Description
applied linear algebra
Explanation & Answer
Attached.
Problem 1:
a
Consider a linear transformation T =
c
b
. Since T maps e1 to −e2 and e2 to e1 + 3e2 , we have
d
e a b e1 ae1 + be2 −e2
T 1 =
=
=
e2 c d e2 ce1 + de2 e1 + 3e2
which gives a = 0, b = −1, c = 1, d = 3 .
0 −1
.
3
Hence, the linear transformation is T =
1
Problem 2:
1 1 s 1 s + 1 0 s
=
= for s , it implies that the transformation maps the x-axis
0 1 0 0 s + 1 0 0
Since A =
to the x-axis itself.
t
1 1 0 1 0 + 1 t t
=
= for t , and the set : t is the line y = x in
0 1 t 0 0 + 1 t t
t
2
Hence, the transformation maps the y-axis to the line y = x in
.
Since A =
Problem 3:
The unit square is described as below:
Problem 4:
2
.
We compute
0 2
A1 =
0 1
1 2
A1 =
0 1
0 0 0
0 2 0 0 0
= , A1 =
= ,
1 0 0
1 1 1 1 1
0 1 2
1 2 0 1 2
= , A1 =
=
1 0 1
1 1 1 1 2
Hence, A1 maps the unit square to (0,0), (0,1), (2,1) and (2,2), as sketched below
We compute
0 0
A2 =
0 −1
1 0
A2 =
0 −1
1 0 0
0 0
= , A2 =
0 0 0
1 −1
1 1 0
1 0
= , A2 =
0 0 −1
1 −1
1 0 1
=
,
0 1 0
1 1 1
=
0 1 −1
Hence, A2 maps the unit square to (0,0), (1,0), (0,-1) and (1,-1), as sketched below
We compute
0 �...
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