We are given the following problem involving a set of polynomials:

Problem Setup:

Let $S = \{ x^3 - x^2 - x + 1, -x + 1, 2x^3 - x^2 - x, 15x^3 - 8x^2 - 13x + 12 \}$.

We need to address the following parts:

(a) Determine a maximal linearly independent subset $A$ of $S$.

To determine the maximal linearly independent subset, we treat these polynomials as vectors in a vector space of polynomials of degree at most 3. We will express the polynomials as vectors in terms of their coefficients:

1. $p_1(x) = x^3 - x^2 - x + 1$ corresponds to the vector $\begin{pmatrix} 1 \\ -1 \\ -1 \\ 1 \end{pmatrix}$.
2. $p_2(x) = -x + 1$ corresponds to the vector $\begin{pmatrix} 0 \\ 0 \\ -1 \\ 1 \end{pmatrix}$.
3. $p_3(x) = 2x^3 - x^2 - x$ corresponds to the vector $\begin{pmatrix} 2 \\ -1 \\ -1 \\ 0 \end{pmatrix}$.
4. $p_4(x) = 15x^3 - 8x^2 - 13x + 12$ corresponds to the vector $\begin{pmatrix} 15 \\ -8 \\ -13 \\ 12 \end{pmatrix}$.

Now, we will check for linear dependence by forming a matrix with these vectors as columns and reducing it to row echelon form.

(b) Hence calculate $\dim(\text{span}(S))$ and justify whether $\text{span}(S) = \mathcal{P}_3$.

The dimension of the span of $S$ is simply the number of linearly independent vectors in $S$. We will determine this by the rank of the matrix from part (a). To check whether $\text{span}(S) = \mathcal{P}_3$, we need to verify if the dimension of the span equals 4, which is the dimension of the vector space of polynomials of degree at most 3, $\mathcal{P}_3$.

(c) Is $A$ a minimal spanning set for $\text{span}(S)$? Justify your answer.

A minimal spanning set has no redundant vectors. If the set $A$ found in part (a) is linearly independent and spans the same space as $S$, then it is a minimal spanning set.

I will now perform the matrix analysis to solve part (a).### (a) Maximal Linearly Independent Subset $A$ of $S$:

The matrix formed by the coefficients of the polynomials in $S$ has full rank $4$. This means all four polynomials are linearly independent. Therefore, the maximal linearly independent subset $A$ is the entire set $S$: $A = \{ x^3 - x^2 - x + 1, -x + 1, 2x^3 - x^2 - x, 15x^3 - 8x^2 - 13x + 12 \}.$

(b) Dimension of $\text{span}(S)$:

Since all four vectors are linearly independent, the dimension of $\text{span}(S)$ is 4: $\dim(\text{span}(S)) = 4.$ This shows that $\text{span}(S) = \mathcal{P}_3$, because $\mathcal{P}_3$ is the space of all polynomials of degree at most 3, which also has dimension 4.

(c) Is $A$ a minimal spanning set for $\text{span}(S)$?

Yes, $A$ is a minimal spanning set for $\text{span}(S)$, because it is both linearly independent and spans the entire space. There are no redundant vectors in $A$.

Would you like further details or have any questions? Here are some related questions:

1. What does it mean for a set of vectors to be linearly independent?
2. How is the dimension of a vector space defined?
3. Can the span of fewer than 4 polynomials ever equal $\mathcal{P}_3$?
4. What is the significance of row-reduction in determining linear independence?
5. How would the analysis change if one of the polynomials in $S$ were dependent?

Tip: When finding the dimension of a span, always compare it with the dimension of the ambient space to understand the relationship.