What if we want to choose a different basis for and/ or ? That is to say, what if we want to take vectors expressable in one basis to the same vectors expressed in another basis? Well, say that we are given two bases: and in . If given an expression of we want to plug in and get an expression in terms of , we write this as
So we get an matrix with entries , whose column is the expression of in the basis as a column vector. So symbolically, we can think of this as . So , i.e. the element of represented by a column vector in new basis is represented by in the old basis . We can think of this more conceptually as being a linear map—i.e. a matrix transformation—by . Breaking this down, we have the identity mapping taking the vector to itself (from one basis to another, but the same vector(s)); thus we are taking things from , here, to the targets of . We can additionally do the same for in reverse! Let , i.e. . Hence: That is to say, .
Note. Matrix of identity from to would be which is the matrix such that is the identity matrix.
If and we change basis using a single new basis for both domain and codomain, then a remarkable simplification occurs. Let be an endomorphism (a linear map from a space to itself). With as above, and using the same basis change for both domain and codomain, we have (since converts back to , the inverse direction of ). Thus the formula becomes:
Theorem 5.2.1 (Change of basis for endomorphisms). Let , let and be two bases of , and let be the change-of-basis matrix. If and , then
Proof. We have established that where . But since applying then (or vice versa) gives the identity transformation on coordinate vectors. Explicitly: if converts old coordinates to new, and converts new to old, then composing identity maps gives .
Thus . ∎
This formula tells us that the matrices representing the same linear operator in different bases are related by conjugation by an invertible matrix.
5.2.1 Similar matrices
Definition 5.2.1 (Similar matrices). Two matrices and over a field are similar if there exists an invertible matrix such that We write .
Proposition 5.2.1 (Similarity is an equivalence relation). The similarity relation on is an equivalence relation:
- Reflexive: (take ).
- Symmetric: If , then (if , then ).
- Transitive: If and , then (if and , then ).
As we can see, similarity classes of matrices correspond exactly to linear operators on : two matrices are similar if and only if they represent the same linear operator in different bases.
Note. From the categorical perspective, similar matrices represent the same morphism in —they are simply different coordinate descriptions of one intrinsic object. This is why similarity-invariant properties (trace, determinant, eigenvalues, rank) are so important: they are properties of the operator itself, not of any particular matrix representation.
5.2.2 Invariants under similarity
What properties of a matrix are preserved under similarity? These are called similarity invariants—they depend only on the underlying linear operator, not on the choice of basis.
Definition 5.2.2 (Trace of a matrix). The trace of an matrix is the sum of its diagonal entries:
Lemma 5.2.1 (Cyclic property of trace). For any matrices and such that both and are defined and square, we have .
Proof. Let be and be . Then Similarly, These double sums are identical (just reorder the summation), so . ∎
Proposition 5.2.2 (Trace is a similarity invariant). If , then . In particular, the trace of a linear operator is well-defined: for any basis .
Proof. If , then using the cyclic property: ∎
Proposition 5.2.3 (Determinant is a similarity invariant). If , then . Hence the determinant of a linear operator is well-defined.
Proof. If , then by the multiplicativity of determinants: ∎
Proposition 5.2.4 (Eigenvalues are similarity invariants). (Cited here ahead of the formal introduction of eigenvalues in a later lecture; for now, a scalar is an eigenvalue of if for some nonzero , equivalently .) If , then and have the same eigenvalues (with the same algebraic multiplicities). In particular, the eigenvalues of a linear operator are intrinsic to the operator, not dependent on the choice of basis.
Proof. A scalar is an eigenvalue of if and only if . If , then Thus , so by the previous result. Hence and have identical characteristic polynomials, and therefore the same eigenvalues with the same multiplicities. ∎
Note. Rank is also a similarity invariant: since multiplying by invertible matrices doesn’t change rank. More sophisticated invariants include the minimal polynomial, the Jordan normal form, and the rational canonical form—these completely classify operators up to similarity over algebraically closed fields. We will encounter these when we study eigenvalues in depth.
Example 5.2.1. Let and over . Both have , , and eigenvalue with multiplicity 2. Yet ! The reason: , so for any . Since , they cannot be similar.
This shows that trace, determinant, and eigenvalues are necessary but not sufficient conditions for similarity. The difference lies in the geometric multiplicity of the eigenvalue 2: for , the eigenspace has dimension 2; for , has dimension 1. Geometric multiplicity is also a similarity invariant!
Question 1.
- Let be the linear map with matrix with respect to the standard basis. Find the matrix of with respect to the basis .
- Prove that an matrix is similar to a diagonal matrix if and only if there exists a basis of consisting of eigenvectors of .
- Show that but in general . Give a counterexample.
Show solution
(1) First, find the change-of-basis matrix from to the standard basis . The columns of are the vectors of expressed in : Then .
The matrix of with respect to is:
Computing first: .
Then .
So . We can verify: and . ✓
(2) () Suppose is similar to a diagonal matrix , so for some invertible . Let be the -th column of . Then , which gives . So each column of is an eigenvector of . Since is invertible, its columns form a basis of .
() Suppose is a basis of eigenvectors with . Let be the matrix with columns . Then where , so .
(3) We proved above. For the determinant counterexample: Then , , but , so .