14.19.1 Tensor Map Product Basis Change Response

Understanding how tensor maps respond to basis changes in product spaces through algebraic transformations.

Tensor Map Product Basis Change Response is the specific rule describing how the matrix representing $f \otimes g$ transforms when independent changes of basis are applied to each of the four spaces $V$ , $V^{'}$ , $W$ , and $W^{'}$ , showing that the response factors cleanly into the separate responses of the matrices of $f$ and $g$ to their own individual changes of basis.

Setting Up Four Independent Changes of Basis

Change of Basis Matrices for Each Space

Let $P$ and $P^{'}$ be the invertible change of basis matrices for $V$ and $V^{'}$ respectively, and let $Q$ and $Q^{'}$ be the invertible change of basis matrices for $W$ and $W^{'}$ . Under these four independent changes, the matrix of $f$ responds as

\tilde{F} = P^{'} F P^{- 1}

and the matrix of $g$ responds as

\tilde{G} = Q^{'} G Q^{- 1}

with the two changes on the domain side, $P$ and $Q$ , and the two on the codomain side, $P^{'}$ and $Q^{'}$ , all independent of one another.

The Combined Response

General Formula

Combining the four independent changes, the matrix of $f \otimes g$ responds as

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

which is the fully general version of the response rule, allowing the domain-side change $P \otimes Q$ and the codomain-side change $P^{'} \otimes Q^{'}$ to differ from one another, in contrast with a pure conjugation where domain and codomain changes coincide.

Derivation From the Two Separate Responses

This formula follows from substituting the individual responses of $\tilde{F}$ and $\tilde{G}$ into their Kronecker product and applying the mixed-product identity $(A \otimes B) (C \otimes D) = (A C) \otimes (B D)$ twice, once to combine the two left-multiplications and once to combine the two right-multiplications by inverses, confirming that the response of the tensor product's matrix is entirely determined by, and factors through, the two individual responses.

Special Cases of the Response

Only the Domain Bases Change

If the codomain bases are left fixed, so that $P^{'}$ and $Q^{'}$ are both identity matrices, the response reduces to

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

a pure right multiplication, matching the ordinary response of a matrix representing a linear map when only its domain basis is changed.

Only One Factor's Basis Changes

If only the basis of $V$ and $V^{'}$ is changed, with $Q$ and $Q^{'}$ fixed at the identity, the response becomes

\tilde{F} \otimes G = (P^{'} \otimes I) (F \otimes G) {(P \otimes I)}^{- 1}

leaving the block corresponding to $G$ untouched in structure while the outer blocks are permuted and recombined according to the change of basis matrix acting on the $V$ -side alone.

Similarity Transformations Coincide on Each Factor

If $V$ equals $V^{'}$ and $W$ equals $W^{'}$ , with the same change of basis matrix used for domain and codomain on each factor, so that $P = P^{'}$ and $Q = Q^{'}$ , the response becomes an ordinary similarity transformation,

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

showing that eigenvalues of $f \otimes g$ , when $f$ and $g$ are endomorphisms, are preserved under this particular response, since similarity transformations never alter the eigenvalues of a matrix.

Consistency Check Against Direct Computation

Verifying the Response on a Basis Tensor

Tracking a single basis tensor $e_{i} \otimes f_{j}$ through the change of basis and applying $f \otimes g$ either before or after re-expressing it in the new basis produces the same vector expressed in two coordinate systems, confirming that the basis change response formula is exactly the condition needed to keep the abstract map $f \otimes g$ consistent regardless of which of the four bases is used to record its action.