MATH 5718, ASSIGNMENT 9 – DUE: 28 APR 2015
LUKE NELSEN
[6B2] Suppose e1 , . . . , em is an orthonormal list of vectors in V . Let v ∈ V . Prove that
||v||2 = |hv, e1 i|2 + · · · + |hv, em i|2
if and only if v ∈ span(e1 , . . . , em ).
Proof. We prove the reverse direction and then the contrapositive of the forward direction.
Suppose v ∈ span(e1 , . . . , em ). Then since e1 , . . . , em is an orthonormal basis for span(e1 , . . . , em ), we
have ||v||2 = |hv, e1 i|2 + · · · + |hv, em i|2 .
Now suppose v ∈
/ span(e1 , . . . , em ) and note that this implies v 6= 0. Then since e1 , . . . , em , v is a linearly
independent list in span(e1 , . . . , em , v), by Gram-Schmidt we have an orthonormal basis e1 , . . . , em , em+1
for span(e1 , . . . , em , v). Thus ||v||2 = |hv, e1 i|2 + · · · + |hv, em i|2 + |hv, em+1 i|2 . Now, if hv, em+1 i = 0,
⊥
then v ∈ {em+1 }⊥ . But by construction, em+1 ∈ span(e1 , . . . , em )⊥ and thus v ∈ span(e1 , . . . , em )⊥ .
And since span(e1 , . . . , em ) is a subspace of the finite vector space span(e1 , . . . , em , v), this implies that
v ∈ span(e1 , . . . , em ). Since this contradicts our assumption, we have hv, em+1 i 6= 0 and thus ||v||2 6=
|hv, e1 i|2 + · · · + |hv, em i|2 .
[6B7] Find a polynomial q ∈ P2 (R) such that p( 21 ) =
R1
0
p(x)q(x) dx for every p ∈ P2 (R).
Proof. Since ϕ(p) = p( 12 is a linear functional on the finite-dimensional P2 (R), by the Riesz Representation Theorem we have that there exists a unique q ∈ P2 (R) such that for all p ∈ P2 (R) we have
R1
P3
p( 21 = 0 p(x)q(x) dx. The q is given by i=1 ϕ(ei )ei , where e1 , e2 , e3 is an orthonormal basis of P2 (R).
So we use Gram-Schmidt on the basis 1, x, x2 to obtain an orthonormal basis:
1
=1
e1 (x) =
||1||
√
√
x − 12
x − hx, 1i1
e2 (x) =
=
= 2 3x − 3
1
||x − hx, 1i1||
||x − 2 ||
√
√
√
√
2
2
√ 2
√
√
x2 − x + 61
x − hx , 1i1 − hx2 , 2 3x − 3i(2 3x − 3)
√
√
√
√
=
=
6
5x
−
6
5x
+
5.
e3 (x) =
1
||x2 − x + 6 ||
||x2 − hx2 , 1i1 − hx2 , 2 3x − 3i(2 3x − 3)||
Then since e1 ( 12 ) = 1, e2 ( 12 ) = 0 and e3 ( 12 ) = −
q(x) =
√
5
2 ,
we have
3
X
3
ϕ(ei )ei = −15x2 + 15x − .
2
i=1
1
[6B11] Suppose h·, ·i1 and h·, ·i2 are inner products on V such that hv, wi1 = 0 if and only if hv, wi2 = 0.
Prove that there is a positive number c such that hv, wi1 = chv, wi2 for every v, w ∈ V .
Proof. We assume that V is finite-dimensional; so there is an orthonormal basis e1 , . . . en with respect to
h·, ·i1 . Now note that for all 1 ≤ i < j ≤ n,
hei − ej , ei + ej i1 = hei , ei i1 + hei , ej i1 − hej , ei i1 − hej , ej i1 = 1 + 0 − 0 − 1 = 0
and thus ei , ej are orthogonal with respect to h·, ·i1 . Thus ei , ej are orthogonal with respect to h·, ·i2 for all
1 ≤ i < j ≤ n and we have
0 = hei − ej , ei + ej i2 = hei , ei i2 + hei , ej i2 − hej , ei i2 − hej , ej i2 = hei , ei i2 − hej , ej i2
. Thus hei , ei i2 = hej , ej i2 for all i, j and thus there exists c > 0 such that hei ,P
ei i = c for all i = 1,
P.n. . , n.
Let v, w ∈ V . Now, there are unique a1 , . . . an , b1 , . . . , bn ∈ F such that v = i=1 ai ei and w = j=1 bj ej .
Then
n
n
n
n
X
X
X
X
X
X
hv, wi1 = h
ai ei ,
bj ej i1 =
ai bj hei , ej i1 =
ai bi hei , ei i1 +
ai bj hei , ej i1 =
ai bi
i=1
j=1
and
hv, wi2 =
X
i,j
ai bj hei , ej i2 =
i,j
n
X
i=1
ai bi hei , ei i2 +
i=1
X
i6=j
i6=j
ai bj hei , ej i2 =
n
X
i=1
ai bi c = chv, wi1 .
i=1
[6C11] In R4 , let U = span((1, 1, 0, 0), (1, 1, 1, 2)). Find u ∈ U such that ||u − (1, 2, 3, 4)|| is minimized.
Proof. We know that u = PU (1, 2, 3, 4); so we find an orthonormal basis of U . Since (1, 1, 0, 0), (1, 1, 1, 2) is
a basis of U , we proceed by Gram-Schmidt:
1
(1, 1, 0, 0)
= √ (1, 1, 0, 0)
e1 =
||(1, 1, 0, 0)||
2
(1, 1, 1, 2) − h √12 (1, 1, 0, 0), (1, 1, 1, 2)i √12 (1, 1, 0, 0)
1
(0, 0, 1, 2)
e2 =
= √ (0, 0, 1, 2)
=
||(0, 0, 1, 2)||
||(1, 1, 1, 2) − h √12 (1, 1, 0, 0), (1, 1, 1, 2)i √12 (1, 1, 0, 0)||
5
So we have
1
1
1
1
PU (1, 2, 3, 4) = h(1, 2, 3, 4), √ (1, 1, 0, 0)i √ (1, 1, 0, 0) + h(1, 2, 3, 4), √ (0, 0, 1, 2)i √ (0, 0, 1, 2)
2
2
5
5
11
3 3 11 22
3
=
( , , , ),
= (1, 1, 0, 0) + (0, 0, 1, 2)
2
5
2 2 5 5
which is the desired vector.
R
1
[6C12] Find p ∈ P3 (R) such that p(0) = 0, p0 (0) = 0, and 0 |2 + 3x − p(x)|2 dx is as small as possible.
R1
Proof. To minimize 0 |2 + 3x − p(x)|2 dx = ||2 + 3x − p(x)||2 , we minimize ||2 + 3x − p(x)||. Observe that
U = {p ∈ P3 (R) : p0 (0) = p(0) = 0} = {a1 x3 + a2 x2 : a1 , a2 ∈ R}. Our solution is PU (2 + 3x); we apply
Gram-Schmidt to the basis x3 , x2 of U :
√
x3
e1 =
= 7x3
3
||x ||
√
√
√
√
− 67 x3 + x2
x2 − hx2 , 7x3 i · 7x3
√
√
e2 =
=
= −7 5x3 + 6 5x2
7 3
2
2
2
3
3
|| − 6 x + x ||
||x − hx , 7x i · 7x ||
So we have
√
√
√
√
√
√
PU (2 + 3x) = h2 + 3x, 7x3 i · 7x3 + h2 + 3x, −7 5x3 + 6 5x2 i · (−7 5x3 + 6 5x2 )
√
√
11 √ √ 3
4√
203 3
=(
7) 7x + (
5)(−7 5x3 + 6 5x2 )
=
−
x + 24x2 ,
10
5
10
which is the desired polynomial.
2