Abstract Linear Algebra 28 | Equivalent Matrices

Equivalent matrices capture when two different matrix representations actually describe the same linear transformation, even after changing the bases of the domain and codomain. The key insight is that a linear map can look like different matrices because the coordinates depend on the chosen bases. In the earlier example, the derivative map from polynomial space P3 to P2 had multiple matrix representations—different matrices, same underlying linear map—because different polynomial bases were used.

That observation leads to a practical question: given a matrix A representing a linear map with respect to bases B (for V) and C (for W), can one find a “nicer” matrix A~ that is still a valid representation of the same map, perhaps with more zeros? The answer is not always—certain matrices, like the zero matrix, can’t represent a nonzero linear map. But in the cases where it is possible, the new matrix arises from changing bases in both vector spaces.

Geometrically, changing bases corresponds to multiplying by invertible “change-of-basis” matrices. If V has dimension n and W has dimension m, then A is an m×n matrix. Switching to new bases introduces invertible matrices S and T so that the new representation A~ is related to the old one by the formula

A~ = S · A · T,

with S invertible (size m×m) and T invertible (size n×n). The transcript emphasizes that every invertible matrix can serve as a change-of-basis matrix, meaning equivalence is exactly the set of matrices reachable through these invertible left and right multiplications.

This motivates the definition of equivalent matrices without mentioning the linear map at all: two matrices of the same shape are equivalent if they can be connected by invertible matrices via A~ = S A T. The notation often uses a tilde symbol (∼) to denote this relation.

Finally, the equivalence relation properties are checked to justify the notation. Reflexivity holds because choosing identity matrices leaves A unchanged. Symmetry holds because if A~ = S A T with invertible S and T, then multiplying by their inverses reverses the relationship. Transitivity holds because composing two such transformations corresponds to multiplying the invertible matrices together. With an equivalence relation in place, matrices split into equivalence classes, and matrices in different classes cannot represent the same linear map. The next step—teased for a later installment—is how to characterize these equivalence classes more concretely.