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I am trying to do the following exercise:

Let $m\in \mathbb N \backslash\{1,2\}$ and $A=(a_{ij})\in M_n(\mathbb Z)$ with $a_{ii} = 0$ for all $1\leq i\leq n$ and $a_{ij} = 1$ or $a_{ij} = m$ for all $1\leq i\neq j \leq n$. Show that $\text{rk } A = n -1$ or $\text{rk } A = n$.

Proof attempt:
Define $C:= A - B$, where $B\in M_n(\mathbb Z)$ with $b_{ij} = 1$ for all $1\leq i,j \leq n$. We then have $\det(C \mod (m-1)) = (-1)^n \neq 0 \implies \det C \neq 0$, hence $\text{rk } C = n$. I feel like I am close but don't know what's the final step. I could use $\text{rk } A = \text{rk } (C + B) \leq \text{rk }C + \text{rk }B = n$ but that is obvious in the first place. I too know that $\text{rk }A = n \Leftrightarrow A$ invertible $\Leftrightarrow \det A \in \{-1,1\}$ since $A$ has only entries from $\mathbb Z$.

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Here's one way: $$ n = \text{rk }(C) = \text{rk } (A + (-B)) \leq \text{rk}A + \text{rk }B = \text{rk}(A) + 1 $$ Thus, conclude $\text{rk}(A) \geq n-1$.

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    Note: it is **not** true that $\text{rk }A = n \iff \det(A) \in \{\pm 1\}$. – 2017-01-26
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    It is, but only for $A\in GL_n(\mathbb Z)$. It is obvious for me now, I made the mistake of thinking $\text{rk } B = 0$. – 2017-01-26
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    https://en.wikipedia.org/wiki/Unimodular_matrix – 2017-01-26
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    @lappen68 a matrix $A$ can have integer entries and rank $n$ without being an element of $GL_n(\Bbb Z)$. The strong connection between rank and invertibility is lost when we take matrix entries from a ring rather than a field. – 2017-01-26
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    That is good to know, big mistake of mine. Thanks! – 2017-01-26
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    However it is still true that $\det A \neq 0 \Longleftrightarrow A$ invertible when we take entries from $\mathbb Z$, is it? – 2017-01-26
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    That depends what you mean by "invertible". If you mean "$A$ has an inverse with rational/real entries", then yes. If you mean "$A$ has an inverse with integer entries", then no. – 2017-01-26
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    I see, only the direction $A$ has an Inverse with arbitrary entries $\implies \det A = 0$ always holds, right? Thanks a lot btw! – 2017-01-26
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    If $A$ has an inverse of any kind, then $\det(A) \neq 0$. – 2017-01-26