What happens to the matrix of a linear map when we choose different bases? Let and be two bases of , the “old” and the “new” one. Write each new basis vector in terms of the old basis:
Definition 5.3.1 (Change-of-basis matrix). The matrix , whose -th column is the coordinate vector of in the old basis , is the change-of-basis matrix from to . Comparing with the definition of the matrix of a linear map, the matrix of the identity map of with the new basis used in the domain and the old basis in the target.
Symbolically, In words:
Proposition 5.3.2 (Change of coordinates). If a vector has coordinate vector in the new basis , then its coordinate vector in the old basis is . The matrix is invertible, with inverse , and .
Proof. The first statement is Proposition 5.2.4 applied to , with the basis in the domain and in the target. By Proposition 5.2.9, and likewise in the other order. ∎
Note the direction: has the new basis vectors as its columns, but it converts new coordinates into old ones.
Theorem 5.3.3 (Change of basis). Let be linear, let be bases of and bases of , and put Then .
Proof. Write and apply Proposition 5.2.9, using the bases : In the symbolic notation, with : ∎
Here is the inverse of the change-of-basis matrix from to . The most important case is an operator, with the same basis used on both sides.
Corollary 5.3.4 (Change of basis for operators). Let , let and be bases of , and let . If and , then
Proof. Apply the theorem with , and . Then by Proposition 5.3.2. ∎
So the matrices representing the same linear operator in different bases are related by conjugation by an invertible matrix. Still, the whole point of linear algebra is to work with linear maps in coordinate-free language as much as possible, and to choose bases only when they make a computation transparent.
5.3.1 Similar Matrices
Definition 5.3.5 (Similar matrices). Two matrices are similar, written , if there exists an invertible matrix such that
Proposition 5.3.6 (Similarity is an equivalence relation). Similarity is an equivalence relation on .
Proof. It is reflexive: . It is symmetric: if , then . It is transitive: if and , then . ∎
Proposition 5.3.7 (Similar matrices represent the same operator). Two matrices are similar if and only if there are an -dimensional space , an operator and bases of with and .
Proof. If and are matrices of one operator in two bases, they are similar by Corollary 5.3.4. Conversely, suppose . Take and , whose matrix in the standard basis is . The columns of form a basis of by Proposition 5.2.14, and because its columns are the vectors of written in the standard basis. By the corollary, . ∎
So a similarity class of matrices records exactly what one operator looks like in all possible bases. A property of matrices that is unchanged by similarity, a similarity invariant, is therefore a property of the operator itself, shared by all the matrices that represent it. We now meet the first such invariant, the trace; the determinant follows in Section 5.4.
5.3.2 The Trace
Definition 5.3.8 (Trace of a matrix). The trace of a square matrix is the sum of its diagonal entries:
The trace is a linear map .
Lemma 5.3.9 (Cyclic property of the trace). If and , then .
Proof. We have These are the same finite sum, taken in two different orders. ∎
Proposition 5.3.10 (Trace is a similarity invariant). If , then .
Proof. If , then by the cyclic property applied to and , ∎
Definition 5.3.11 (Trace of an operator). Let be finite-dimensional. The trace of an operator is for any basis of . It does not depend on : by Corollary 5.3.4 the matrices of in two bases are similar, and similar matrices have the same trace.
Proposition 5.3.12 (Properties of the trace). Let be finite-dimensional, and . Then , , , and .
Proof. Fix a basis and take matrices. The map is a ring isomorphism (Corollary 5.2.10), the trace of matrices is linear, by Lemma 5.3.9, and has trace . ∎
Note. In Chapter 9 the trace of an operator is described without any choice of basis, as a contraction of tensors; that description makes its basis-independence evident.