We have matrix $A\in M_{n-1\times n}(\mathbb Z)$ so that the sum of entries in each row is zero. Prove that $\det(AA^T)=nk^2.$ Problem: We have matrix $A\in M_{n-1\times n}(\mathbb Z)$ so that the sum of elements in each row is zero. Prove that $\det(AA^T)=nk^2$, where $k\in \mathbb Z$.
What have I considered so far:


*

*First I thought, since sum of all elements in each row is zero, zero is eigenvalue of $A$ but $A\in M_{(n-1)\times n}(\mathbb Z)$ confused me.

*I see that $AA^T$ will be $(n-1)$ by $(n-1)$, so I tried calculating $AA^T$ but I failed to see any connection.


I did some research and found this question but we had no mention of it, which makes me believe I am on the wrong track.
Thank you all for your help.
 A: Let $A = (B,\mathbf{b})$, where $B \in M_{n-1\times n-1}(\mathbb{Z})$ is a "left square part" of matrix $A$, i.e. $B_i^j = A_i^j$, and $\mathbf{b} \in \mathbb{Z}^{n-1}$ is a "right" part of $A$, i.e. $b_i = A_i^n$. As sum of each row is zero, we have $b_i = -\sum_{k=1}^{n-1}B_i^k$. Let $\mathbf{e}\in \mathbb{Z}^{n-1}$ be a vector with each union component, i.e. $e_i = 1$. Then $\mathbf{b} = -B\mathbf{e}$. Now we may multiply $A$ with $A^{\top}$ as block matrices:
$$
AA^{\top} = 
(B\;\;-B\mathbf{e})
\begin{pmatrix}
B^\top \\
-\mathbf{e}^\top B^\top
\end{pmatrix} = BB^\top + B\mathbf{e}\mathbf{e}^\top B^\top = B(I + \mathbf{e}\mathbf{e}^\top)B^\top
$$
where $I$ is identity matrix. As bouth $B$ and $I + \mathbf{e}\mathbf{e}^\top$ are square matrices we now have
$$
\det(AA^\top) = \det\left(B(I + \mathbf{e}\mathbf{e}^\top)B^\top\right) = \det (B) \det(I + \mathbf{e}\mathbf{e}^\top)\det(B^\top) = \det(I + \mathbf{e}\mathbf{e}^\top)\left(\det B\right)^2.
$$
$B \in M_{n-1\times n-1}(\mathbb{Z})$ thus $\det B \in \mathbb{Z}$.
Now all we need to proof is that $\det(I + \mathbf{e}\mathbf{e}^\top) = n$. This matrix (denote it with $E_{n-1}$ where $n-1$ is dimension of the space or count of raws in $E$) looks like
$$
E_{n-1} = \begin{pmatrix}
2 & 1 & \dots & 1 \\
1 & 2 & \dots & 1\\
\vdots & \vdots & \ddots & \vdots \\
1 & 1 & \dots & 2
\end{pmatrix}.
$$
Let $E^i_n$ be a matrix $E_n$ in which we replaced $2$ in $i$-th row with $1$. It's easy to see that $\det{E^1_n} = 1$ (using Gaussian process). Thus if $i$ is even $\det E_n^i = 1$ and if i is odd $\det E^i_n = -1$, i.e. $\det E^i_n = (-1)^i$. Now let's suppose we know that $\det E_{n-1} = n$. We may decompose $\det E_n$ using Laplace expansion:
$$
\det E_n = 2\det E_{n-1} + \sum_{i=1}^{n-1} (-1)^{i}\det E^i_{n-1} = 2n - (n-1)  = n+1.
$$
QED.
A: Rather than partitioning $A$ as in the other answer, we may let a row vector and $A$ adjoin instead. Let $e=(1,\ldots,1)^T\in\mathbb R^n$. Then
\begin{align}
&\pmatrix{e^T\\ A}\left(\begin{array}{c|c}e&\begin{array}{c}0\\ \hline I_{n-1}\end{array}\end{array}\right)=\pmatrix{n&\ast\\ 0&\ast},\tag{1}\\
&\pmatrix{e^T\\ A}\pmatrix{e&A^T}=\pmatrix{n&0\\ 0&AA^T},\tag{2}
\end{align}
where the RHS of $(1)$ is an integer matrix. It follows from $(1)$ that $\det\pmatrix{e^T\\ A}=nk$ for some integer $k$, and from $(2)$ that $(nk)^2=n\det(AA^T)$. Thus $\det(AA^T)=nk^2$.
