Question 1.12.1: In the vector space P of all polynomials, consider the subsp...

To begin, we observe that every element of H = P_{2}is a polynomial function of the form p(t ) = a_{0} + a_{1}t + a_{2}t^{2}. In particular, every element of P_{2} is a linear combination of the functions 1, t , and t^{2}, and therefore the set B = {1, t , t^{2}} spans H.

In addition, to determine whether the set B is linearly independent, we consider the equation

c_{0} +c_{1}t +c_{2}t^{2} = 0       (1.12.2)

and ask whether or not this equation has a nontrivial solution. Keeping in mind that the ‘0’ on the right-hand side represents the zero function in P_{2}, the function that is everywhere equal to zero, we can see that if at least one of c_{0},c_{1},  or  c_{2} is nonzero, we will be guaranteed to have either a nonzero constant function, a linear function, or a quadratic function, thus making c_{0} +c_{1}t +c_{2}t^{2} not identically zero. This shows that (1.12.2) has only the trivial solution, and therefore the set B = {1, t , t^{2}} is linearly independent. Having shown that B is a linearly independent spanning set for H = P_{2}, we can conclude that B is a basis for H.

On the other hand, the set {1, t , t^{2},4−3t } is not a basis for H since we can observe that the element 4−3t is a linear combination of the elements 1 and t : 4−3t =4·1−3· t . This shows that the set {1, t , t^{2},4−3t } is linearly dependent and thus cannot be a basis.