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In linear algebra, understanding linear independence is crucial for determining whether a set of vectors forms a basis for a vector space. Vectors are considered linearly independent if no vector in the set can be written as a linear combination of the others. This means the only solution to the equation \( c_1 \cdot v_1 + c_2 \cdot v_2 + \, \ldots \, + c_n \cdot v_n = 0 \) is when all coefficients \( c_1, c_2, \, \ldots \, , c_n \) are zero. In simpler terms, there's no redundancy among the vectors; each vector brings a unique direction or dimension to the space.
To illustrate, consider the vectors \( p_1(x) = x+1 \) and \( p_2(x) = x-1 \). To show they are linearly independent, set up the equation \( c_1(x+1) + c_2(x-1) = 0 \). By simplifying, we get the expression \((c_1 + c_2)x + (c_1 - c_2) = 0 \). For this equation to be true for all values of \(x\), both \( (c_1 + c_2) \) and \( (c_1 - c_2) \) need to be zero, leading us to conclude \( c_1 = 0 \) and \( c_2 = 0 \). Hence, \( p_1(x) \) and \( p_2(x) \) prove their linear independence.

A basis of a vector space is a collection of vectors that are linearly independent and span the entire space. "Spanning the space" means any vector in the space can be expressed as a linear combination of the basis vectors. When we have a basis, we can represent every element in that space uniquely as a linear combination of the basis vectors, using the linear independence property.
Let's take the simple vector space \( P_1 \), which consists of all polynomials of degree at most 1. The vectors \( p_1(x) = x+1 \) and \( p_2(x) = x-1 \) form a basis for this space. Because these vectors are linearly independent and can combine to produce any first-degree polynomial such as \( ax + b \), they effectively span the space. Thus, any polynomial in \( P_1 \) can be uniquely represented by combinations of \( p_1(x) \) and \( p_2(x) \), reinforcing their role as a basis of this vector space.

A linear combination involves taking two or more vectors and combining them using addition and scalar multiplication to form a new vector. Understanding linear combinations is fundamental when working with bases, as every vector in a space can be a linear combination of the basis vectors.
For the exercise concerning \( p(x) = 5x + 2 \), it can be expressed as a linear combination of the basis vectors \( p_1(x) = x + 1 \) and \( p_2(x) = x - 1 \). To find this combination, solve the equation \( a(x + 1) + b(x - 1) = 5x + 2 \). Simplifying gives \( (a + b)x + (a - b) \). By comparing coefficients, we have the equations \( a + b = 5 \) and \( a - b = 2 \). Solving these simultaneously, we find \( a = \frac{7}{2} \) and \( b = \frac{3}{2} \), assuring that \( p(x) = \frac{7}{2}p_1(x) + \frac{3}{2}p_2(x) \). This clear breakdown shows how effective linear combinations are in vector space expressions.