Linear Independence And Span Analysis In Polynomial Vector Spaces

In the realm of linear algebra, understanding concepts like linear independence and the span of a set of vectors is crucial. These concepts are not limited to traditional vector spaces like $\mathbb{R}^n$; they extend to other vector spaces, including polynomial spaces. This article delves into these ideas within the context of the polynomial vector space $P_2$, which consists of polynomials of degree at most 2. We will analyze a specific subset $S$ of $P_2$, given by $S = \{1 + x + 2x^2, 1 - x, -2 + kx - x^2\}$, and explore the values of $k$ for which $S$ is linearly independent. Furthermore, we will demonstrate, using the definition of the span, that $S$ does not span the entire vector space $P_2$.

To determine the values of $k$ for which the subset $S = \{1 + x + 2x^2, 1 - x, -2 + kx - x^2\}$ is linearly independent, we need to investigate the linear combination of these polynomials that equals the zero polynomial. Let $c_1$, $c_2$, and $c_3$ be scalar coefficients. We consider the equation:

$c_1(1 + x + 2x^2) + c_2(1 - x) + c_3(-2 + kx - x^2) = 0$

This equation must hold true only when $c_1 = c_2 = c_3 = 0$ for $S$ to be linearly independent. Expanding and collecting like terms, we get:

$(c_1 + c_2 - 2c_3) + (c_1 - c_2 + kc_3)x + (2c_1 - c_3)x^2 = 0$

For this polynomial to be the zero polynomial (i.e., equal to 0 for all values of $x$), the coefficients of each power of $x$ must be zero. This leads to the following system of linear equations:

1. $c_1 + c_2 - 2c_3 = 0$
2. $c_1 - c_2 + kc_3 = 0$
3. $2c_1 - c_3 = 0$

We can represent this system of equations as an augmented matrix and use Gaussian elimination to solve for the coefficients:

$\begin{bmatrix} 1 & 1 & -2 & 0 \\ 1 & -1 & k & 0 \\ 2 & 0 & -1 & 0 \end{bmatrix}$

Subtracting the first row from the second row and subtracting twice the first row from the third row, we get:

$\begin{bmatrix} 1 & 1 & -2 & 0 \\ 0 & -2 & k + 2 & 0 \\ 0 & -2 & 3 & 0 \end{bmatrix}$

Now, subtracting the second row from the third row, we obtain:

$\begin{bmatrix} 1 & 1 & -2 & 0 \\ 0 & -2 & k + 2 & 0 \\ 0 & 0 & 1 - k & 0 \end{bmatrix}$

For the subset $S$ to be linearly independent, the only solution to this system must be the trivial solution ($c_1 = c_2 = c_3 = 0$). This occurs if and only if the coefficient matrix has a non-zero determinant. From the row-echelon form, we can see that this requires $1 - k \neq 0$, which implies $k \neq 1$. Therefore, the subset $S$ is linearly independent for all values of $k$ except $k = 1$. If $k=1$ the system has infinite solutions and the subset $S$ is not linearly independent. In summary, to show the linear independence of the given set $S$, we set up a linear combination of the polynomials equal to the zero polynomial. This leads to a system of linear equations, which we solve using an augmented matrix and Gaussian elimination. The condition for linear independence is that the only solution to the system is the trivial solution, which occurs when the determinant of the coefficient matrix is non-zero. In this case, the subset $S$ is linearly independent for all values of $k$ except $k = 1$. This means that if $k$ is any value other than 1, the polynomials in $S$ do not form a linearly dependent set and can be used as a basis for a subspace of $P_2$. Understanding linear independence is essential in linear algebra as it helps in identifying sets of vectors (or polynomials in this case) that can uniquely represent other vectors within their span. The process of Gaussian elimination is a fundamental tool in determining the solutions of systems of linear equations and plays a crucial role in various applications, including network analysis, optimization problems, and computer graphics. Moreover, this concept extends beyond polynomial spaces and is applicable to any vector space, making it a cornerstone of linear algebra theory and practice. Understanding these concepts is important for more advanced topics in linear algebra and its applications in various fields such as engineering, physics, and computer science.

To show that $S = \{1 + x + 2x^2, 1 - x, -2 + kx - x^2\}$ does not span $P_2$, we need to demonstrate that there exists at least one polynomial in $P_2$ that cannot be written as a linear combination of the polynomials in $S$. Recall that $P_2$ is the vector space of all polynomials of degree at most 2. A general polynomial in $P_2$ can be written as $ax^2 + bx + c$, where $a$, $b$, and $c$ are scalars.

The span of $S$, denoted as $\text{span}(S)$, is the set of all possible linear combinations of the polynomials in $S$. That is,

$\text{span}(S) = \{c_1(1 + x + 2x^2) + c_2(1 - x) + c_3(-2 + kx - x^2) \mid c_1, c_2, c_3 \in \mathbb{R}\}$

where $c_1$, $c_2$, and $c_3$ are scalar coefficients. We want to show that there exists a polynomial $ax^2 + bx + c$ in $P_2$ that is not in $\text{span}(S)$. This means that there are no values of $c_1$, $c_2$, and $c_3$ that satisfy the equation:

$c_1(1 + x + 2x^2) + c_2(1 - x) + c_3(-2 + kx - x^2) = ax^2 + bx + c$

Expanding and collecting like terms, we get:

$(c_1 + c_2 - 2c_3) + (c_1 - c_2 + kc_3)x + (2c_1 - c_3)x^2 = ax^2 + bx + c$

Equating the coefficients of the corresponding powers of $x$, we obtain the following system of linear equations:

1. $c_1 + c_2 - 2c_3 = c$
2. $c_1 - c_2 + kc_3 = b$
3. $2c_1 - c_3 = a$

We can represent this system as an augmented matrix:

$\begin{bmatrix} 1 & 1 & -2 & c \\ 1 & -1 & k & b \\ 2 & 0 & -1 & a \end{bmatrix}$

Now, we perform row operations to analyze the system. Subtract the first row from the second row and subtract twice the first row from the third row:

$\begin{bmatrix} 1 & 1 & -2 & c \\ 0 & -2 & k + 2 & b - c \\ 0 & -2 & 3 & a - 2c \end{bmatrix}$

Subtract the second row from the third row:

$\begin{bmatrix} 1 & 1 & -2 & c \\ 0 & -2 & k + 2 & b - c \\ 0 & 0 & 1 - k & a - b - c \end{bmatrix}$

The system is inconsistent (i.e., has no solution) if the last row implies $0 = a - b - c$ when $1-k = 0$ (i.e. $k=1$) and $a - b - c \neq 0$. This means that for $k = 1$, if we choose $a$, $b$, and $c$ such that $a - b - c \neq 0$, then the polynomial $ax^2 + bx + c$ is not in $\text{span}(S)$. For example, let's take $a = 1$, $b = 0$, and $c = 0$. Then, the polynomial $x^2$ is in $P_2$. However, if $k = 1$, the system becomes:

$\begin{bmatrix} 1 & 1 & -2 & 0 \\ 1 & -1 & 1 & 0 \\ 2 & 0 & -1 & 1 \ \end{bmatrix}$

After row reduction, the system becomes:

$\begin{bmatrix} 1 & 1 & -2 & 0 \\ 0 & -2 & 3 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$

The last row represents the equation $0 = 1$, which is a contradiction. Therefore, the polynomial $x^2$ cannot be written as a linear combination of the polynomials in $S$ when $k = 1$. This shows that $S$ does not span $P_2$ when $k=1$. We can generalize this. For $S$ to span $P_2$, the system must be consistent for all $a, b, c$. This requires that the coefficient matrix has a pivot in every row, which means $1-k \neq 0$. When $k = 1$, the third row of the reduced matrix becomes $\begin{bmatrix} 0 & 0 & 0 & a - b - c \end{bmatrix}$. If we choose $a, b, c$ such that $a - b - c \neq 0$, the system is inconsistent, and the polynomial $ax^2 + bx + c$ is not in the span of $S$. Therefore, $S$ does not span $P_2$ when $k=1$. In summary, to demonstrate that $S$ does not span $P_2$, we show that there exists a polynomial in $P_2$ that cannot be expressed as a linear combination of the polynomials in $S$. We set up a general polynomial $ax^2 + bx + c$ and try to solve for the coefficients $c_1, c_2,$ and $c_3$ in the linear combination. By forming a system of linear equations and analyzing the augmented matrix, we find that for specific values of $k$ (in this case, $k = 1$), there are polynomials in $P_2$ that cannot be reached by any linear combination of the polynomials in $S$. This conclusion highlights the importance of the span concept in determining whether a set of vectors can generate the entire vector space. Understanding the span is essential in various applications, including computer graphics, where the ability to represent objects using a basis of vectors is fundamental. Moreover, this understanding forms the basis for many concepts in advanced mathematics, such as functional analysis and Hilbert spaces. Linear algebra is critical in various fields, including engineering, physics, and computer science, and understanding the limitations of a set of vectors' ability to span a space is just as important as understanding when they can.

In this article, we investigated the linear independence and span of a subset $S$ within the polynomial vector space $P_2$. We found that the subset $S = \{1 + x + 2x^2, 1 - x, -2 + kx - x^2\}$ is linearly independent for all values of $k$ except $k = 1$. Furthermore, we demonstrated that $S$ does not span the entire vector space $P_2$ when $k=1$, highlighting the limitations of the set in generating all polynomials of degree at most 2. These concepts are fundamental in linear algebra and have broad applications in various fields, reinforcing the importance of understanding vector spaces, linear independence, and spans.