Linear Algebra, Theory And Applications, 2012a

230 LINEAR TRANSFORMATIONS
2. If x ∼ y then y ∼ x. (Symmetric)
3. If x ∼ y and y ∼ z, thenx ∼ z. (Transitive)
Definition 9.3.7 [x] denotes the set of all elements of S which are equivalent to x and [x]
is called the equivalence class determined by x or just the equivalence class of x.
Also recall the notion of equivalence classes.
Theorem 9.3.8 Let ∼ be an equivalence class defined on a set S and let H denote the set
of equivalence classes. Then if [x] and [y] are two of these equivalence classes, either x ∼ y
and [x] =[y] or it is not true that x ∼ y and [x] ∩ [y] =∅.
Theorem 9.3.9 In the vector space of n × n matrices, define
A ∼ B
if there exists an invertible matrix S such that
A = S −1 BS.
Then ∼ is an equivalence relation and A ∼ B if and only if whenever V is an n dimensional
vector space, there exists L ∈L(V,V ) and bases {v 1 , ··· ,v n } and {w 1 , ··· ,w n } such that
A is the matrix of L with respect to {v 1 , ··· ,v n } and B is the matrix of L with respect to
{w 1 , ··· ,w n }.
Proof: A ∼ A because S = I works in the definition. If A ∼ B ,thenB ∼ A, because
A = S −1 BS
implies B = SAS −1 . If A ∼ B and B ∼ C, then
A = S −1 BS, B = T −1 CT
and so
A = S −1 T −1 CTS =(TS) −1 CTS
which implies A ∼ C. This verifies the first part of the conclusion.
Now let V be an n dimensional vector space, A ∼ B so A = S −1 BS and pick a basis for
V,
β ≡{v 1 , ··· ,v n }.
Define L ∈L(V,V )by
Lv i ≡ ∑ a ji v j
j
where A =(a ij ) . Thus A is the matrix of the linear transformation L. Consider the diagram
F n −→
B F n
q γ ↓ ◦ q γ ↓
V −→
L V
q β ↑ ◦ q β ↑
F n −→
A F n
where q γ is chosen to make the diagram commute. Thus we need S = qγ
−1 q β which requires
q γ = q β S −1