Matrices equipped with the rank distance form a metric space Right now, I'm taking a basic introductory analysis course.   I came across this on wikipedia and I'm Really struggling to prove it is a metric space. (Need help with the 3rd axiom of metric space the most and making the 2nd one more rigorous/less handwaving)  

The set of all $m$ by $b$ matrices over some field is a metric space with respect to the rank distance $d(X,Y)=\operatorname{rank}(Y-X) $

Facts that I know:
So I know that in order to prove something is a metric space


*

*$d(X,Y)=0$ iff $X=Y$   (have to prove both ways)

*$d(X,Y)=d(Y,X)$  (Symmetry)

*$d(X,Y)\leqslant d(X,Z) +d(Z,Y)$ (Triangle inequality).


Also, it has been a long time since I have taken linear algebra", but I now that to find the rank, you reduce a matrix to its row echelon form and the number of nonzero rows is the rank (or I think number of linearly independent rows, again it has been awhile).
My attempt:


*

*If $Y=X$, We have $\operatorname{rank}(Y-X)=\operatorname{rank}(Y-Y)=\operatorname{rank}(O)$ (where $O$ is the null matrix) which is clearly equal to $0$.  Thus $d(X,Y)=0$.  Proving other direction if $d(X,Y)=0$, we have $\operatorname{rank}(Y-X)=0$.  In this case $Y$ must necessarily equal $X$.  If $Y$ does not equal $X$, rank must be at least $1$.  

*Why must $d(X,Y)=d(Y,X)$?  The matrix $X-Y$ and $Y-X$ only differ by signs.  That doesn't change rank.  Hence $\operatorname{rank}(Y-X)=\operatorname{rank}(X-Y)$.  (feel like I hand waved this explanation a bit).

*Why must $d(X,Y)\leqslant d(X,Z)+ d(Z,Y)$? Need to show: $$\operatorname{rank}(Y-X)\leqslant\operatorname{rank}(Z-X)+ \operatorname{rank}(Y-Z)$$where $X,Y,Z$ are $m \times n$ matrices.
This one I'm not sure at all how to approach this.  Why must $\operatorname{rank}(Y-X)\leqslant \operatorname{rank}(Z-X)+\operatorname{rank}(Y-Z)$?  I'm not sure why this must be the case and how to formally show this.
 A: Your answers for the first and second points are correct. The second is not hand-wavy by any means, you have said the crucial thing, namely that $X-Y$ is a scalar multiple of $Y-X$, so their rank is the same.
As for the third one, if $Z - X = a$ and $Y - Z = b$, then you essentially have to prove that $\operatorname{rank}(a+b) \leq \operatorname{rank}(a) + \operatorname{rank}(b)$.
However, note that if some set of vectors span the columns of $a$, and another set of vectors span the columns of $b$, then the union of these vectors spans the columns of $a+b$. From here, I urge you to use the definition of rank to understand why this statement follows.
EDIT : Suppose that $\{e_i\}$ span the columns of $a$, and $\{f_j\}$ span the columns of $b$. My claim is that $\{e_i\} \cup \{f_j\}$ span the columns of $a+b$.
It's easy to see why. Suppose $A$ is a column of $a+b$, then we know that  $A= A_a + A_b$, where $A_a$ and  $A_b$ are the corresponding columns of $a$ and $b$ which were added to give the column $A$ of $a+b$.
Now, $A_a$ is spanned by the $\{e_i\}$, so there exist constants $c_i$ such that $A_a = \sum_{i} c_ie_i$. Simiarly, since $A_b$ is spanned by $\{f_j\}$, we get that $A_b = \sum_j d_jf_j$ for some constants $d_j$.
Adding these two, gives that $A = \sum_i c_ie_i + \sum_j {d_jf_j}$. Hence, you can see that $A$ is spanned by $\{e_i\} \cup \{f_j\}$. Hence, the rank of $A$ is less than the size of $\{e_i\} \cup \{f_j\}$, which is less than $\operatorname{rank} a + \operatorname{rank} b$. Hence, the inequality follows.
