Math 416 Homework 11. Solutions.
1. Let V be an inner product space, and let x, y ∈ V . Show that if x ⊥ y, then
kx + yk2 = kxk2 + kyk2 .
Solution: We compute
kx + yk2 = hx + y, x + yi = hx, xi + hx, yi + hy, xi + hy, yi = hx, xi + hy, yi = kxk2 + kyk2 .
2. For each of the following inner products, write down an orthonormal basis for P3 (R) using
the Gram–Schmidt Algorithm applied to the basis (1, x, x2 , x3 ).
R1
(a) hf, gi = 0 f (t)g(t) dt,
R2
(b) hf, gi = 0 f (t)g(t) dt,
R2
(c) hf, gi = −1 f (t)g(t) dt,
Solution:
(a) Let us first notice that
Z
0
1
xn dx =
1
.
n+1
We compute v1 = 1. Then
hw2 , v1 i
1/2
v1 = x −
1 = x − 1/2.
hv1 , v1 i
1
hw3 , v1 i
hw3 , v2 i
v2 −
v1
v3 = w 3 −
hv2 , v2 i
hv1 , v1 i
hx2 , x − 1/2i
1/3
= x2 −
(x − 1/2) −
hx − 1/2, x − 1/2i
1
1/12
1
= x2 −
(x − 1/2) − = x2 − x + 1/6.
1/12
3
v2 = w 2 −
Finally,
hw4 , v3 i
hw4 , v2 i
hw4 , v1 i
v3 −
v2 −
v1
hv3 , v3 i
hv2 , v2 i
hv1 , v1 i
1/120
3/40
1
= x3 −
v3 −
v2 −
1/180
1/12
4
3
9
= x3 − (x2 − x + 1/6) − (x − 1/2) − 1/4
2
10
1
3 2 3
3
=x − x + x− .
2
5
20
v4 = w 4 −
So we have
v1 = 1,
v2 = x − 1/2,
v3 = x2 − x + 1/6,
v4 = x3 − 3x2 /x + 3x/5 − 1/20.
We can rescale all of these and make them unit vectors, and then the set is orthonormal.
(b) But I am
(c) le tired
3. Let β be a basis for a subspace W of an inner product space V , and let z ∈ V . Show that
z ∈ W ⊥ if and only if hz, vi = 0 for all v ∈ β.
Solution: The forward direction is easy. Since z ∈ W ⊥ means that hz, wi = 0 for all w ∈ W ,
then since β ⊆ W , this means that hz, wi = 0 for all w inβ.
For the opposite direction, notice that if hz, vi i = 0 for all i ∈ I, then
*
+
X
X
z,
vi =
hz, vi i = 0,
i∈I
i∈I
where I is any set that indexes the elements of β. From this we can deduce that z is orthogonal
to the span of β, which is W .
4. Let V be a finite-dimensional vector space, and S a nonempty subset of V . Show that S ⊥ is
a subspace of V .
Solution: Recall that
S ⊥ = {x ∈ V : hx, yi = 0 ∀y ∈ S}.
Let y, z ∈ S ⊥ . This means that hx, yi = hx, zi = 0 for all x ∈ S. Then we see that
hx, ay + bzi = a hx, yi + b hx, zi = 0
for all x ∈ S, so that ay + bz ∈ S ⊥ . Since S ⊥ is a subset of V that is closed under addition
and scalar multiplication, it is a subspace of V .
5. Let V be a vector space, and W a finite-dimensional subspace. Show that (W ⊥ )⊥ = W .
Solution: Let us write Y = W ⊥ , and recall that x ∈ Y if and only if hx, wi = 0 for all x ∈ W .
Therefore, if w ∈ W , then w ∈ Y ⊥ . This shows that W ⊆ Y ⊥ = (W ⊥ )⊥ .
Conversely, now assume that x 6∈ W . Let (v1 , . . . , vk ) be an orthonormal basis for W (this
exists since W is finite-dimensional). Then if we write
w=
k
X
i=1
hx, vi i vi ,
then w ∈ W , and z = x − w ∈ W ⊥ . From this, we see that
hx, zi = hw + z, zi = hw, zi + hz, zi = 0 + kzk2 > 0.
So there is a z ∈ W ⊥ such that hx, zi 6= 0. This means that x 6∈ (W ⊥ )⊥ . Since x 6∈ W =⇒
x 6∈ (W ⊥ )⊥ , this means that (W ⊥ )⊥ ⊆ W , and we are done.
6. Let S be any finite subset of V . Show that Span(S) = (S ⊥ )⊥ .
Solution: A restatement of problem #3 above is that
S ⊥ = Span(S)⊥ ,
namely: a vector is orthogonal to S if and only if it is orthogonal to Span(S).
Now, let W = Span(S). Since S is finite, this means that W is finite dimensional. By the
argument above, W ⊥ = S ⊥ , and by Problem #4, (W ⊥ )⊥ = W , so
(S ⊥ )⊥ = (W ⊥ )⊥ = W = Span(S).
7. Let β = (v1 , . . . , vn ) be an orthonormal basis for V . Show that for any x, y ∈ V ,
hx, yi =
n
X
hx, vi i hy, vi i.
i=1
Solution: From Theorem 6.?, we have
x=
n
X
hx, vi i vi ,
y=
i=1
n
X
hy, vi i vi ,
i=1
so
hx, yi =
* n
X
hx, vi i vi ,
i=1
=
n
X
+
hy, vj i vj
j=1
n X
n
X
hx, vi i hy, vj i hvi , vj i .
i=1 j=1
Since hvi , vj i =
6 0 only if i = j, this double sum becomes a single sum, and we get
n
X
i=1
hx, vi i hy, vi i.
8. Recall homework problem 1.10. Let us make P (R) into an inner product space using the inner
product
Z
1
hp, qi =
p(x)q(x) dx.
−1
Recall that we say a function is even if ∀x, f (−x) = f (x) and odd if ∀x, f (−x) = −f (x).
Show that W1 corresponds to the set of odd polynomials and W2 the set of even polynomials.
Show that W2 = W1⊥ and W1 = W2⊥ .
Solution: We first show that if p(x) is odd and q(x) is even, then hp, qi = 0.
First notice that if f (x) is odd, then
Z 1
Z 0
Z 1
Z 0
Z 1
f (x) dx =
f (x) dx +
f (x) dx = −
f (−x) dx +
f (x) dx
−1
−1
0
−1
0
Z 1
Z 1
=−
f (x) dx +
f (x) dx = 0.
0
0
Moreover, notice that if p(x) is even and q(x) is odd, then p(x)q(x) is odd:
p(−x)q(−x) = (−p(x))(q(x)) = −p(x)q(x).
Finally, notice that a polynomial is odd (resp. even) if it contains only odd-power (resp. evenpower) terms.
What this means is that if p ∈ W1 and q ∈ W2 , then hp, qi = 0. From this we can deduce, for
example, that W1 ⊆ W2⊥ .
Now, assume that p 6∈ W1 , so that p(x) is not an odd polynomial. This means that we can
write p(x) = p1 (x) + p2 (x), where p1 is odd and p2 is even, and p2 6= 0. But then we have
hp, p2 i = hp1 + p2 , p2 i = hp1 , p2 i + hp2 , p2 i = hp2 , p2 i > 0,
so that p 6∈ W2⊥ . Therefore p ∈ W1 if and only if p ∈ W2⊥ , and thus means that W1 = W2⊥ .
The proof for W2 = W1⊥ is similar.
9. Let T be a linear operator on an inner product space V . Let U1 = T + T ∗ and U2 = T T ∗ .
Show that U1 , U2 are both self-adjoint.
Solution: We write
hU1 x, yi = h(T + T ∗ )x, yi = hT x + T ∗ x, yi = hT x, yi + hT ∗ x, yi
= hx, T ∗ yi + hx, T yi = hx, T y + T ∗ yi = hx, (T + T ∗ )yi = hx, U1 yi .
Therefore U1∗ = U1 .
We proceed in a similar manner for U2 . We have
hU2 x, yi = hT T ∗ x, yi = hT ∗ x, T ∗ yi = hx, T T ∗ yi = hx, U2 yi ,
so U2∗ = U2 .
10. Let V be an inner product space, and choose y, z ∈ V . Define T : V → V by
T (x) = hx, yi z.
Show that T is linear. Then show that T ∗ exists, and compute an explicit formula for it.
Compute the rank of T and T ∗ .
Solution: First we show that T is linear. We have
T (a1 x1 + a2 x2 ) = ha1 x1 + a2 x2 , yi z = (a1 hx1 , yi + a2 hx2 , yi)z = a1 T (x1 ) + a2 T (x2 ).
Since T : V → V is linear, then T ∗ : V → V exists, is unique, is linear, and satisfies
hT x, wi = hx, T ∗ wi .
So we compute
D
E
hT x, wi = hhx, yi z, wi = hx, yi hz, wi = x, hz, wi y = hx, hw, zi yi .
Therefore T ∗ w = hw, zi y.
Now, we want to compute the rank of T . Notice that if q ∈ R(T ), then q is a scalar multiple
of the vector z. This means that T has rank at most one. Moreover, if we choose x = y,
then T (y) = hy, yi z, and as long as y 6= 0, this is a non-zero number. So R(T ) is exactly
one-dimensional. Therefore it has rank one.
The rank of T ∗ is the same, which we can see in two ways. First, a similar argument to the
above will show that T ∗ has rank one (T ∗ w is a scalar multiple of y, etc.). Also, recall that
we proved in class that T and T ∗ have the same rank, so T ∗ also has rank one.