Linear Algebra, 2020a

410 Chapter Five. Similarity
Focus first on the bottom equation. There are two cases: either b 2 = 0 or x = 1.
In the b 2 = 0 case the first equation gives that either b 1 = 0 or x = 3. Since
we’ve disallowed the possibility that both b 2 = 0 and b 1 = 0, we are left with
the first diagonal entry λ 1 = 3. With that, (∗)’s first equation is 0·b 1 +2·b 2 = 0
and so associated with λ 1 = 3 are vectors having a second component of zero
while the first component is free.
(
3 2
0 1
)(
b 1
0
) (
= 3 ·
To get a first basis vector choose any nonzero b 1 .
( )
1
⃗β 1 =
0
The other case for the bottom equation of (∗) isλ 2 = 1. Then (∗)’s first
equation is 2 · b 1 + 2 · b 2 = 0 and so associated with this case are vectors whose
second component is the negative of the first.
(
3 2
0 1
)(
)
b 1
= 1 ·
−b 1
(
b 1
0
)
)
b 1
−b 1
Get the second basis vector by choosing a nonzero one of these.
( )
1
⃗β 2 =
−1
Now draw the similarity diagram
R 2 wrt E 2
id
⏐
↓
R 2 wrt B
t
−−−−→
T
R 2 wrt E 2
t
−−−−→
D
id
⏐
↓
R 2 wrt B
and note that the matrix Rep B,E2 (id) is easy, giving this diagonalization.
( ) ( ) −1 ( )( )
3 0 1 1 3 2 1 1
=
0 1 0 −1 0 1 0 −1
The rest of this section expands on that example by considering more closely
the property of Lemma 2.4, including seeing a streamlined way to find the λ’s.
The section after that expands on Example 2.3, to understand what can prevent
diagonalization. Then the final section puts these two together, to produce a
canonical form that is in some sense as simple as possible.