Why do the determinants of the types $\det|f(i)+g(j)|$ and $\det|f(i)g(j)|$ vanish? I find it very amusing that $$\det|f(i)+g(j)|_{n\times n}=0, n \ne 2~~~~~~(1)$$ and
$$\det|f(i)g(j)|_{n\times n}=0,n \ne 2.~~~~~(2)$$ Here, $f(x)$ and $g(x)$
are arbitrary functions and $i,j\in [1,n].$
This appears to be the characteristics of determinants. Have you experienced this? is it something trivial and warranted?!
 A: With $a = (f(1), \ldots , f(n))^T$  and $b = (g(1), \ldots, g(n))^T$ the matrix
$$
 M_1 = (a_i b_j) = a b^T
$$
has zero as an eigenvalue if $n \ge 2$: $M_1 c = 0$ for every vector $c$ orthogonal to $b$. Therefore the determinant $(2)$ is zero.
Similarly, $(1)$ is the determinant of the matrix
$$
M_2 = a e^T + e b^T
$$
with $e = (1\ldots, 1)^T$. If $n\ge 3$ then there exists a vector $c$ orthogonal to both $e$ and $b$, so that $M_2 c = 0$.
Generally, a square matrix of the form $uv^T$ has the rank (at most) one, the sum of $k$ rank-one matrices has rank at most $k$, and $\det M = 0$ if the rank of $M$ is less than its dimension.
A: A solution which uses only the elementary properties of determinants:
We have
$$\begin{vmatrix}
f(1)+g(1) & f(1)+g(2)& \cdots & f(1)+g(n)\\
f(2)+g(1) & f(2)+g(2) & \cdots & f(2)+g(n)\\
\vdots & \vdots  & \ddots & \vdots \\
f(n)+g(1) & f(n)+g(2)  & \cdots & f(n)+g(n)\\
\end{vmatrix} = \begin{vmatrix} f + g(1)e & f + g(2)e & \cdots & f+g(n)e\end{vmatrix}$$
where $f = \begin{bmatrix} f(1) \\ f(2) \\ \vdots \\ f(n)\end{bmatrix}$ and $e = \begin{bmatrix} 1 \\ \vdots \\ 1\end{bmatrix}$.
Subtract the first column from the rest to obtain
$$\begin{vmatrix} f + g(1)e & (g(2)-g(1))e & \cdots & (g(n)-g(1))e\end{vmatrix} = 0$$
since the columns are linearly dependent for $n \ge 3$.
For the second one we have
$$\begin{vmatrix}
f(1)g(1) & f(1)g(2)& \cdots & f(1)g(n)\\
f(2)g(1) & f(2)g(2) & \cdots & f(2)g(n)\\
\vdots & \vdots  & \ddots & \vdots \\
f(n)g(1) & f(n)g(2)  & \cdots & f(n)g(n)\\
\end{vmatrix} = \begin{vmatrix} g(1)f & g(2)f & \cdots & g(n)f\end{vmatrix} = 0$$
since the columns are linearly dependent for $n \ge 2$.
A: Yes they are true, but it is equation $(1)$ that only works for $n\gt2$ rather than equation $(2)$.
There is something called rank, and other things called eigenvalues, that you can look up, but you don't get taught until university.
