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

Question

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.

Answer 1

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 $$ \begin{align} \operatorname{tr}(\phi \otimes \psi) &= \sum_{i=1}^n \sum_{j=1}^m (\alpha_i \otimes \beta_j)(\phi \otimes \psi)(e_i \otimes \beta_j) \\&= \sum_{i=1}^n \sum_{j=1}^m (\alpha_i \otimes \beta_j)(\phi(e_i) \otimes \psi(f_j)) \\&= \sum_{i=1}^n \sum_{j=1}^m \alpha_i(\phi(e_i)) \beta_j(\psi(f_j)) \\ & = \left(\sum_{i=1}^n \alpha_i(\phi(e_i)) \right) \left(\sum_{j=1}^m \beta_j(\psi(f_j)) \right) = \operatorname{tr}(\phi)\operatorname{tr}(\psi). \end{align} $$

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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 $\det\Phi = \det(\psi)^m$, which was what we wanted. The proof for $\det \Psi$ is similar. The conclusion follows.

Answer 2

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

$$ \operatorname{tr}_V(|v\rangle\langle f|) = \langle f| v\rangle $$

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

$$ \left\langle\phi_{1} \otimes \phi_{2}, \psi_{1} \otimes \psi_{2}\right\rangle_{V_1\otimes V_2}=\left\langle\phi_{1}, \psi_{1}\right\rangle_{V_1}\cdot \left\langle\phi_{2}, \psi_{2}\right\rangle_{V_2} $$

And then immediately we have

$$ \operatorname{tr}_{E\otimes F}(|u\otimes v\rangle \langle f\otimes g|) \overset{\text{def}}{=} \langle f\otimes g|u\otimes v\rangle_{E\otimes F} \overset{\text{def}}{=} \langle f|u\rangle_E\langle g|v \rangle_F \overset{\text{def}}{=} \operatorname{tr}_E(|u\rangle\langle f|)\operatorname{tr}_F(| v \rangle\langle g|) $$

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:

$$ A\otimes B = (A\otimes I_m)\cdot (I_n \otimes B) $$