Math 333: Homework 5: Due: Tuesday, Dec. 1, 2015 in... 1. [5pt] A basis for V = IR is S =

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Math 333: Homework 5: Due: Tuesday, Dec. 1, 2015 in class
1. [5pt] A basis for V = IR3 is
S = {u1 , u2 , u3 } = {(2, 1, 0), (1, 1, 1), (4, 1, 1)}
Use the Gram-Schmidt process to derive an orthogonal basis S? .
2. [5pt] Let V = IR3 and
W = span {u1 , u2 } = span {(1, 1, 1), (0, 1, 1)}
be a subspace defined by its orthogonal basis vectors uk . Find a matrix
A 2 IR3⇥3 such that
T (v) ⌘ Av = projW v
An example of this procedure is given in the posted notes.
3. [5pt] We are given the data points
Z = {(x1 , y1 ), (x2 , y2 ), (x3 , y3 )} = {( 1, 2), (0, 1), (1, 3)}
and want a least squares fit to the quadratic model
y = ax2 + c
Thus, x = (a, c) must be the least squares solution of
Ax = b
where
2
x21
A = 4 x22
x23
3
1
1 5
1
2
3
y1
b = 4 y2 5
y3
Find x and plot the data along with y = ax2 + c.
4. [5pt] Let x̂ be the least squares solution of Ax = b and define the residual
vector r ⌘ Ax̂ b. Prove k r k2 = < b, r >. To do so, start noting
k r k2 = (Ax̂
b)T (Ax̂
b) · · ·
Write this out as a formal proof explaining each step.
5. [5pt] Let T : X ! IR3 be a linear transformation with basis S = {u1 , u2 , u3 }.
Given
0
1
0
1
0
1
1
5
1
T (u1 ) = @ 2 A T (u2 ) = @ 7 A T (u3 ) = @ 1 A
1
1
1
Find a basis for the range R(T ). DO NOT put T (uk ) in the rows of any
matrix.
1
6. [25pt] In each the the four problems below a linear transformation T :
X ! Y is defined. For each, clearly define the kernel ker(T ) and the
range R(T ) of each transformation. When T has an inverse, define its
formula and its domain.
a) T : P5 ! IR (average)
T (u) =
b) T : P2 ! P2
Z
1
u(x)dx
1
T (u) = u(x + 1)
u(x
c) T : M22 ! IR
T (u) = trace(u)
,
d) T : P2 ! P2
T (u) = x
e) T : P2 ! P2
u=
du
dx
T (u) = u(2x + 1)
2
1)

a
c
b
d
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