Prove that $d$ is a metric in $M_n(\mathbb{R})$ Let $M_n\mathbb{R}$ the vector space of the $n x n$ matrices with real coefficients. For $A=(a_{ij}) , B= (b_{ij}) \in M_n(\mathbb{R})$ define $d(A,B)= \displaystyle\sum_{i,j=1}^n{} | a_{ij} - b_{ij} |  $.
Prove that $d$ is a metric in $M_n(\mathbb{R})$. Recall that a matrix $N \in M_n(\mathbb{R})$ is called nilpotent if there is $k \in \mathbb{N}$ such that $N^k= 0$. Prove that the set of nilpotent matrices in $M_n(\mathbb{R})$ is a closed set in $M_n(\mathbb{R})$
I need a hint to start proving the axioms
Thanks
 A: Hint 1: For the first question, the fact that the entries are arranged in a square array makes no difference. Think $\mathbb{R}^{n^2}.$
Hint 2: For the second question, first fix $k=k_0.$ Is it clear that the set of $N$ such that $N^{k_0} = 0$ is closed. Now, show that only a finite number of $k_0$ are needed.
A: To show that $d$ is a metric, you need to show that $d(A,B) \geq 0$ for all $n \times n$ matrices $A$ and $B$. Additionally, you must also show $d(A,B) = 0 \iff A = B$. You also need to show $d(A,B) \leq d(A,C ) + d(C, B)$ where $A,  B, C$ are any $n \times n$ matrices, and that $d(A,B) = d(B,A)$.
Showing $d(A,B) \geq 0$ shouldn't be too bad. Simply observe that the definition of $d(A,B)$ is a sum consisting of non-negative terms. What can you conclude about a sum of non-negative terms?
To show that $d(A,B) = d(B,A)$, it might help to know that the following property holds when working with absolute values $|a-b| = |b-a|$ since $|a-b| = |-(b-a)| = |-1||b-a| = |b-a|$.
To show the triangle inequality, it will help to know that $|a_{ij} - b_{ij}| \leq |a_{ij} - c_{ij}| + |c_{ij} - b_{ij}|$.
