Problem 21 - Differential Equations and Linear Algebra

Question

determine whether the given set of vectors is linearly independent in \(P_{2}(\mathbb{R})\). $$p_{1}(x)=1-3 x^{2}, \quad p_{2}(x)=2 x+x^{2}, \quad p_{3}(x)=5$$.

Step-by-Step Solution

Answer

The given set of polynomials \(p_1(x) = 1-3x^2\), \(p_2(x) = 2x+x^2\), and \(p_3(x) = 5\) is linearly independent in \(P_2(\mathbb{R})\) since the only coefficients that make the linear equation \(c_1p_1(x) + c_2p_2(x) + c_3p_3(x) = 0\) true are \(c_1 = c_2 = c_3 = 0\).

Step 1: Write the equation for checking linear independence

Write the equation to check for linear independence using the given polynomials and scalar coefficients: $$c_1(1-3x^2) + c_2(2x+x^2) + c_3(5) = 0$$

Step 2: Combine terms and collect coefficients

Combine the polynomial terms and collect the coefficients of each term: $$c_3 = 5c_3, \quad c_2 = c_2, \quad x(c_2+4c_1) = c_2+4c_1, \quad x^2(-3c_1 - c_2) = 0$$

Step 3: Create a system of linear equations

Using the coefficients found in Step 2, create a system of linear equations to solve for the values of \(c_1\), \(c_2\), and \(c_3\): \[ \begin{cases} 5c_3 = 0 \\ c_2 + 4c_1 = 0 \\ -3c_1 - c_2 = 0 \end{cases} \]

Step 4: Solve the system of linear equations

First, from the first equation, we find that $$c_3 = 0$$ Next, use the second and third equations to solve for \(c_1\) and \(c_2\): \[ \begin{cases} c_2 = -4c_1 \\ -3c_1 - (-4c_1) = 0 \end{cases} \] Using substitution, we get: $$c_1 = 0$$ Now substituting the value of \(c_1\) back into the equation for \(c_2\) we get: $$c_2 = 0$$

Step 5: Determine linear independence

Since the coefficients that make the linear equation true are \(c_1 = c_2 = c_3 = 0\), we can conclude that the set of polynomials: $$p_{1}(x)=1-3 x^{2}, \quad p_{2}(x)=2 x+x^{2}, \quad p_{3}(x)=5$$ is linearly independent in \(P_{2}(\mathbb{R})\).

Key Concepts

System of Linear Equations

In the context of linear algebra, a system of linear equations is a collection of one or more linear equations involving the same set of variables. In this exercise, we examine the linear independence of polynomial vectors, which leads us to form such a system.

• The equation: \[ c_1(1-3x^2) + c_2(2x+x^2) + c_3(5) = 0 \] maps to the components of each variable.
• Collect these components to assemble the equations: \[ \begin{cases} 5c_3 = 0 \ c_2 + 4c_1 = 0 \ -3c_1 - c_2 = 0 \end{cases} \]

This system is key to solving for the coefficients \(c_1\), \(c_2\), and \(c_3\). When each equation equals zero, it suggests the combination of polynomials is linearly independent.
To verify, solve iteratively or using substitution. This exercise illustrates how systems of linear equations underpin the concept of linear independence.

Polynomial Vectors

Polynomial vectors are unique mathematical objects where each element of the vector is a polynomial. In our exercise, we're working with a vector space composed of polynomials of degree at most 2, represented as \( P_{2}(\mathbb{R}) \).
The polynomials given are:

• \( p_{1}(x) = 1 - 3x^2 \)
• \( p_{2}(x) = 2x + x^2 \)
• \( p_{3}(x) = 5 \)

These polynomials can be envisioned as vectors in a three-dimensional space where each dimension corresponds to a basis polynomial, particularly \( 1 \), \( x \), and \( x^2 \).
Understanding polynomial vectors helps visualize how these individual polynomials combine linearly. When confirming linear independence, the goal is to see if a non-trivial combination of these vectors results in the zero vector. This means checking if the only solution for the coefficients \( c_1, c_2, c_3 \) that satisfies the equation is zero.
In our problem, because no non-trivial combination yields the zero vector, we confirm these polynomial vectors are linearly independent.

Coefficients

Coefficients play a crucial role in determining linear independence. They are the scalars (often denoted by \( c_1, c_2, c_3 \) in our scenario) that are multiplied by each polynomial vector.
Starting with the equation:\[ c_1(1 - 3x^2) + c_2(2x + x^2) + c_3(5) = 0 \]
Our task is to derive the coefficients within this combination that would result in the zero polynomial vector through a system of equations.

• The equation \( 5c_3 = 0 \) immediately tells us \( c_3 = 0 \).
• The other equations work in tandem to show that \( c_1 \) and \( c_2 \) must also be zero.

Through these analyses, we find all coefficients are zero.
This zero-result confirms the linear independence of our polynomial vectors. Thus, understanding coefficients helps in checking whether different polynomial vectors can combine to form something new and distinct or simply revert back to zero.