Let V span4 x2 4 x2 x2 Find a basis for V SolutionSince V

Let V = span{4 + x^2, 4 - x^2, x^2}. Find a basis for V.

Solution

Since V = span{ 4+x², 4 –x²,x²}, an arbitrary vector v in V has to be a linear combination of the vectors 4+x², 4 –x² and x². Let v = a(4 +x² )+b(4 –x²) +cx² where a, b, and c are arbitrary scalars. Then v = x² (a-b+c)+( 4a+4b +c)*1. Apparently, v is a linear combination of 1 and x².Thus, every linear combination of the vectors 4 +x² , 4 –x² and x² is also a linear combination of the vectors 1 and x² . Also 1 and x² are linearly independent vectors and the vectors 4 +x² , 4 –x² and x² are linear combination of the vectors 1 and x². Further 1 = 1/8( 4 +x²) + 1/8(4 –x²) + 0.x² and x² = (4+x²)–(4-x²) –x².This shows that 1 and x² are also linear combinations of the vectors 4 +x², 4 –x² and x² . Therefore, {1,x²} is a basis for V.