Prove that $text{tr} (phi otimes psi) = text {tr} phi text {tr} psi $.

Let $E,F$ be two vector spaces of dimension $n$ and $m$ respectively and let $phi : E rightarrow E$,
$psi : F rightarrow F $ be two linear transformations. Prove that $text{tr} (phi otimes psi) = text {tr} phi text {tr} psi $ and $text {dt}(phi otimes psi) = (text {dt} phi)^m (text{det} psi)^n$.

Looking at one of my old books I found this exercise and I found it interesting because it involves one of the fundamental concepts of linear algebra: trace and determinant.

I have been trying to do the first equality by direct method, however I have not had a result yet. I am clear that $ text{tr} (AB) = text{tr} (BA) $, and $ text{im} (phi otimes psi) = text{im} phi otimes text{im} psi $, but could this help me?
I am in need of help with this exercise please. I am not very familiar with the subject.

Mathematics Asked on November 21, 2021

2 Answers

2 Answers

One approach to compute the trace is as follows:

Suppose we have bases ${e_1,dots,e_n},{f_1,dots,f_m}$ and associated dual bases ${alpha_1,dots,alpha_n},{beta_1,dots,beta_m}$ respectively. It follows that the sets ${e_i otimes f_j}$ and ${alpha_i otimes beta_j}$ form a basis and associated dual basis for $E otimes F$. It follows that $$ begin{align} operatorname{tr}(phi otimes psi) &= sum_{i=1}^n sum_{j=1}^m (alpha_i otimes beta_j)(phi otimes psi)(e_i otimes beta_j) \&= sum_{i=1}^n sum_{j=1}^m (alpha_i otimes beta_j)(phi(e_i) otimes psi(f_j)) \&= sum_{i=1}^n sum_{j=1}^m alpha_i(phi(e_i)) beta_j(psi(f_j)) \ & = left(sum_{i=1}^n alpha_i(phi(e_i)) right) left(sum_{j=1}^m beta_j(psi(f_j)) right) = operatorname{tr}(phi)operatorname{tr}(psi). end{align} $$

For the determinant, use the fact that for maps $Phi,Psi$ over $E otimes F$, $det(Phi_1 circ Phi_2) = det(Phi_1) det(Phi_2)$. Now, define $Phi = phi otimes operatorname{id}_F$, so that $$ Phi(x otimes y) = phi(x) otimes y. $$ Similarly, define $Psi = operatorname{id}_E otimes psi$. It suffices to show that $det(Phi) = det(phi)^m$, and $det(Psi) = det(psi)^n$.

To show that $det(Phi) = det(phi)^m$, note that the spaces $V_i = {x otimes beta_i : x in E}$ are invariant subspaces of $Phi$. So, we can write $Phi$ as a direct sum of maps $$ Phi = overbrace{phi oplus cdots oplus phi}^m. $$ It follows that $detPhi = det(psi)^m$, which was what we wanted. The proof for $det Psi$ is similar. The conclusion follows.

Answered by Ben Grossmann on November 21, 2021

Writing $operatorname{tr}(phiotimes psi)$ is technically not 100% correct. By definition, the trace on a K-vector space $V$ is the unique linear map $operatorname{tr}_Vcolon (Votimes V^*) to K$ which sends $(votimes f)to f(v)$, i.e. by extending through linearity $operatorname{tr}_Vbig(sum_i gamma_i (v_iotimes f_i)big) = sum_igamma_if_i(v_i)$. Or, if you like bra-ket notation we have

$$ operatorname{tr}_V(|vranglelangle f|) = langle f| vrangle $$

However $phiotimes psi in (Eotimes E^*) otimes (Fotimes F^*)$ which is not of the form $Votimes V^*$, but the space is isomorphic to $(Eotimes F) otimes (Eotimes F)^*$. By definition, the canonical/induced inner product on the tensor product is:

$$ leftlanglephi_{1} otimes phi_{2}, psi_{1} otimes psi_{2}rightrangle_{V_1otimes V_2}=leftlanglephi_{1}, psi_{1}rightrangle_{V_1}cdot leftlanglephi_{2}, psi_{2}rightrangle_{V_2} $$

And then immediately we have

$$ operatorname{tr}_{Eotimes F}(|uotimes vrangle langle fotimes g|) overset{text{def}}{=} langle fotimes g|uotimes vrangle_{Eotimes F} overset{text{def}}{=} langle f|urangle_Elangle g|v rangle_F overset{text{def}}{=} operatorname{tr}_E(|uranglelangle f|)operatorname{tr}_F(| v ranglelangle g|) $$

So this property is pretty much equivalent to the definition of the inner product on the tensor product space. The determinant follows from the mixed product property:

$$ Aotimes B = (Aotimes I_m)cdot (I_n otimes B) $$

Answered by Hyperplane on November 21, 2021

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