Show that the determinant of the matrix $E$ is $0$ Let
$E=\begin{pmatrix}
(1-n)e_1 & e_2 & ... & e_n\\
e_1 & (1-n)e_2 & ... &e_n\\
. & . & . &.\\
. & . & . &.\\
e_1 & e_2 & ... & (1-n)e_n
\end{pmatrix}$
an $n \times n$ matrix. I need to prove that the determinant is zero. I thought about using guassian elimination, and somehow $e_{nn}$ (position) will lead to zero. So, as a triangular matrix with a zero in the diagonal means $Det(E)=0$, then it is solved.
However, I would like to know if there's a better way to prove it, or a hint, and if what I thought is correct.
 A: HINT: If you sum all the rows, which is a non-trivial linear combination of them, you get... And this would imply the determinant is zero because any non-trivial linear combination of the rows of an invertible (i.e. non-zero determinant) matrix cannot be this value.
A: First $(1)$, factor out the coefficients. Second $(2)$, add all rows to the first row:
$$\begin{vmatrix}
(1-n)e_1 & e_2 & ... & e_n\\
e_1 & (1-n)e_2 & ... &e_n\\
. & . & . &.\\
. & . & . &.\\
e_1 & e_2 & ... & (1-n)e_n
\end{vmatrix} $$$$ 
\overset 1=(e_1\cdots e_n)\begin{vmatrix}
(1-n) & 1 & ... & 1\\
1 & (1-n) & ... &1\\
. & . & . &.\\
. & . & . &.\\
1 & 1 & ... & (1-n)
\end{vmatrix} $$$$\overset 2 =(e_1\cdots e_n)\begin{vmatrix}
(1-n)+(n-1) & (1-n)+(n-1) & ... & (1-n)+(n-1)\\
1 & (1-n) & ... &1\\
. & . & . &.\\
. & . & . &.\\
1 & 1 & ... & (1-n)
\end{vmatrix} \\= (e_1\cdots e_n)\begin{vmatrix}
0 & 0 & ... & 0\\
1 & (1-n) & ... &1\\
. & . & . &.\\
. & . & . &.\\
1 & 1 & ... & (1-n)
\end{vmatrix}$$$$=0$$
A: Consider the vector $v = [1,1,\ldots, 1]^T$. You can calculate component wise that
$E^Tv = 0$ which implies that $E$ is singular so $\det(E^T) = 0$ and since $\det(E)=\det(E^T)$ we have that
$$ \det(E)=\det(E^T) = 0$$
as desired.
