Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 191
Subsection Five.III.1: Self-Composition
Five.III.1.8 For the zero transformation, no matter what the space, the chain of rangespaces is V ⊃
{⃗0} = {⃗0} = · · · and the chain of nullspaces is {⃗0} ⊂ V = V = · · · . For the identity transformation
the chains are V = V = V = · · · and {⃗0} = {⃗0} = · · · .
Five.III.1.9
(a) Iterating t 0 twice a + bx + cx 2 ↦→ b + cx 2 ↦→ cx 2 gives
a + bx + cx 2 t2 0
↦−→ cx 2
and any higher power is the same map. Thus, while R(t 0 ) is the space of quadratic polynomials
with no linear term {p + rx ∣ 2 p, r ∈ C}, and R(t 2 0) is the space of purely-quadratic polynomials
{rx ∣ 2 r ∈ C}, this is where the chain stabilizes R ∞ (t 0 ) = {rx ∣ 2 n ∈ C}. As for nullspaces, N (t 0 )
is the space of purely-linear quadratic polynomials {qx ∣ q ∈ C}, and N (t 2 0) is the space of quadratic
polynomials with no x 2 term {p + qx ∣ p, q ∈ C}, and this is the end N ∞ (t 0 ) = N (t 2 0).
(b) The second power ( ( ) ( )
a t
↦−→
1 0 t
↦−→
1 0
b)
a 0
is the zero map. Consequently, the chain of rangespaces
( )
R 2 0 ∣∣
⊃ { p ∈ C} ⊃ {⃗0 } = · · ·
p
and the chain of nullspaces
( )
q ∣∣
{⃗0 } ⊂ { q ∈ C} ⊂ R 2 = · · ·
0
each has length two. The generalized rangespace is the trivial subspace and the generalized nullspace
is the entire space.
(c) Iterates of this map cycle around
a + bx + cx 2 t
↦−→
2
b + cx + ax
2 t
↦−→
2
c + ax + bx
2 t
↦−→
2
a + bx + cx2 · · ·
and the chains of rangespaces and nullspaces are trivial.
P 2 = P 2 = · · · {⃗0 } = {⃗0 } = · · ·
Thus, obviously, generalized spaces are R ∞ (t 2 ) = P 2 and N ∞ (t 2 ) = {⃗0 }.
(d) We have
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎝ a b⎠ ↦→
c
and so the chain of rangespaces
⎛ ⎞
⎝ a a
b
⎠ ↦→
⎝ a a⎠ ↦→
a
⎝ a a
a
⎠ ↦→ · · ·
R 3 ⊃ { ⎝ p p⎠ ∣ p, r ∈ C} ⊃ { ⎝ p p⎠ ∣ p ∈ C} = · · ·
r
p
and the chain of nullspaces
⎛ ⎞
⎛ ⎞
0
{⃗0 } ⊂ { ⎝0⎠ ∣ 0
r ∈ C} ⊂ { ⎝q⎠ ∣ q, r ∈ C} = · · ·
r
r
each has length two. The generalized spaces are the final ones shown above in each chain.
Five.III.1.10
Each maps x ↦→ t(t(t(x))).
Five.III.1.11 Recall that if W is a subspace of V then any basis B W for W can be enlarged to make
a basis B V for V . From this the first sentence is immediate. The second sentence is also not hard: W
is the span of B W and if W is a proper subspace then V is not the span of B W , and so B V must have
at least one vector more than does B W .
Five.III.1.12 It is both ‘if’ and ‘only if’. We have seen earlier that a linear map is nonsingular if and
only if it preserves dimension, that is, if the dimension of its range equals the dimension of its domain.
With a transformation t: V → V that means that the map is nonsingular if and only if it is onto:
R(t) = V (and thus R(t 2 ) = V , etc).
⎛
⎞