Why do the determinants of the types $\det|f(i)+g(j)|$ and $\det|f(i)g(j)|$ vanish?

Question

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?!

Answer 1

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.

Answer 2

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.

Answer 3

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 $n \ge 2$.