Question

In each case show that the condition \(a \mathbf{u}+b \mathbf{v}+c \mathbf{w}=\mathbf{0}\) in \(V\) implies that \(a=b=c=0 .\) a. \(V=\mathbb{R}^{4} ; \mathbf{u}=(2,1,0,2), \mathbf{v}=(1,1,-1,0)\) \(\mathbf{w}=(0,1,2,1)\) $$ \begin{aligned} \text { b. } V &=\mathbf{M}_{22} ; \mathbf{u}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right], \mathbf{v}=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \\ \mathbf{w} &=\left[\begin{array}{rr} 1 & 1 \\ 1 & -1 \end{array}\right] \\ \text { c. } V &=\mathbf{P} ; \mathbf{u}=x^{3}+x, \mathbf{v}=x^{2}+1, \mathbf{w}=x^{3}-x^{2}+x+1 \end{aligned} $$ d. \(V=\mathbf{F}[0, \pi] ; \mathbf{u}=\sin x, \mathbf{v}=\cos x, \mathbf{w}=1-\) the constant function

Short answer

For all cases, the condition implies that \(a = b = c = 0\), proving linear independence.

Step-by-step solution

Show linear independence for vectors in \(\mathbb{R}^4\)

First, we start with the equation \(a\mathbf{u} + b\mathbf{v} + c\mathbf{w} = \mathbf{0}\) for \(\mathbb{R}^4\). Substitute the vectors: \(a(2,1,0,2) + b(1,1,-1,0) + c(0,1,2,1) = (0,0,0,0)\). This gives us the system of equations:1. \(2a + b = 0\)2. \(a + b + c = 0\)3. \(-b + 2c = 0\)4. \(2a + c = 0\).Solve this system to determine the values of \(a, b,\) and \(c\).

Solve the system for \(\mathbb{R}^4\)

Using the system of equations:1. From equation 3, \(-b + 2c = 0\) gives \(b = 2c\).2. Substituting \(b = 2c\) in equation 2, \(a + 2c + c = 0\), gives \(a + 3c = 0\), so \(a = -3c\).3. Substitute \(a = -3c\) and \(b = 2c\) in equation 1: \(2(-3c) + 2c = 0\), solve to find \(-6c + 2c = 0\), thus \(-4c = 0\), giving \(c = 0\).4. Substitute \(c = 0\) back to get \(b = 0\) and \(a = 0\).Thus, \(a = b = c = 0\), proving linear independence.

Show linear independence for matrices in \(\mathbf{M}_{22}\)

Substitute the matrices into the equation \(a\mathbf{u} + b\mathbf{v} + c\mathbf{w} = \mathbf{0}\) to get:\[a\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + b\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} + c\begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}\].This leads to the system of equations:1. \(a + c = 0\)2. \(b + c = 0\)3. \(b + c = 0\)4. \(a - c = 0\).Solve to determine \(a, b,\) and \(c\).

Solve the system for \(\mathbf{M}_{22}\)

From equation 1 and 4, we add them: \((a + c) + (a - c) = 0 \Rightarrow 2a = 0\), giving \(a = 0\).Substituting \(a = 0\) in equation 1: \(c = 0\). Use \(c = 0\) in equation 2: \(b + 0 = 0\), hence \(b = 0\).Therefore, \(a = b = c = 0\), proving linear independence in matrix space.

Show linear independence for polynomials in \(\mathbf{P}\)

Consider \(a(x^3 + x) + b(x^2 + 1) + c(x^3 - x^2 + x + 1) = 0\). Expand to obtain:1. Cubic terms: \((a + c)x^3\)2. Quadratic terms: \(bx^2 - cx^2\) which simplifies to \((b - c)x^2\)3. Linear terms: \(ax + cx\) giving \((a + c)x\)4. Constant terms: \(b + c = 0\).Equate coefficients to zero to solve.

Solve coefficient equations for \(\mathbf{P}\)

Equating coefficients to zero gives:1. Cubic: \(a + c = 0\)2. Quadratic: \(b - c = 0\)3. Linear: \(a + c = 0\) -- same as cubic4. Constant: \(b + c = 0\).Since cubic and linear are similar, use any (say cubic): \(a + c = 0\). From quadratic: \(b = c\), and from constant: \(b = -c\). Both imply \(c = 0\), then \(a = 0\) and \(b = 0\). Thus, \(a = b = c = 0\), confirming independence.

Demonstrate linear independence for functions in \(\mathbf{F}[0, \pi]\)

Consider the equation: \(a \sin x + b \cos x + c = 0\) for all \(x\). Differentiate to get: \(a \cos x - b \sin x = 0\). Consider special values of \(x\) to solve for \(a, b, \) and \(c\).

Solve for \(\mathbf{F}[0, \pi]\)

At \(x = 0\), the equation becomes \(c = 0\). Differentiate gives: \(a \cos x - b \sin x = 0\). At \(x = \pi/2\), \(a \cdot 0 - b \cdot 1 = 0\), yielding \(b = 0\). At \(x=0\), substitute back into original for \(a \sin 0 + 0 + 0 = 0\), so \(a = 0\). Thus, \(a = b = c = 0\), asserting the functions are linearly independent.

Key concepts

Linear Independence

Understanding linear independence is crucial in linear algebra. It tells us whether a set of vectors (or functions) can be expressed uniquely in terms of others in the set. If a group of vectors is linearly independent, this means that the only solution to the equation formed, by any linear combination of these vectors equaling the zero vector, is when all coefficients are zero (i.e., if \(a\mathbf{u} + b\mathbf{v} + c\mathbf{w} = \mathbf{0}\) implies \(a = b = c = 0\)). This property is highly valued as it implies that none of the vectors in the set can be constructed from a combination of the others.
Linear independence lays the foundation for many concepts in linear algebra. One crucial aspect is its role in defining bases in vector spaces. A basis is a set of linearly independent vectors that spans the entire space, ensuring that each vector in the space can be expressed as a unique combination of the basis vectors. This is why checking linear independence is often the first step when dealing with vector spaces.

Vector Spaces

Vector spaces are a fundamental concept in linear algebra, providing a framework where vectors can be added together and multiplied by scalars, i.e., numbers. Formally, a vector space over a field \(F\) is a set \(V\) along with two operations: vector addition and scalar multiplication. These operations must satisfy certain axioms, such as commutativity, associativity, the existence of an additive identity, and distributivity, among others.
A key feature of vector spaces is their ability to be spanned by a set of vectors. If vectors are linearly independent and span the vector space, they form a basis. This means that any vector in the space can be written as a linear combination of these basis vectors. Vector spaces are used across various fields, including physics and computer science, as they provide a means to construct and manipulate data objects consistently and predictably.

Systems of Equations

Systems of equations are a collection of two or more equations with the same set of unknowns. Solving these systems is a common task in linear algebra, often involving techniques like substitution, elimination, or matrix methods such as Gaussian elimination. The solution to a system of equations is the set of values for the unknowns that satisfy all the equations simultaneously.
When a problem involves determining linear independence for vectors or functions, solving systems of equations becomes pivotal. For example, in our exercise, we formed equations by substituting given vectors into a linear combination equating to zero. The task was to show that the only solution is the trivial one, where all coefficients are zero, thus proving linear independence.

Matrices

Matrices are rectangular arrays of numbers, symbols, or expressions, arranged in rows and columns. They are an essential tool in linear algebra, lending themselves to a wide range of applications, from solving systems of equations to transformations in geometry. Matrices allow us to compactly represent and manipulate data structures.
Matrices also play a role in determining linear independence. In the context of the exercise, a matrix can be used to represent a system of linear equations. Each row can correspond to an equation, and each column represents the coefficients of a particular vector or variable. Solving for linear independence often involves performing operations on matrices such as row reduction, where we use elementary row operations to simplify the matrix, allowing us to easily obtain solutions to the system of equations and check for triviality.

Polynomials

Polynomials are expressions consisting of variables and coefficients, involving terms in the form \(ax^n\), where \(a\) is a coefficient and \(n\) is a non-negative integer. They can be seen as vectors in a polynomial vector space, whereby each term is equivalent to an element in a vector.
In the exercise's context, we considered polynomial vectors in determining linear independence. Similar methods of forming a system of equations using coefficients apply here as well. By comparing coefficients of similar powers of \(x\) from both sides of a polynomial equation, we extract systems of equations. Solving for all coefficients equalling zero indicates the polynomials are linearly independent. This approach exemplifies how abstract concepts in mathematics can have structured methods of solution, paralleling more intuitive numerical techniques.