Matrices equipped with the rank distance form a metric space

Solution 1:

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.