Problem 3. Consider the space of polynomials of degree less than three. One basi
ID: 3116904 • Letter: P
Question
Problem 3. Consider the space of polynomials of degree less than three. One basis of P3 is 2 Define polynomials by and define A. Verify that Q is a basis and find the change of basis matrices Spo and Sçp B. Let f(x)-2+3x+5x2. Find the coordinates [f(x)lp. C. Find the coordinates [f (x)]e of f(x) with respect to the basis 2. Write f(x) as a linear combination of qi (x), q2(x) and q3(x). Expand and simplify this expression to check that it's correct. D. Let g(x) = 3-5(x-3)-3(z-3)2 Find the coordinates [g(x)p of g(x) with respect to the basis P by using the change of basis matrix. Check that this is correct.Explanation / Answer
3. A. Let M =
1
-3
9
0
1
-6
0
0
1
It may be observed that the entries in the columns of A are the scalar multiples of 1, and the coefficients of x, x2 in q1(x),q2(x) and q3(x). The RREF of M is I3. It implies that q1(x),q2(x),q3(x)are linearly independent and span P3. Thus, Q is a basis for P3. Now, let M1=
1
-3
9
1
0
0
0
1
-6
0
1
0
0
0
1
0
0
1
The RREF of M1 is
1
0
0
1
3
9
0
1
0
0
1
6
0
0
1
0
0
1
Thus, the change of basis matrix from P to Q is SPQ =
1
3
9
0
1
6
0
0
1
Also, the change of basis matrix from Q to P is SQP =
1
-3
9
0
1
-6
0
0
1
B. The coordinate vector for f(x) relative to P i.e. [f(x)]P is (2,3,5)T.
C. Let M2 =
1
-3
9
2
0
1
-6
3
0
0
1
5
The RREF of M2 is
1
0
0
56
0
1
0
33
0
0
1
5
Hence [f(x)]Q = (56,33,5)5. Then f(x) = 56q1(x)+33q2(x)+5q3(x) = 56*1+33*(x-3)+5*(x2-6x+9) = 56 +33x-99+5x2 -30x +45 = 2+3x+5x2 =f(x).
D. We have g(x) = 3-5(x-3)-3(x-3)2 so that [g(x)]Q = (3,-5,-3)T . Then [g(x)]P = SQP[g(x)]Q = (-9,13,-3)T.
1
-3
9
0
1
-6
0
0
1
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