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North Carolina State University MA 405 Department of Mathematics Practice Test 3
ANSWERS
- We need to compute L(1), L(1 + x), L(1 + x + x^2 ) and find their coordinates relative to the ordered basis {x^2 , x, 1 }. This will give us the three columns of the matrix A representing L. We have:
L(1) = 1 + x · 0 = 1 = 0 · x^2 + 0 · x + 1 · 1 , L(1 + x) = (1 + x) + x · 1 = 1 + 2x = 0 · x^2 + 2 · x + 1 · 1 , L(1 + x + x^2 ) = (1 + x + x^2 ) + x(1 + 2x) = 1 + 2x + 3x^2 = 3 · x^2 + 2 · x + 1 · 1.
Hence, the matrix representing L is A =
- (i) We compute:
‖p‖^2 = 〈p, p〉 =
0
(1 + x)^2 dx =
‖q‖^2 = 〈q, q〉 =
0
(x^3 )^2 dx =
which gives ‖p‖ =
3, ‖q‖ = 1/
(ii) The distance between p and q is ‖p − q‖. We have:
‖p − q‖^2 = 〈p − q, p − q〉 = 〈p, p〉 − 2 〈p, q〉 + 〈q, q〉
and
〈p, q〉 =
0
(1 + x)x^3 dx =
Using the results of part (i), we find
‖p − q‖^2 =
and ‖p − q‖ =
1
2
(iii) If the angle between p and q is θ, then
cos θ =
〈p, q〉 ‖p‖‖q‖
p and q are not orthogonal, because the angle θ 6 = π/2 or, equivalently, 〈p, q〉 6 = 0.
- (i) Let us call the given basis vectors v 1 , v 2 , v 3. The first step of the Gram–Schmidt process is to set
u 1 =
v 1 ‖v 1 ‖
Next, we compute the orthogonal projection p 1 of v 2 onto u 1 :
p 1 = 〈v 2 , u 1 〉u 1 =
Then
v 2 − p 1 =
(^) , u 2 = v^2 −^ p^1 ‖v 2 − p 1 ‖
The orthogonal projection p 2 of v 3 onto Span{u 1 , u 2 } is
p 2 = 〈v 3 , u 1 〉u 1 + 〈v 3 , u 2 〉u 2 =
Then
v 3 − p 2 =
(^) , u 3 = v^3 −^ p^2 ‖v 3 − p 2 ‖
(ii) For every orthonormal basis {u 1 , u 2 , u 3 } and vector v, we have v = 〈v, u 1 〉u 1 + 〈v, u 2 〉u 2 + 〈v, u 3 〉u 3.
We compute:
〈v, u 1 〉 = −
2 , 〈v, u 2 〉 =
, 〈v, u 3 〉 =
