I mean, the question is simple, asking me to show $(A\otimes B)^* = A^*\otimes B^*$, where * is just the hermitian adjoints. However, it's hard for me to use the definition, aka put them into (x,y) just like the $+$ case, in fact, I am not even sure about the dimension since the kronecker product will be $n^2$ if the original matrices are square with dim $n$.
Thanks for the help.
$(A \otimes B)^*$ is the unique linear operator satisfying $$ \langle (A \otimes B)^* y, x\rangle = \langle y , (A\otimes B) x\rangle, $$ for all $x, y \in H\otimes K$.
Now take any two product vectors in $v_1 \otimes w_1, v_2 \otimes w_2 \in H \otimes K$. Then for a product vector we find that $$ \begin{aligned} \langle v_1 \otimes w_1, (A\otimes B)(v_2\otimes w_2)\rangle &= \langle v_1 \otimes w_1, (Av_2) \otimes (Bw_2)\rangle \\ &= \langle v_1, A v_2\rangle \langle w_1, B w_2\rangle \\ &= \langle A^* v_1,v_2\rangle \langle B^* w_1,w_2\rangle \\ &= \langle (A^* \otimes B^*)(v_1 \otimes w_1), v_2 \otimes w_2\rangle. \end{aligned} $$
So we see that for any product vector the correct identity holds. Then by linearity of the inner product and the fact that we can write any vector as a linear combination of product vectors, we can extend this to all vectors in $H \otimes K$.