Matrix Multipication and Inner Product

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:

$$(NA^*)^*NA^*=trace\left((NA^*)^*NA^*\right)$$

• No, the last step is illegal. The trace of a matrix is a scalar, while the product on the LHS is an $n\times n$ matrix. Equating them makes no sense. Further, $\text{trace}(A) = 0 \nRightarrow A = 0_{n\times n}$, take $A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. – stochasticboy321 Nov 3 '15 at 16:01
• $M=0 \Rightarrow \,trace M=0$ but not the inverse. In you case : $(NA^*)^*NA^*=0 \Rightarrow \,trace[(NA^*)^*NA^*]=0$ is correct. And, since this is the square of the Frobenius norm of the matrix $NA^*$, this matrix is $0$. – Emilio Novati Nov 3 '15 at 16:08