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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.

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  • $\begingroup$ 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 $A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. $\endgroup$ – stochasticboy321 Nov 3 '15 at 16:01
  • $\begingroup$ I see. Thank you. So How can the solution involve a trace in the middle? $\endgroup$ – Alan Nov 3 '15 at 16:02
  • $\begingroup$ The solution you'ce written in the first paragraph under 'solution' is pretty complete, and doesn't contain the trace. Do you perhaps have an alternate solution involving the trace, or do you just want to shove a trace in there if you could? If the former, you should add it to the question. $\endgroup$ – stochasticboy321 Nov 3 '15 at 16:05
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    $\begingroup$ $M=0 \Rightarrow \,trace M=0$ 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 $0$. $\endgroup$ – Emilio Novati Nov 3 '15 at 16:08

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