Let's break down the problem and solve it step by step.

Problem Summary:

1. Vector space $P_n$: The space of all real polynomials of degree ≤ $n$.

   • Basis: $\{1, x, x^2, ..., x^n\}$
   • Dimension: $n+1$

2. Given:

   • $n = 4$, so dimension $\text{dim}(P_4) = 5$.
   • General polynomial $p(x) = a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$ with basis elements $\{1, x, x^2, x^3, x^4\}$.

3. Question (Part ii):
   Is the set $\{2t^2 + 1, t^2 - 1, t + 2\}$ a basis of $P_2$?

Step 1: Basics of $P_2$

The space $P_2$ consists of all polynomials of degree at most 2. Its standard basis is: $\{1, t, t^2\}$ Thus, any element of $P_2$ can be written as a linear combination of these basis vectors.

Step 2: Verifying the Given Set

The set to test is $\{2t^2 + 1, t^2 - 1, t + 2\}$. For this set to be a basis of $P_2$, it must satisfy two conditions:

1. Spanning: The given vectors must span $P_2$.
2. Linear Independence: The vectors must be linearly independent.

To test linear independence, assume a linear combination of the vectors equals zero: $a_1(2t^2 + 1) + a_2(t^2 - 1) + a_3(t + 2) = 0$ We simplify this equation and collect terms based on powers of $t$.

Step 3: Solve the Linear Combination

Expand the terms: $a_1(2t^2 + 1) = 2a_1t^2 + a_1, \quad a_2(t^2 - 1) = a_2t^2 - a_2, \quad a_3(t + 2) = a_3t + 2a_3$ Combine all terms: $2a_1t^2 + a_2t^2 + a_3t + a_1 - a_2 + 2a_3 = 0$ Group by powers of $t$: $(2a_1 + a_2)t^2 + a_3t + (a_1 - a_2 + 2a_3) = 0$ For this equation to hold for all $t$, the coefficients of $t^2$, $t$, and the constant term must all equal zero:

1. Coefficient of $t^2$: $2a_1 + a_2 = 0$
2. Coefficient of $t$: $a_3 = 0$
3. Constant term: $a_1 - a_2 + 2a_3 = 0$

Step 4: Solve the System of Equations

From the equations:

1. $2a_1 + a_2 = 0 \Rightarrow a_2 = -2a_1$
2. $a_3 = 0$
3. $a_1 - a_2 + 2a_3 = 0$

Substitute $a_2 = -2a_1$ into the third equation: $a_1 - (-2a_1) + 2(0) = 0$ Simplify: $a_1 + 2a_1 = 0 \Rightarrow 3a_1 = 0 \Rightarrow a_1 = 0$ Thus: $a_1 = 0, \; a_2 = -2a_1 = 0, \; a_3 = 0$

Step 5: Conclusion

Since $a_1 = a_2 = a_3 = 0$ is the only solution to the linear combination, the set $\{2t^2 + 1, t^2 - 1, t + 2\}$ is linearly independent.

Additionally, there are 3 vectors, which is equal to the dimension of $P_2$. Therefore, the set forms a basis of $P_2$.

Final Answer:

Yes, $\{2t^2 + 1, t^2 - 1, t + 2\}$ is a basis of $P_2$.

Follow-Up Questions:

1. What is the definition of a vector space basis?
2. How can you determine the dimension of a polynomial vector space $P_n$?
3. What does it mean for a set of vectors to be linearly independent?
4. How do you test if a given set spans a vector space?
5. Can you extend this result to higher-degree polynomial spaces?

Tip: In problems involving polynomial vector spaces, always start by comparing the number of given vectors to the dimension of the space.