Let $V_1,V_2,W_1,W_2$ be all finite dimensional and $T_1\in \mathcal L(V_1,W_1), T_2\in \mathcal L(V_2,W_2).$ Show that $(T_1\otimes T_2)^T = T_1^T\otimes T_2^T$.
I'm having a hard time trying to prove this. I'll post here how far I could reach:
Recall that given $T:V\rightarrow W$ linear, its transpose is by definition the following function: $T^T: W^*\rightarrow V^*, f\mapsto f\circ T.$
Since $T_1\otimes T_2: V_1\otimes V_2 \rightarrow W_1\otimes W_2$, then:
$$ (T_1\otimes T_2)^T: (W_1\otimes W_2)^* \rightarrow (V_1\otimes V_2)^* $$
$$ \mbox{ is given by } (T_1\otimes T_2)(f) = f\circ(T_1\otimes T_2) \mbox{ for every $f\in (W_1\otimes W_2)^*$}.$$
Also, we have by definition:
$$ T_1^T \otimes T_2^T:W_1^*\otimes W_2^* \rightarrow V_1^*\otimes V_2^*$$
$$ \mbox{and satisfies } T_1^T\otimes T_2^T(f_1\otimes f_2) = T_1^T(f_1)\otimes T_2^T(f_2) = (f_1\circ T_1)\otimes (f_2\circ T_2)$$
for every $f_1\in W_1*, f_2\in W_2^*.$
I can't see how those two linear functions could be equal. I suppose that I must use somehow the isomorphism : $(V_1\otimes V_2)^* \cong V_1^* \otimes V_2^*$ (and also for $W_i$) but I dunno exactly how. In an earlier exercise we have constructed such a isomorphism by the following function:
$$\Gamma: V_1^*\otimes V_2^* \rightarrow (V_1\otimes V_2)^*, \mbox{given by }\Gamma(f_1\otimes f_2)(v_1\otimes v_2) = f_1(v_1)f_2(v_2). $$
Any help would be much appreciated.
Best Answer
Following your notation, you have two isomorphisms $\Gamma_V:V_1^*\otimes V_2^*\to (V_1\otimes V_2)^*$, and $\Gamma_W$ with similar domain and range. You have $$\begin{align}T_1^T\otimes T_2^T&:W_1^*\otimes W_2^*\to V_1^*\otimes V_2^* \\ (T_1\otimes T_2)^T&: (W_1\otimes W_2)^*\to (V_1\otimes V_2)^*. \end{align} $$So writing $(T_1\otimes T_2)^T=T_1^T\otimes T_2^T$ is an abuse of notation for saying that these maps are equal modulo isomorphisms. What you can prove is that $$\Gamma_V\circ (T_1^T\otimes T_2^T) = (T_1\otimes T_2)^T\circ\Gamma_W. $$