Problem 36 - Differential Equations and Linear Algebra

Question

Determine all values of the constant \(\alpha\) for which \(\left\\{1+\alpha x^{2}, 1+x+x^{2}, 2+x\right\\}\) is a basis for \(P_{2}(\mathbb{R})\)

Step-by-Step Solution

Answer

For all \(\alpha \neq -1, 1, \frac{1}{2}\), the polynomials \(\left\{1+\alpha x^2, 1+x+x^2, 2+x\right\}\) form a linearly independent set and thus a basis for \(P_{2}(\mathbb{R})\).

Step 1: Set Up the Linear Combination

We want to check if the polynomials in the set are linearly independent. We will set up a linear combination of the polynomials and try to find the scalars that satisfy the equation: \(c_1(1+\alpha x^2) + c_2(1+x+x^2) + c_3(2+x) = 0\) We will rewrite this equation as a system of linear equations in terms of the coefficients.

Step 2: Rewrite the Linear Combination as a System of Linear Equations

Equating the coefficients of x: \(\alpha c_1 + c_2 + c_3 = 0\) Equating the coefficients of \(x^2\): \(\alpha c_1 + c_2 = 0\) Equating the constant terms: \(c_1 + 2c_2 + 2c_3 = 0\) Now, we have a system of linear equations: \[\begin{cases}\alpha c_1 + c_2 + c_3 = 0\\ \alpha c_1 + c_2 = 0\\ c_1 + 2c_2 + 2c_3 = 0\end{cases}\]

Step 3: Solve the System of Linear Equations for c1, c2, and c3

From the 2nd equation, we have: \(c_2 = -\alpha c_1\) Substitute the expression for \(c_2\) into the 1st and 3rd equations: \(\alpha c_1 - \alpha^2 c_1 + c_3 = 0\) \(c_1 - 2\alpha c_1 + 2c_3 = 0\)

Step 4: Simplify the System of Linear Equations

Now we simplify the first equation: \((1-\alpha^2)c_1 + c_3 = 0\) And simplify the third equation: \((1-2\alpha)c_1 + 2c_3 = 0\) To be linearly independent, the only solution should be \(c_1 = c_2 = c_3 = 0\). Now, only two linear equations are left. We want these two equations to be consistent only when having \(c_1 = c_3 = 0\). As we require \(\alpha\) to be a constant number, we can compare the ratios of coefficients of \(c_1\) and \(c_3\) to determine the values for \(\alpha\).

Step 5: Determine the Values of \(\alpha\)

From the simplified system of linear equations, divide the first equation by the second equation: \[\frac{1-\alpha^2}{1-2\alpha} = \frac{(1-\alpha^2)c_1 + c_3}{(1-2\alpha)c_1 + 2c_3} = \frac{c_1}{2c_3}\] Now, set the numerator (c_1) and the denominator (2c_3) to zero independently, as this would give us the trivial solution: Numerator: \(1 - \alpha^2 = 0 \Rightarrow \alpha = \pm 1\) Denominator: \(1 - 2\alpha = 0 \Rightarrow \alpha = \frac{1}{2}\) Since we are looking for values of \(\alpha\) for which the polynomials are linearly independent, we want to exclude the values of \(\alpha\) for which there is a non-trivial solution. Therefore, we should exclude \(\alpha = \pm 1\) and \(\alpha = \frac{1}{2}\). For all other values of \(\alpha\), the polynomials \(\left\{1+\alpha x^2, 1+x+x^2, 2+x\right\}\) form a linearly independent set and thus a basis for \(P_{2}(\mathbb{R})\).

Key Concepts

Linear Independence

Linear independence is a fundamental concept in linear algebra, which helps us determine if a set of vectors or functions can form a basis for a given space. In simple terms, a set is linearly independent if no vector in the set can be written as a combination of the other vectors. When it comes to polynomials, this means that no polynomial in the set can be expressed as a combination of the others using scalar coefficients.

For instance, in our exercise, we are verifying if the set of polynomials \(\{1+\alpha x^2, 1+x+x^2, 2+x\}\) is linearly independent. We achieve this by setting up a linear combination of the polynomials equated to zero:

• \(c_1(1+\alpha x^2) + c_2(1+x+x^2) + c_3(2+x) = 0\)

Here, \(c_1, c_2,\) and \(c_3\) are scalars. For linear independence, the only solution to this equation must be \(c_1 = c_2 = c_3 = 0\). Thus, we seek the values of \(\alpha\) that fulfill this criteria. This process checks if the polynomials can serve as a basis for a polynomial space by being linearly independent.

System of Linear Equations

A system of linear equations consists of two or more linear equations with the same set of variables. Solving such systems is crucial in checking conditions like linear independence. In our task, breaking down the linear combination of polynomials resulted in the following system of equations:

• \(\alpha c_1 + c_2 + c_3 = 0\)
• \(\alpha c_1 + c_2 = 0\)
• \(c_1 + 2c_2 + 2c_3 = 0\)

These equations stem from equating the coefficients of like terms from the polynomials in the set. Solving this system involves expressing one of the variables in terms of the others and substituting back into the other equations. This substitution allows us to simplify the system and find conditions under which there are only trivial solutions (i.e., zero solutions).

In our scenario, we derived that certain values of \(\alpha\) lead to non-trivial solutions, indicating the polynomials are not linearly independent for those values. This process powerfully demonstrates how systems of equations enable us to analyze conditions needed for independence and thereby determining basis qualification.

Polynomial Spaces

Polynomial spaces expand our understanding of vector spaces to polynomials, where each polynomial can be considered a vector. In this context, the space of all polynomials with degree less than or equal to a certain value \(n\), say \(P_n(\mathbb{R})\), includes all real polynomials with degrees up to \(n\). For example, \(P_2(\mathbb{R})\) consists of all quadratic polynomials.

To form a basis for such a space, we need a set of linearly independent polynomials whose linear combinations can express any polynomial in \(P_n(\mathbb{R})\). The number of polynomials in this set must match the dimension of the space, which is \(n+1\) for \(P_n(\mathbb{R})\). So, for \(P_2(\mathbb{R})\), we need 3 linearly independent polynomials.

In our exercise, we were tasked with checking if the set \(\{1+\alpha x^2, 1+x+x^2, 2+x\}\) spans \(P_2(\mathbb{R})\) by being a linearly independent basis. By solving the linear combination set up in the system of equations, we assure that for values of \(\alpha\) excluding \(\pm 1\) and \(\frac{1}{2}\), this particular set indeed qualifies as a basis for \(P_2(\mathbb{R})\). This showcases the elegance of polynomial spaces in algebra, where polynomials behave analogously to vectors.