please finish part a by solving the system of linear equations

Sure, let's complete the solution to part (a) by solving the system of linear equations to show that \(\beta' = \{x + 1, x - 1, x^2 + 1\}\) spans \(P_2(\mathbb{R})\).

We previously derived the following system of equations from setting a general polynomial equal to a linear combination of elements in \(\beta'\):

Given any \(p(x) = \alpha_0 + \alpha_1 x + \alpha_2 x^2\), we express it as:
\[ p(x) = a(x + 1) + b(x - 1) + c(x^2 + 1) \]

Expanding the right-hand side:
\[ a(x + 1) + b(x - 1) + c(x^2 + 1) = ax + a + bx - b + cx^2 + c \]
\[ = cx^2 + (a + b)x + (a - b + c) \]

For this to equal \(\alpha_0 + \alpha_1 x + \alpha_2 x^2\), we match coefficients:
\[ \alpha_2 = c \]
\[ \alpha_1 = a + b \]
\[ \alpha_0 = a - b + c \]

This gives us the system of linear equations:
1. \( c = \alpha_2 \)
2. \( a + b = \alpha_1 \)
3. \( a - b + c = \alpha_0 \)

Substitute \( c = \alpha_2 \) into the third equation:
\[ a - b + \alpha_2 = \alpha_0 \]

Now we have:
1. \( c = \alpha_2 \)
2. \( a + b = \alpha_1 \)
3. \( a - b + \alpha_2 = \alpha_0 \)

Solve the second and third equations simultaneously:
From \( a + b = \alpha_1 \), we get:
\[ a = \alpha_1 - b \]

Substitute \( a = \alpha_1 - b \) into \( a - b + \alpha_2 = \alpha_0 \):
\[ (\alpha_1 - b) - b + \alpha_2 = \alpha_0 \]
\[ \alpha_1 - 2b + \alpha_2 = \alpha_0 \]
\[ -2b = \alpha_0 - \alpha_1 - \alpha_2 \]
\[ b = \frac{\alpha_1 + \alpha_2 - \alpha_0}{2} \]

Now substitute \( b \) back into \( a = \alpha_1 - b \):
\[ a = \alpha_1 - \frac{\alpha_1 + \alpha_2 - \alpha_0}{2} \]
\[ a = \frac{2\alpha_1 - (\alpha_1 + \alpha_2 - \alpha_0)}{2} \]
\[ a = \frac{\alpha_1 - \alpha_2 + \alpha_0}{2} \]

Thus, we have:
\[ a = \frac{\alpha_1 - \alpha_2 + \alpha_0}{2} \]
\[ b = \frac{\alpha_1 + \alpha_2 - \alpha_0}{2} \]
\[ c = \alpha_2 \]

Therefore, for any polynomial \( \alpha_0 + \alpha_1 x + \alpha_2 x^2 \in P_2(\mathbb{R}) \), we can write it as a linear combination of \(\beta'\) as follows:
\[ p(x) = \frac{\alpha_1 - \alpha_2 + \alpha_0}{2} (x + 1) + \frac{\alpha_1 + \alpha_2 - \alpha_0}{2} (x - 1) + \alpha_2 (x^2 + 1) \]

This shows that \(\beta' = \{x + 1, x - 1, x^2 + 1\}\) spans \(P_2(\mathbb{R})\), completing the proof that \(\beta'\) is a basis for \(P_2(\mathbb{R})\).

Would you like further details on any specific part of this explanation or have any questions?

Here are 8 related questions you might find interesting:

1. How do you determine if a set of vectors is linearly independent?
2. What is the significance of a basis in a vector space?
3. How do you find the transition matrix between two different bases?
4. What are the properties of transition matrices?
5. How do you solve a system of linear equations?
6. Can you provide an example of changing coordinates using a transition matrix?
7. How do you determine the dimension of a vector space?
8. What are some applications of basis and transition matrices in real-world problems?

**Tip:** Always verify the linear independence of a set of vectors by setting up and solving the corresponding linear combination equal to the zero vector.

Learn how to solve a system of linear equations to determine if a set of vectors spans a vector space. This problem explores linear algebra concepts, including vector spaces, linear independence, and the basis theorem. Suitable for advanced undergraduate students and beyond.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation