) with \(u_1\) and \(u_2\)
Finally, let's evaluate \(u_5\). We will try to write \(u_5\) as a linear combination of \(u_1\) and \(u_2\). If no combination exists, the trio of vectors is linearly independent.
Checking the relationship between the components, we find that \(u_5\) is not a linear combination of \(u_1\) and \(u_2\).
Therefore, \(u_5\) is linearly independent with \(u_1\) and \(u_2\).
06
Identify the basis for R³
We have found a subset of three linearly independent vectors in \(\left\\{u_{1}, u_{2}, u_{3}, u_{4}, u_{5}\right\\}\) which forms a basis for R³. The basis for R³ consists of:
\(\left\\{u_{1}, u_{2}, u_{5}\right\\}\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
basis for R³
In the realm of linear algebra, a **basis for** \(\mathbb{R}^3\) refers to a set of vectors that are both linearly independent and span the vector space \(\mathbb{R}^3\). This means any vector in \(\mathbb{R}^3\) can be expressed as a linear combination of vectors from the basis.
A basis provides a simplest form of representation for vectors and is fundamental for understanding the structure of vector spaces. For three-dimensional space, a basis must consist of exactly three vectors, because \(\mathbb{R}^3\) is a three-dimensional space, which implies that it takes three vectors to span the space.
Importantly, these vectors must be **linearly independent**, meaning that no vector in the basis can be written as a combination of others in that basis. In this exercise, the vectors \(u_1\), \(u_2\), and \(u_5\) form a basis for \(\mathbb{R}^3\) because:
A basis provides a simplest form of representation for vectors and is fundamental for understanding the structure of vector spaces. For three-dimensional space, a basis must consist of exactly three vectors, because \(\mathbb{R}^3\) is a three-dimensional space, which implies that it takes three vectors to span the space.
Importantly, these vectors must be **linearly independent**, meaning that no vector in the basis can be written as a combination of others in that basis. In this exercise, the vectors \(u_1\), \(u_2\), and \(u_5\) form a basis for \(\mathbb{R}^3\) because:
- The combination of these vectors can span \(\mathbb{R}^3\).
- No vector can be written as a mixture of the other two, making them linearly independent.
linear combination
A **linear combination** in mathematics is an expression constructed from a set of terms by multiplying each term by a constant and adding the results. In terms of vectors, linear combinations involve adding two or more vectors together, each multiplied by a scalar (or constant), to create a new vector.
Suppose we have vectors \(u_1, u_2,\) and a constant \(a, b\), the linear combination might look like this: \(a u_1 + b u_2\).
The main idea is that, when you calculate with the right scalars, you can compose new vectors that cover a whole vector space such as \(\mathbb{R}^3\).
For instance, in the solution, we saw that \(u_3\) could be represented as a linear combination of \(u_1\) and \(u_2\): \(u_3 = -4u_1 + u_2\). This representation implies that \(u_3\) does not add any new dimensionality beyond what \(u_1\) and \(u_2\) already cover.
Hence, understanding and using linear combinations is essential for determining linear dependency and finding a basis, as it helps decide which vectors carry unique information and which depend on others.
Suppose we have vectors \(u_1, u_2,\) and a constant \(a, b\), the linear combination might look like this: \(a u_1 + b u_2\).
The main idea is that, when you calculate with the right scalars, you can compose new vectors that cover a whole vector space such as \(\mathbb{R}^3\).
For instance, in the solution, we saw that \(u_3\) could be represented as a linear combination of \(u_1\) and \(u_2\): \(u_3 = -4u_1 + u_2\). This representation implies that \(u_3\) does not add any new dimensionality beyond what \(u_1\) and \(u_2\) already cover.
Hence, understanding and using linear combinations is essential for determining linear dependency and finding a basis, as it helps decide which vectors carry unique information and which depend on others.
vectors in R³
Vectors in \(\mathbb{R}^3\) are fundamental objects of study in linear algebra. Each vector has three components, corresponding to the three dimensions in space.
A vector is often notated as \((x, y, z)\), where \(x\), \(y\), and \(z\) represent the vector's components on the \(x\), \(y\), and \(z\) axes, respectively.
These coordinates give the vector a direction and a magnitude, visually representing a point or arrow in 3D space.
Consider the vectors presented in our exercise like \(u_1=(2,-3,1)\) and \(u_2=(1,4,-2)\). These vectors are expressed in the standard form showing their influence on each axis: \(x\), \(y\), and \(z\).
Such representations make vectors powerful tools for solving real-world problems, in fields ranging from physics to computer graphics.
When dealing with \(\mathbb{R}^3\), most often we explore how vectors interact through operations like addition, subtraction, and scalar multiplication.
Thus, learning about vectors in \(\mathbb{R}^3\) sets the foundation for tackling more complex problems in higher dimensions and varying applications. Understanding these interactions is essential for proficiency in analyzing and manipulating multi-dimensional spaces efficiently.
A vector is often notated as \((x, y, z)\), where \(x\), \(y\), and \(z\) represent the vector's components on the \(x\), \(y\), and \(z\) axes, respectively.
These coordinates give the vector a direction and a magnitude, visually representing a point or arrow in 3D space.
Consider the vectors presented in our exercise like \(u_1=(2,-3,1)\) and \(u_2=(1,4,-2)\). These vectors are expressed in the standard form showing their influence on each axis: \(x\), \(y\), and \(z\).
Such representations make vectors powerful tools for solving real-world problems, in fields ranging from physics to computer graphics.
When dealing with \(\mathbb{R}^3\), most often we explore how vectors interact through operations like addition, subtraction, and scalar multiplication.
Thus, learning about vectors in \(\mathbb{R}^3\) sets the foundation for tackling more complex problems in higher dimensions and varying applications. Understanding these interactions is essential for proficiency in analyzing and manipulating multi-dimensional spaces efficiently.