# 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

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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