Wednesday, June 11, 2014

Linear Algebra: #10 Similar Matrices; Changing Bases

  • 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 : VV be a linear mapping and let {u1, . . . , un}, {v1, . . . , vn} be two bases for V. Assume that A is the matrix for f with respect to the basis {v1, . . . , vn} and furthermore A' is the matrix for f with respect to the basis {u1, . . . , un}.

Linear Algebra: #10 Similar Matrices; Changing Bases equation pic 1
 Then we have A' = C−1AC.

Proof
From the definition of A', we have
Linear Algebra: #10 Similar Matrices; Changing Bases equation pic 2

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

Linear Algebra: #10 Similar Matrices; Changing Bases equation pic 3

Therefore we have A' = C−1AC.

Note that we have written here vk (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

Linear Algebra: #10 Similar Matrices; Changing Bases equation pic 4

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

Which mapping does the matrix C represent? From the equations

Linear Algebra: #10 Similar Matrices; Changing Bases equation pic 5

we see that it represents a mapping g : VV such that g(vi) =  ui for all i, expressed in terms of the original basis {v1, . . . , vn}. 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.

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