91Ó°ÊÓ

To prove that \(P_n\) is a subspace of \(F[x]\), we need to verify that \(P_n\) is closed under vector addition and scalar multiplication. Let's consider two polynomials \(p(x)\) and \(q(x)\) in \(P_n\), and a scalar \(c\in F\). For vector addition, we have \((p(x) + q(x))\). Since \(p(x)\) and \(q(x)\) both have degree less than \(n\), their sum will also have a degree less than \(n\). Therefore, \(p(x) + q(x) \in P_n\). For scalar multiplication, we have \((c\cdot p(x))\). Since \(p(x)\) has a degree less than \(n\), and multiplying by a constant doesn't change the degree of a polynomial, the result will still have a degree less than \(n\). Therefore, \(c\cdot p(x) \in P_n\). Since \(P_n\) is closed under vector addition and scalar multiplication, it is a subspace of the vector space \(F[x]\).

In order to find a basis for \(P_n\), we need to find a set of linearly independent vectors that span the space. Recall that the standard basis for the vector space of all polynomials is given by \(\{1, x, x^2, x^3, \cdots\}\). Since we are dealing with polynomials of degree less than \(n\), our basis for \(P_n\) will contain n elements. Therefore, a basis for \(P_n\) is given by: $$B = \{1, x, x^2, x^3, \cdots, x^{n-1}\}$$ This basis is linearly independent because no element in the set can be expressed as a linear combination of the others, and it spans the space because any polynomial of degree less than \(n\) can be written as a linear combination of these basis elements.

The dimension of a vector space is the number of elements in its basis. From Step 2, we know that the basis for \(P_n\) has n elements. Therefore, the dimension of \(P_n\) is: Dimension\((P_n) = n\).