Conditions for the value of a determinant to be zero

Question

The theory states that the value of a determinant will be zero if it contains a row or column full of zeroes or if it has two identical rows or two rows proportional to each other.

Similarly, can we say that the value of the determinant is zero only if it satisfies the above mentioned conditions?

Could someone please explain me this?

Answer 1

At first, some definitions:

A function $f\colon X^n\to\mathbb{R}$ is multilinear when is linear respect all their variables.

and other

A function $f\colon X^n\to\mathbb{R}$ is alternating when: $$ f(x_1,\ldots,x_i,\ldots,x_j,\ldots,x_n)=-f(x_1,\ldots,x_j,\ldots,x_i,\ldots,x_n) $$

At least

A determinant function $f\colon X^n\to\mathbb{R}$ is a multilinear alternating function.

So if we have $x_i=x_k$ for some $i,k\leq n$:

$$ f(x_1,\ldots,x_i,\ldots,x_k,\ldots,x_n)=-f(x_1,\ldots,x_k,\ldots,x_i,\ldots,x_n) $$ $$ =-f(x_1,\ldots,x_i,\ldots,x_k,\ldots,x_n) \mbox{ since }x_i=x_k$$ so $$f(x)=-f(x) \Leftrightarrow f(x)=0, \mbox{ where } x=(x_1,\ldots,x_n)$$ So, If $x$ has two equal elements, $f(x)=0$.

If $x$ has some $0$: $$f(x_1,\ldots,0,\ldots,x_n)=f(x_1,\ldots,0+0,\ldots,x_n)$$ $$f(x_1,\ldots,0,\ldots,x_n)=f(x_1,\ldots,0,\ldots,x_n)+f(x_1,\ldots,0,\ldots,x_n)\mbox{ since } f \mbox{ is linear }$$ $$f(x)=f(x)+f(x)$$ $$f(x)=0$$