I am wondering about the difference between matrix multiplication and inner product.
It is regarding the following question:
Let $N \in M^{\Bbb C}_{n \times n}$ be a normal matrix. Prove that if for the matrix $A \in M^{\Bbb C}_{n \times n}$ $A*N=0$, then$ A*N^*=0$.Then sign * stands for the Conjugate transpose.
I know that be definition, the inner product of $(A,B)$ is $(A,B)=tr(B^*A)$. I am wondering where can I use this fact in the question.
Solution
$A*N=0 \Rightarrow (A*N)^*=N^*A^*=0$ therefore, since $AN=0$, $ANN^*A^*=0$. Since $N$ is normal, We can replace $NN^*$ be $N^*N$, and get $AN^*NA=0$ and then $(NA^*)^*NA^*=0$.
Now is the part I don't understand. In the solution I have the answer involves the $trace$. How can I insert a $trace$ here if an inner product is not involved? Or inner product is always involved in matrix multiplication?
$(NA^*)^*NA^*=0 \Rightarrow trace(NA^*)^*NA^*=0 $ and therefore $(NA^*,NA^*)=0$ where $(,)$ implies inner product.
Is it legal to do that:
$$(NA^*)^*NA^*=trace\left((NA^*)^*NA^*\right)$$
Thanks for your help.