Problem 27 - Advanced Engineering Mathematics

Question

In Problems \(25-28\), determine whether the given vectors are linearly independent or linearly dependent. $$ 1,(x+1),(x+1)^{2} \text { in } P_{2} $$

Step-by-Step Solution

Answer

The vectors are linearly independent.

Step 1: Understanding Vectors in Context

We are given three vectors: \(1\), \((x+1)\), and \((x+1)^2\), which are to be checked for linear dependence or independence in the polynomial space \(P_2\). The space \(P_2\) includes all polynomials of degree up to 2. These vectors can be represented as polynomials.

Step 2: Setting Up a Linear Combination

To determine linear dependence, we set up a linear combination equation: \(a \cdot 1 + b \cdot (x+1) + c \cdot (x+1)^2 = 0\), where \(a\), \(b\), and \(c\) are scalar coefficients. Our goal is to find whether non-trivial solutions exist for \(a\), \(b\), and \(c\) such that the linear combination equals the zero polynomial.

Step 3: Simplifying the Linear Combination

Expand the linear combination:\[ a + b(x+1) + c(x^2 + 2x + 1) = 0 \]This becomes:\[ a + bx + b + cx^2 + 2cx + c = 0 \]Collecting the like terms gives us:\[ cx^2 + (b+2c)x + (a+b+c) = 0 \]

Step 4: Solving for Coefficients

Since this equation must hold for all \(x\), each coefficient of \(x^2\), \(x\), and the constant term must be zero. This gives us the system of equations:1. \(c = 0\) (for \(x^2\))2. \(b + 2c = 0\) (for \(x\))3. \(a + b + c = 0\) (for the constant term)

Step 5: Analyzing the System of Equations

From the equation \(c = 0\), substitute \(c = 0\) into \(b + 2c = 0\), yielding \(b = 0\). Substituting \(b = 0\) and \(c = 0\) into \(a + b + c = 0\) results in \(a = 0\). These are trivial solutions, indicating no non-trivial solutions exist.

Step 6: Conclusion on Dependency

Since the only solution to the system of equations \(a = 0\), \(b = 0\), \(c = 0\) is the trivial solution, the vectors \(1\), \((x+1)\), and \((x+1)^2\) are linearly independent.

Key Concepts

Polynomials

A polynomial is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. For example, in the problem, we have the polynomials represented by the vectors \(1\), \((x+1)\), and \((x+1)^2\). These have degrees ranging from 0 to 2, respectively.
Polynomials are fundamental elements in algebra, functioning under certain operations such as addition, subtraction, and multiplication similar to numbers. In this particular problem, our task is within the polynomial space, specifically \(P_2\), which contains polynomials of degree up to 2.

• Constant polynomials such as \(1\) have no variables, meaning their degree is 0.
• Linear polynomials like \((x+1)\) contain a variable raised to the first power, giving them a degree of 1.
• Quadratic polynomials, such as \((x+1)^2\), involve variables raised to the second power.

Understanding polynomials involves recognizing their structure and how to identify their degree. In vector spaces, they can serve as basic elements or 'vectors' to analyze various mathematical properties, such as linear independence.

Linear Combinations

To explore linear combinations, consider how we can form new expressions by combining given polynomials using scalar coefficients. In our exercise, we assess whether these polynomials can sum up to form the zero polynomial. We achieve this using a linear combination of the form \(a \cdot 1 + b \cdot (x+1) + c \cdot (x+1)^2 = 0\).
Linear combinations are utilized to test for linear dependency. This entails rearranging and solving for the coefficients \(a\), \(b\), and \(c\). Here’s what we do:

• Expand and rearrange the expression: every term involving \(x\), \(x^2\), and the constants are gathered separately.
• Set each type of term (e.g., those with \(x^2\)) equal to zero to ensure the equation holds for all values of \(x\).

Linear independence implies that the only solution for \(a\), \(b\), and \(c\) is the trivial one: all coefficients are zero. This outcome shows that no polynomial in \(P_2\) can be re-expressed as a combination of the others, confirming their linear independence.

Vector Spaces

A vector space is a collection of objects, called vectors, which can be added together and multiplied (scaled) by numbers, called scalars. In this context, the polynomials \(1\), \((x+1)\), and \((x+1)^2\) form elements of a vector space known as \(P_2\).
Vector spaces must satisfy specific rules, or axioms, related to vector addition and scalar multiplication, like commutativity, associativity, and distributivity, to name a few. These rules help structure how polynomials behave when combined in a vector-like manner. Here are some things to know about vector spaces:

• Subspace: A subset of a vector space is also a vector space if it satisfies all vector space properties.
• Linear Dependence: If any vector can be represented as a combination of others, the group is dependent.
• Linear Independence: The polynomials form a set where none can be written as a linear combination of the others, showing independence.

This problem illustrates how polynomial vectors interact within a vector space, emphasizing their independence through testing possible combinations. Recognizing such independence is crucial in many areas of linear algebra and applied mathematics.