442 CHAPTER 18. LINEAR TRANSFORMATIONS
Definition 18.5.5 In the special case where V =W and only one basis is used for V =W,this becomes
q−11 q2A2q−1
2 q1 = A1.
Letting S be the matrix of the linear transformation q−12 q1 with respect to the standard basis
vectors in Fn,S−1A2S = A1. (18.11)
When this occurs, A1 is said to be similar to A2 and A 7→ S−1AS is called a similaritytransformation.
Here is some terminology.
Definition 18.5.6 Let S be a set. The symbol, ∼ is called an equivalence relation on S if itsatisfies the following axioms.
1. x∼ x for all x ∈ S. (Reflexive)
2. If x∼ y then y∼ x. (Symmetric)
3. If x∼ y and y∼ z, then x∼ z. (Transitive)
Definition 18.5.7 [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.
With the above definition one can prove the following simple theorem which you shoulddo if you have not seen it.
Theorem 18.5.8 Let ∼ be an equivalence class defined on a set, S and let H denote theset of equivalence classes. Then if [x] and [y] are two of these equivalence classes, eitherx∼ y and [x] = [y] or it is not true that x∼ y and [x]∩ [y] = /0.
Theorem 18.5.9 In the vector space of n×n matrices, define
A∼ B
if there exists an invertible matrix S such that
A = S−1BS.
Then ∼ is an equivalence relation and A∼ B if and only if whenever V is an n dimensionalvector space, there exists L ∈L (V,V ) and bases {v1, · · · ,vn} and {w1, · · · ,wn} such thatA is the matrix of L with respect to {v1, · · · ,vn} and B is the matrix of L with respect to{w1, · · · ,wn}.
Proof: A ∼ A because S = I works in the definition. If A ∼ B , then B ∼ A, becauseA = S−1BS implies B = SAS−1. If A∼ B and B∼C, then A = S−1BS, B = T−1CT and so
A = S−1T−1CT S = (T S)−1 CT S
which implies A∼C. This verifies the first part of the conclusion.It was pointed out in the above definition that if A,B are matrices which come from
a single linear transformation, then they are similar. Suppose now that A,B are similar.