Is the alternatization of a m-form on a m-dimensional vector space non-zero if there exists a basis on which the m-form is non-zero? Let $V$ be a $m$-dimensional real vector space, and let $e_1,\ldots, e_m$ be a basis such that the $m$-form $f$ has the property $$f(e_1,\ldots,e_m)\neq0$$Is $\operatorname{Alt}(f)\neq0$? Or does there exist an example of $f$ with the properties above such that $\operatorname{Alt}(f)=0$?
I ran into this problem as a subproblem of a larger proof I'm writing up, so I'm hoping that the statement is true; otherwise, I'd have to restart.

Taking $m=2$, the bilinear form
$$f(x, y) = x^TAy$$
where $A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, and $x,y \in \Bbb R^2$ gives a counterexample.
