• Linear Algebra: #10 Similar Matrices; Changing Bases

Definition
Let A and A' be n ×n matrices. If a matrix C ∈ GL(n, F) exists, such that A' = C⁻¹AC then we say that the matrices A and A' are similar.


Theorem 27
Let f : V → V be a linear mapping and let {u₁, . . . , u_n}, {v₁, . . . , v_n} be two bases for V. Assume that A is the matrix for f with respect to the basis {v₁, . . . , v_n} and furthermore A' is the matrix for f with respect to the basis {u₁, . . . , u_n}.

 Then we have A' = C⁻¹AC.

Proof
From the definition of A', we have

for all i = 1, . . . , n. On the other hand we have

Therefore we have A' = C⁻¹AC.

Note that we have written here v_k (in terms of the summation from l = 1, ...., n), and then we have said that the resulting matrix (which we call C^*) is, in fact, C⁻¹. To see that this is true, we begin with the definition of C itself. We have

That is, CC^* = I_n, and therefore C^* = C⁻¹.

Which mapping does the matrix C represent? From the equations

we see that it represents a mapping g : V → V such that g(v_i) =  u_i for all i, expressed in terms of the original basis {v₁, . . . , v_n}. So we see that a similarity transformation, taking a square matrix A to a similar matrix A' = C⁻¹AC is always associated with a change of basis for the vector space V.

Much of the theory of linear algebra is concerned with finding a simple basis (with respect to a given linear mapping of the vector space into itself), such that the matrix of the mapping with respect to this simpler basis is itself simple — for example diagonal, or at least trigonal.