Does t2 43t 5 5t2 2t form a basis for P2 Why or why notSo

Does {t^2 - 4,3t + 5, 5t^2 - 2t} form a basis for P_2? Why or why not?

Solution

Yes.

(t²-4, 3t+5, 5t²-2t) forms a basis for p2.

It has three elements and can be shown to be a spanning set and we can say it is linearly independent set.

Proving the set is linearly independent set:

a(t²-4) + b(3t+5) + c(5t²-2t) = 0

t²(a+5c) + t(3b-2c) + (-4a+5b) = 0

this means that

a+5c = 0 ------(1)

3b-2c = 0 ------(2)

-4a+5b = 0 ------(3)

By solving eq(1) and (2)

4a + +20c = 0

-4a +5b = 0

---------------------------

5b + 20c = 0 ------(4)

---------------------------

by solving eq (2) and (4)

30b - 20c = 0

5b + 20c = 0

--------------------

35b = 0

---------------------

this implies b = 0

plugging b into the eq(4) gives us c = 0 and plugging c = 0 into the first equation gives us a = 0.

thus a = b = c = 0.

therefore (t²-4, 3t+5, 5t²-2t) is linearly independent set.

Hence (t²-4, 3t+5, 5t²-2t) forms a basis for P2.