Linear Algebra Exercises-n-Answers.pdf

98 Linear Algebra, by Hefferon
The resulting matrix
⎛
⎞
1 1 1 1 . . . 1
(
∫
0 1 2 3 . . . n ( 2)
Rep B,B ( ) =
0 0 1 3 . . . n 3)
⎜
⎝
⎟
.
⎠
0 0 0 . . . 1
is Pascal’s triangle (recall that ( n
r)
is the number of ways to choose r things, without order and
without repetition, from a set of size n).
Three.III.1.18
is the n×n identity matrix.
Three.III.1.19
Where the space is n-dimensional,
⎛ ⎞
1 0 . . . 0
0 1 . . . 0
Rep B,B (id) = ⎜ ⎟
⎝ . ⎠
0 0 . . . 1
Taking this as the natural basis
( )
B = 〈 β ⃗ 1 , β ⃗ 2 , β ⃗ 3 , β ⃗ 1 0
4 〉 = 〈 ,
0 0
( )
0 1
,
0 0
B,B
( )
0 0
,
1 0
the transpose map acts in this way
⃗β 1 ↦→ β ⃗ 1 β2 ⃗ ↦→ β ⃗ 3 β3 ⃗ ↦→ β ⃗ 2 β4 ⃗ ↦→ β ⃗ 4
( )
0 0
〉
0 1
so that representing the images with respect to the codomain’s basis and adjoining those column
vectors together gives this.
⎛ ⎞
1 0 0 0
Rep B,B (trans) = ⎜0 0 1 0
⎟
⎝0 1 0 0⎠
0 0 0 1
Three.III.1.20 (a) With respect to the basis of the codomain, the images of the members of the
basis of the domain are represented as
⎛ ⎞
⎛ ⎞
⎛ ⎞
⎛ ⎞
0
0
0
0
Rep B ( β ⃗ 2 ) = ⎜1
⎟
⎝0⎠ Rep B( β ⃗ 3 ) = ⎜0
⎟
⎝1⎠ Rep B( β ⃗ 4 ) = ⎜0
⎟
⎝0⎠ Rep B(⃗0) = ⎜0
⎟
⎝0⎠
0
0
1
0
and consequently, the matrix representing the transformation is this.
⎛ ⎞
0 0 0 0
⎜1 0 0 0
⎟
⎝0 1 0 0⎠
0 0 1 0
⎛
⎞
0 0 0 0
(b) ⎜1 0 0 0
⎟
⎝0 0 0 0⎠
⎛
0 0 1 0
⎞
0 0 0 0
(c) ⎜1 0 0 0
⎟
⎝0 1 0 0⎠
0 0 0 0
Three.III.1.21
(a) The picture of d s : R 2 → R 2 is this.
B,B
d s(⃗u)
⃗u
⃗v
d s
−→
d s (⃗v)