Linear Algebra, 2020a

268 Chapter Three. Maps Between Spaces
2.2 Example The matrix
(
cos(π/6)
T =
sin(π/6)
) (√ )
−sin(π/6) 3/2 −1/2
= √
cos(π/6) 1/2 3/2
represents, with respect to E 2 , E 2 , the transformation t: R 2 → R 2 that rotates
vectors through the counterclockwise angle of π/6 radians.
( 1
3
) ( √ )
(−3 + 3)/2
t π/6
−→
(1 + 3 √ 3)/2
We can translate T to a representation with respect to these
( )( ) ( )( )
1 0
−1 2
ˆB = 〈 〉 ˆD = 〈
〉
1 2
0 3
by using the arrow diagram above.
R 2 wrt E 2
id
⏐
↓
R 2 wrt ˆB
t
−−−−→
T
R 2 wrt E 2
t
−−−−→
ˆT
id
⏐
↓
R 2 wrt ˆD
The picture illustrates that we can compute ˆT either directly by going along the
square’s bottom, or as in formula (∗) by going up on the left, then across the
top, and then down on the right, with ˆT = Rep E2 , ˆD (id) · T · RepˆB,E 2
(id). (Note
again that the matrix multiplication reads right to left, as the three functions
are composed and function composition reads right to left.)
Find the matrix for the left-hand side, the matrix RepˆB,E 2
(id), in the usual
way: find the effect of the identity matrix on the starting basis ˆB — which is
no effect at all — and then represent those basis elements with respect to the
ending basis E 2 .
( )
1 0
RepˆB,E 2
(id) =
1 2
This calculation is easy when the ending basis is the standard one.
There are two ways to compute the matrix for going down the square’s right
side, Rep E2 , ˆD
(id). We could calculate it directly as we did for the other change
of basis matrix. Or, we could instead calculate it as the inverse of the matrix