Linear Algebra

Section II. Similarity 355
2.4 Corollary A transformation t is diagonalizable if and only if there is a
basis B = 〈 � β1,... , � βn〉 and scalars λ1,... ,λn such that t( � βi) =λi � βi for each i.
Proof. This follows from the definition by considering a diagonal representation
matrix.
⎛
.
.
⎜
.
.
RepB,B(t) = ⎜
⎝RepB(t(
� β1)) ··· RepB(t( � ⎞
⎟
βn)) ⎟
⎠
.
.
.
.
=
⎛
⎞
λ1 0
⎜ .
⎝ . .
.
..
.
⎟
. ⎠
0 λn
This representation is equivalent to the existence of a basis satisfying the stated
conditions simply by the definition of matrix representation. QED
2.5 Example To diagonalize
T =
� �
3 2
0 1
we take it as the representation of a transformation with respect to the standard
basis T =RepE2,E2 (t) and we look for a basis B = 〈 � β1, � β2〉 such that
� �
λ1 0
RepB,B(t) =
0 λ2
that is, such that t( � β1) =λ1 and t( � β2) =λ2.
�
3
0
�
2 �β1 = λ1 ·
1
� β1
�
3
0
�
2
1
We are looking for scalars x such that this equation
� �� � � �
3 2 b1 b1
= x ·
0 1
b2
b2
�β2 = λ2 · � β2
has solutions b1 and b2, which are not both zero. Rewrite that as a linear system.
(3 − x) · b1 + 2· b2 =0
(1 − x) · b2 =0
In the bottom equation the two numbers multiply to give zero only if at least
one of them is zero so there are two possibilities, b2 =0andx =1. Intheb2 =0
possibility, the first equation gives that either b1 =0orx = 3. Since the case
of both b1 =0andb2 = 0 is disallowed, we are left looking at the possibility of
x = 3. With it, the first equation in (∗) is0· b1 +2· b2 = 0 and so associated
with 3 are vectors with a second component of zero and a first component that
is free.
� �� �
3 2 b1
=3·
0 1 0
� �
b1
0
(∗)