Linear Algebra, 2020a

414 Chapter Five. Similarity
with this eigenvector.
( )( ) ( )
2 0 1 2
T⃗e 1 =
= = 2⃗e 1
0 0 0 0
Suppose that T represents a transformation t: C 2 → C 2 with respect to the
standard basis. Then the action of this transformation t is simple.
( ) ( )
x t 2x
↦−→
y 0
Of course, ˆT represents the same transformation but with respect to a different
basis B. We can find this basis. Following the arrow diagram from the lower left
to the upper left
t
V wrt E2 −−−−→ V wrt E2
T
⏐
⏐
id↓
id↓
V wrt B
t
−−−−→
ˆT
V wrt B
shows that P −1 = Rep B,E2 (id). By the definition of the matrix representation
of a map, its first column is Rep E2
(id(⃗β 1 )) = Rep E2
(⃗β 1 ). With respect to the
standard basis any vector is represented by itself, so the first basis element ⃗β 1 is
the first column of P −1 . The same goes for the other one.
(
B = 〈
2
−1
) (
,
Since the matrices T and ˆT both represent the transformation t, both reflect the
action t(⃗e 1 )=2⃗e 1 .
−1
1
)
〉
Rep E2 ,E 2
(t) · Rep E2
(⃗e 1 )=T · Rep E2
(⃗e 1 )=2 · Rep E2
(⃗e 1 )
Rep B,B (t) · Rep B (⃗e 1 )=ˆT · Rep B (⃗e 1 )=2 · Rep B (⃗e 1 )
But while in those two equations the eigenvalue 2’s are the same, the vector
representations differ.
( ) ( )
1 1
T · Rep E2
(⃗e 1 )=T = 2 ·
0 0
( ) ( )
1 1
ˆT · Rep B (⃗e 1 )=ˆT · = 2 ·
1 1
That is, when the matrix representing the transformation is T = Rep E2 ,E 2
(t)
then it “assumes” that column vectors are representations with respect to E 2 .
However ˆT = Rep B,B (t) “assumes” that column vectors are representations with
respect to B and so the column vectors that get doubled are different.