Linear Algebra, 2020a

Section V. Change of Basis 269
for going up Rep ˆD,E 2
(id). That matrix is easy to find and we have a formula
for the 2×2 inverse, so that’s what is in the equation below.
( ) −1 (√ )( )
RepˆB, ˆD (t) = −1 2 3/2 −1/2 1 0
√
0 3 1/2 3/2 1 2
(
(5 − √ 3)/6 (3 + 2 √ )
3)/3
=
(1 + √ √
3)/6 3/3
The matrix is messier but the map that it represents is the same. For instance,
to replicate the effect of t in the picture, start with ˆB,
( ) ( )
RepˆB ( 1 1
)=
3 1
apply ˆT,
(
(5 − √ 3)/6 (3 + 2 √ ) ( )
3)/3
(1 + √ 1
√
3)/6 3/3 1
ˆB, ˆD
and check it against ˆD.
11 + 3 √ ( )
3 −1
·
6 0
+ 1 + 3√ 3
6
ˆB
ˆB
=
(
(11 + 3 √ 3)/6
(1 + 3 √ 3)/6
( ) (
2 (−3 + √ )
3)/2
· =
3 (1 + 3 √ 3)/2
2.3 Example Changing bases can make the matrix simpler. On R 3 the map
⎛ ⎞ ⎛ ⎞
x y + z
⎜ ⎟
⎝y⎠
↦−→
t ⎜ ⎟
⎝x + z⎠
z x + y
is represented with respect to the standard basis in this way.
⎛
⎜
0 1 1
⎞
⎟
Rep E3 ,E 3
(t) = ⎝1 0 1⎠
1 1 0
Representing it with respect to
⎛ ⎞ ⎛ ⎞ ⎛
1 1
⎜ ⎟ ⎜ ⎟ ⎜
1
⎞
⎟
B = 〈 ⎝−1⎠ , ⎝ 1 ⎠ , ⎝1⎠〉
0 −2 1
gives a matrix that is diagonal.
⎛
⎞
−1 0 0
⎜
⎟
Rep B,B (t) = ⎝ 0 −1 0⎠
0 0 2
)
ˆD