Math 416 Homework 10. Solutions.
1. Let T : V → V be linear. Show that R(T ) and N (T ) are T -invariant.
Solution: Let y ∈ R(T ). Then T y ∈ R(T ) by definition, so we are done.
If x ∈ N (T ), then T x = 0. We also know that 0 ∈ N (T ), since T 0 = 0. Thus T x ∈ N (T ).
2. Let T : V → V be linear and finite-dimensional, and recall that < x >T is defined as
< x >T := Span{x, T x, T 2 x, . . . , }.
(a) Show that < x >T is finite-dimensional for any x ∈ V .
(b) Show that < x >T is a T -invariant subspace of V .
Solution:
(a) Since < x >T is defined as the span of a set of vectors, it is a subspace of V . Since it is
subspace of V , it is spanned by any basis of V , and thus is finite dimensional.
(b) Let y ∈< x >T . This means that y ∈ Span{x, T x, T 2 x, . . . , }.
Then T y ∈ Span{T x, T 2 x, . . . , }, and this is clearly a subspace of < x >T .
3. In each case, you are given a vector space V and a linear map T : V → V , and an z ∈ V . In
each case, compute a basis for < z >T .
(a) V = R2 , T is rotation by π/4, z = (1, 0).
(b) V = R5 , T (a, b, c, d, e) = (a + b, b + c, c + d, d + e, e + a), z = (1, 0, 0, 0, 0).
(c) V = P3 (R), T f = f 0 , z = x3 + x2 .
Solution:
(a) We compute
√
√
T z = (1/ 2, 1/ 2).
Since z, T z are independent, Span(z, T z) is two-dimensional. Being a subspace of R2 , it
must be all of R2 , so (z, T z) form a basis.
(b) We compute
z
Tz
T 2z
T 3z
T 4z
= (1, 0, 0, 0, 0),
= (1, 0, 0, 0, 1),
= (1, 0, 0, 1, 2),
= (1, 0, 1, 3, 3),
= (1, 1, 4, 6, 4).
This gives us five vectors. Moreover, it is easy to see that they are all independent (each
new vector has support in a component that the previous ones do not). Therefore < z >T
is all of R5 and these form a basis for it.
(c) The vector space P3 (R) is four-dimensional. We compute
z
Tz
T 2z
T 3z
= x3 + x2 ,
= 3x2 + 2x,
= 6x + 2,
= 6,
and as before, we see that these are all independent (again, moving back up the chain, a
new monomial appears each time). So again < z >T is the entire space, and these four
vectors form a basis for it.
4. Let v, w ∈ C4 given by
v = (1, i, 1 + i, 0),
w = (1 + i, 2 + 3i, 4 + 5i, 6 + 7i).
Compute hv, wi, kvk, kwk, kv + wk. Verify that the triangle inequality and the CauchySchwarz Inequality hold for these vectors.
Solution: We have
hv, wi = 1(1 − i) + i(2 − 3i) + (1 + i)(4 − 5i) = 1 − i + 2i + 3 + 4 − 5i + 4i + 20 = 13.
We compute
kvk2 = 1 · 1 + i · (−i) + (1 + i)(1 − i) = 1 + 1 + 2 = 4,
so kvk = 2. We also have
kwk2 = (1 + i)(1 − i) + (2 + 3i)(2 − 3i) + (4 + 5i)(4 − 5i) + (6 + 7i)(6 − 7i)
= 2 + 13 + 41 + 85 = 141,
√
so kwk = 141.
Cauchy–Schwarz says that we should have
√
√
13 ≤ 2 141 = 564.
Since 13 · 13 = 169, this is true.
Finally, we have
v + w = (2 + i, 2 + 4i, 5 + 6i, 6 + 7i),
and thus
so kv + wk =
√
kv + wk2 = 5 + 20 + 61 + 85 = 170,
170. The Triangle Inequality says that
√
√
170 ≤ 4 + 141.
√
√
By squaring both sides, we see that this is true iff 13 ≤ 8 141, and since 141 > 2 this is
easy to see.
5. Let f, g ∈ C([0, 1]) be given by
f (t) = t2 ,
g(t) = et .
Let us define the inner product as
Z
hf, gi =
1
f (t)g(t) dt.
0
Compute hf, gi, kf k, kgk, kf + gk. Verify that the triangle inequality and the Cauchy-Schwarz
Inequality hold for these vectors. Compute the angle between f and g.
Solution: We compute
2
t
t ,e
Z
1
=
t2 et dt = e − 2,
Z0 1
1
1
t4 dt = , kf k = √ ,
5
5
0
Z 1
1
1 √
e2t dt = (e2 − 1), kgk = √ e2 − 1,
kgk2 =
2
2
0
Z 1
2
e
43
(t2 + et )2 dt =
kf + gk2 =
+ 2e − .
2
10
0
kf k2 =
Up to six digits of accuracy, this gives
hf, gi = 0.718282,
kf k = 0.447214,
kgk = 1.78732,
kf + gk = 2.19797.
Cauchy-Schwarz says that
|hf, gi| ≤ kf k kgk ,
or
0.718282 ≤ 0.447214 ∗ 1.78732 = 0.799315,
which is true.
The triangle inequality says that
kf + gk ≤ kf k + kgk ,
or
2.19797 ≤ 0.447214 + 1.78732 = 2.23453,
which is also true (as it better be!).
Finally, we have that the angle satisfies
cos θ =
hf, gi
0.718282
=
= 0.898622
kf k kgk
0.799315
which is about 0.45 radians, or 26 degrees.
6. The parallelogram law states that, for any x, y in an inner product space V , we have
kx + yk2 + kx − yk2 = 2 kxk2 + 2 kyk2 .
Prove the parallelogram law. Draw a picture of what it corresponds to in R2 .
Solution: We write
kx + yk2 + kx − yk2 = hx + y, x + yi + hx − y, x − yi
= hx, xi + hx, yi + hy, xi + hy, yi + hx, xi − hx, yi − hy, xi + hy, yi
= 2 hx, xi + 2 hy, yi = 2 kxk2 + 2 kyk2 .
7. Let v1 , v2 , . . . vn be an orthogonal set in an inner product space V . Prove that
2
n
n
X
X
a
v
=
|ai |2 kvi k2 .
i i
i=1
i=1
This is known as the Pythagorean Theorem. Draw a picture in R2 to explain why it has this
name.
Solution: We compute:
2 *
+
n
n
n
X
X
X
ai v i =
ai vi ,
aj vj
i=1
=
=
i=1
n
n
XX
j=1
ai aj hvi , vj i =
i=1 j=1
n
X
a2i kvi k2
i=1
n
X
ai ai hvi , vi i
i=1
.
8. Let α > 0. Assume that T : V → V is a linear operator that has the property that kT (x)k ≥
α kxk. Show that T must be one-to-one.
Solution: Let x 6= 0. Then kxk > 0. From the assumptions above, we have that kT xk >
α kxk > 0, which means that T x 6= 0. Since T x 6= 0 for any x 6= 0, this means that T is
one-to-one.