14.19.4 Tensor Map Product Kronecker Representation Change

Tensor Map Product Kronecker Representation Change explains how tensor products transform under linear maps using Kronecker products.

Tensor Map Product Kronecker Representation Change is the study of what happens to the Kronecker product matrix representing $f \otimes g$ when the basis of the combined space $V \otimes W$ is changed by a matrix that is not itself of Kronecker product form, revealing that the recognizable Kronecker structure of the representation is a special feature tied to bases compatible with the tensor decomposition, not a property preserved under arbitrary changes of basis on the combined space.

Two Different Kinds of Basis Change

Changes Compatible With the Tensor Decomposition

A change of basis on $V \otimes W$ given by a matrix of the form $P \otimes Q$ , for invertible $P$ acting on the $V$ -side and invertible $Q$ acting on the $W$ -side, respects the decomposition of $V \otimes W$ as a tensor product, since the new basis vectors are still simple tensors of new basis vectors of $V$ and $W$ individually.

Changes Not Compatible With the Tensor Decomposition

An arbitrary invertible matrix $S$ on the $n m$ -dimensional space $V \otimes W$ need not have this form; generic invertible matrices of size $n m$ are not expressible as $P \otimes Q$ for any pair of smaller matrices, since the set of Kronecker products has strictly lower dimension, as a subvariety, than the full space of invertible matrices once $n$ and $m$ are both greater than one.

Behavior Under Compatible Changes

Kronecker Structure Is Preserved

When the basis change is $P \otimes Q$ on the domain and $P^{'} \otimes Q^{'}$ on the codomain, the transformed matrix

(P^{'} \otimes Q^{'}) (F \otimes G) {(P \otimes Q)}^{- 1}

remains a Kronecker product, equal to $(P^{'} F P^{- 1}) \otimes (Q^{'} G Q^{- 1})$ , by the mixed-product property of the Kronecker product.

Explicit Verification of the Mixed-Product Property

This preservation rests on the identity

(A \otimes B) (C \otimes D) = (A C) \otimes (B D)

applied twice, first to combine the two left factors and then to combine the two right factors of the conjugation, and this identity holds for any conformable matrices, not only invertible ones.

Behavior Under Incompatible Changes

General Similarity Destroys Kronecker Form

If $S$ is an invertible matrix of size $n m$ that is not a Kronecker product, the conjugated matrix

S (F \otimes G) S^{- 1}

is still similar to $F \otimes G$ , and therefore still represents the same linear map $f \otimes g$ in a different basis of $V \otimes W$ , but it need not itself be expressible as a Kronecker product of two smaller matrices, since the new basis vectors of $V \otimes W$ produced by $S$ need not be simple tensors at all.

What Is Preserved Regardless

Even when the Kronecker form is lost, invariants of the underlying linear map that do not depend on any particular matrix representation, such as its rank, its trace, its determinant, and its eigenvalues, remain exactly the same, since these are similarity invariants; only the visibly factored block structure of the matrix, not the map itself, is sensitive to whether the change of basis respects the tensor decomposition.

Illustrative Contrast

A Basis of Entangled Vectors

Replacing the basis input elements ${e_{1} \otimes f_{1}, e_{1} \otimes f_{2}, e_{2} \otimes f_{1}, e_{2} \otimes f_{2}}$ of a four-dimensional $V \otimes W$ with a basis containing a vector such as $e_{1} \otimes f_{1} + e_{2} \otimes f_{2}$ , which is not a simple tensor, produces a change of basis matrix that is not of Kronecker form; the matrix of $f \otimes g$ relative to this new basis is a perfectly ordinary matrix representing the same map, but no block decomposition of it into a Kronecker product of two two-by-two matrices exists in general.

Recovering the Kronecker Form Requires Reverting

The recognizable block structure can only be recovered by returning to a basis built from simple tensors of individually chosen bases of $V$ and $W$ , confirming that the Kronecker representation is a convenience tied to a specific, tensor-compatible choice of basis rather than an intrinsic feature of the map $f \otimes g$ itself.