Closed formula for $\sum_{i=1}^n \det(v_1, \ldots, Av_i, \ldots, v_n)$

Question

Denote $(v_1, \ldots, v_n)$ the matrix that has columns $v_1,\ldots, v_n\in \mathbb{R}^n$. Let $A\in \mathcal{M}_{n\times n}(\mathbb{R})$. Is there a clever way (without expanding LHS and doing calculations) to show that $$\sum_{i=1}^n \det(v_1, \ldots, Av_i, \ldots, v_n)=\text{tr}(A)\det(v_1, \ldots, v_n)\ ?$$

Answer 1

I am not sure that this is what you want, but it is the cleanest solution I can find.

It is easy to check that, if all but one of the $v_j$ are fixed and both the function is viewed as a function of a single vector, it will be a linear function. Additionally, if two of the $v_j$s are the same, say $v_k=v_{\ell}$ the left hand side will be zero. To see this, note that in the sum, there will be two types of summands, those where $i\in \{k,\ell\}$, and those where $i\not\in \{k,\ell\}$. The two terms of the first type will cancel each other out because one is obtained by swapping two of the columns in the other, and the rest of the terms will be zero because determinants are zero when two columns are equal.

It is a theorem that every multi-linear and alternating function is a multiple of the determinant, so the left hand side is equal to $c\det(v_1,\ldots, v_n)$ for some constant $c$. By setting $v_i=e_i$, the $i$th basis vector, we can calculate that $c=\operatorname{tr} A$.