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.