The problem asks whether the set $\{x + 1, x^2 - 1, 2x - 1\}$ is a minimal generating set for the vector space $P_2 = \{ax^2 + bx + c \mid a, b, c \in \mathbb{R}\}$, which represents the space of polynomials of degree at most 2.

Step-by-Step Approach:

1. Dimension of $P_2$:

   • The space $P_2$ consists of all polynomials of the form $ax^2 + bx + c$, where $a, b, c \in \mathbb{R}$. This space has dimension 3, because any polynomial in $P_2$ can be written as a linear combination of the three basis elements $x^2, x,$ and $1$.

2. Check if $\{x + 1, x^2 - 1, 2x - 1\}$ spans $P_2$:

   • A set of vectors spans a space if any element of the space can be written as a linear combination of the vectors in the set.
   • Consider a general polynomial $ax^2 + bx + c \in P_2$. We need to express this as a linear combination of $x + 1$, $x^2 - 1$, and $2x - 1$. That is, we want to find scalars $\alpha, \beta, \gamma$ such that: $ax^2 + bx + c = \alpha(x + 1) + \beta(x^2 - 1) + \gamma(2x - 1)$ Expanding the right-hand side: $\alpha(x + 1) = \alpha x + \alpha$ $\beta(x^2 - 1) = \beta x^2 - \beta$ $\gamma(2x - 1) = 2\gamma x - \gamma$ Adding these up: $\alpha x + \alpha + \beta x^2 - \beta + 2\gamma x - \gamma = \beta x^2 + (\alpha + 2\gamma)x + (\alpha - \beta - \gamma)$ Now compare this with $ax^2 + bx + c$. We get the system of equations: $\beta = a$ $\alpha + 2\gamma = b$ $\alpha - \beta - \gamma = c$

3. Solve the system: From $\beta = a$, substitute $\beta$ into the third equation: $\alpha - a - \gamma = c \quad \text{(Equation 3)}$ Now, solve for $\alpha$ and $\gamma$:

   • From Equation 3, $\alpha = a + \gamma + c$.
   • Substitute this into Equation 2: $(a + \gamma + c) + 2\gamma = b$ Simplify to find $\gamma$: $a + \gamma + c + 2\gamma = b \quad \Rightarrow \quad a + c + 3\gamma = b$ $\gamma = \frac{b - a - c}{3}$ Now substitute $\gamma$ into $\alpha = a + \gamma + c$: $\alpha = a + \frac{b - a - c}{3} + c = \frac{2a + b + 2c}{3}$

Thus, for any polynomial $ax^2 + bx + c$, we can express it as a linear combination of $x + 1$, $x^2 - 1$, and $2x - 1$. Hence, the set $\{x + 1, x^2 - 1, 2x - 1\}$ spans $P_2$.

4. Minimality: The set $\{x + 1, x^2 - 1, 2x - 1\}$ contains three elements, which is equal to the dimension of $P_2$. Therefore, this set is a basis (minimal generating set) for $P_2$.

Conclusion:

Yes, $\{x + 1, x^2 - 1, 2x - 1\}$ is a minimal generating set for $P_2$.

Would you like a further breakdown of the system of equations or any additional details?

Related Questions:

1. What is the general form of a polynomial in $P_2$?
2. How do we determine the dimension of a vector space?
3. How do we verify if a set of vectors spans a vector space?
4. What does it mean for a set of vectors to be linearly independent?
5. How can you find a basis for a polynomial vector space?

Tip:

When determining if a set is a basis, always check for both spanning and linear independence conditions!