Linear Algebra Exercises-n-Answers.pdf

194 Linear Algebra, by Hefferon
and P is the inverse of that.
⎛
⎞
1 0 0 0 0
−1 1 1 −1 1
P = Rep E5,B(id) = (P −1 ) −1 =
⎜−1 0 1 0 0
⎟
⎝ 1 0 −1 1 −1⎠
0 0 0 0 1
(c) The calculation to check this is routine.
Five.III.2.22
(a) The calculation
p
(
N p ) (
N
)
(N p )
1/2 −1/2 u ∣∣
1
{ u ∈ C}
1/2 −1/2 u
2 –zero matrix– C 2
shows that any map represented by the matrix must act on the string basis in this way
⃗β 1 ↦→ ⃗ β 2 ↦→ ⃗0
because the nullspace after one application has dimension one and exactly one basis vector, β ⃗ 2 , is
sent to zero. Therefore, this representation with
(
respect
)
to 〈 β ⃗ 1 , β ⃗ 2 〉 is the canonical form.
0 0
1 0
(b) The calculation here is similar to the prior one.
p
⎛
N p N (N p )
1 ⎝ 0 0 0
⎞ ⎛
0 −1 1⎠ { ⎝ u ⎞
v⎠ ∣ u, v ∈ C}
0 −1 1 v
2 –zero matrix– C 3
The table shows that the string basis is of the form
⃗β 1 ↦→ β ⃗ 2 ↦→ ⃗0
⃗β 3 ↦→ ⃗0
because the nullspace after one application of the map has dimension two — β ⃗ 2 and β ⃗ 3 are both
sent to zero — and one more iteration results in the additional vector being brought to zero.
(c) The calculation
p
⎛
N p ⎞ ⎛
N (N p )
−1 1 −1
1 ⎝ 1 0 1 ⎠ { ⎝ u ⎞
0 ⎠ ∣ u ∈ C}
⎛1 −1 1
⎝ 1 0 1
⎞
0 0 0
−1 0 −1
⎛ −u⎞
2
⎠ { ⎝ u v ⎠ ∣ u, v ∈ C}
−u
3 –zero matrix– C 3
shows that any map represented by this basis must act on a string basis in this way.
⃗β 1 ↦→ β ⃗ 2 ↦→ β ⃗ 3 ↦→ ⃗0
Therefore, this is the canonical form. ⎛
⎝ 0 0 0
⎞
1 0 0⎠
0 1 0
Five.III.2.23 A couple of examples
⎛
( ) ( ) ( )
0 0 a b 0 0
=
⎝ 0 0 0
⎞ ⎛
1 0 0⎠
⎝ a b c
⎞ ⎛
d e f⎠ = ⎝ 0 0 0
⎞
a b c⎠
1 0 c d a b
0 1 0 g h i d e f
suggest that left multiplication by a block of subdiagonal ones shifts the rows of a matrix downward.
Distinct blocks ⎛ ⎞ ⎛
⎞ ⎛
⎞
0 0 0 0 a b c d 0 0 0 0
⎜1 0 0 0
⎟ ⎜ e f g h
⎟
⎝0 0 0 0⎠
⎝ i j k l ⎠ = ⎜ a b c d
⎟
⎝0 0 0 0⎠
0 0 1 0 m n o p i j k l
act to shift down distinct parts of the matrix.
Right multiplication does an analgous thing to columns. See Exercise 17.