Metric in complex matrix

Hi everyone hope you can hep me. Let $A,B\in M_{m\times n}(\mathbb{C})$ two matrix with complex inputs and let's define a function $d:M_{m\times n}(\mathbb{C})\times M_{m\times n}(\mathbb{C})\rightarrow \mathbb{R}$ as $d(A,B)=\{\sum_{k=1}^{m}\sum_{j=1}^{n}|a_{kj}-b_{kj}|^{2} \}^{\frac{1}{2}}$. I was trying to prove that $d$ is a metric in $M_{m\times n}(\mathbb{C})$, at the beginning I thought that I could use Minkowski inequality for normed spaces ($|| x+y ||_{p}\leq ||x||_{p}+||y||_{p}$) but then I realized that that wouldn't work, What can I use? thank you all.

Solution 1:

First $d(A,B) \geq 0$ clearly follows by your definition and $d(A,B)=0$ if and only if $A=B$. The symmetricity property is also easy to prove.

By Minkowski inequality we know that $$ (\sum\limits_{k=1}^m \sum\limits_{j=1}^n |a_{kj}-b_{kj}|^2)^{1/2} + (\sum\limits_{k=1}^m \sum\limits_{j=1}^n |b_{kj}-c_{kj}|^2)^{1/2} \geq (\sum\limits_{k=1}^m \sum\limits_{j=1}^n |a_{kj}-c_{kj}|^2)^{1/2}$$ where $a_{ij}, b_{ij}, c_{ij}$ can be complex numbers, and thus proved. If you are only allowed to use the theorem which only allows real entries, note that $|z_1-z_2|^2 = |re(z_1)-re(z_2)|^2+|im(z_1)-im(z_2)|^2$; hence the complex result is directly proved.