Proving additivity of determinant in each argument

Question

I'm reading through Treil's Linear Algebra Done Wrong and have a question about proving a remark he leaves as an exercise on p.77. Here it is in my own words:

Consider a function on $n$ vectors $D(v_1,v_2,\ldots,v_n)$. It has the following properties.

1. $D(v_1,\ldots,\alpha v_k,\ldots,v_n) = \alpha D(v_1,\ldots,v_k,\ldots,v_n)$, for any $\alpha \in \mathbb{F}$, and
2. $D(v_1,\ldots,v_j + \alpha v_k, \ldots, v_k, \ldots, v_n) = D(v_1,\ldots,v_j,\ldots,v_k,\ldots,v_n)$ for $j \neq k$ and any $\alpha$.

Using only these two properties, I'm supposed to show that $D$ is additive in each argument, that is,

$$D(v_1,\ldots,u_k + v_k, \ldots, v_n) = D(v_1,\ldots,u_k,\ldots,v_n) + D(v_1,\ldots,v_k,\ldots,v_n).$$

So far, I've checked that additivity holds for $u_k \in \operatorname{span}(\{v_i\}_{i=1, i\neq k}^n)$ and $u_k = \alpha v_k$, with the hope of checking each case of $u_k$ or something, but I don't think that's going anywhere. I think there is something clever I'm not seeing — tips or solutions appreciated!

Answer 1

You are close! You just need to show that nothing else can happen besides the cases you've covered.

We first establish a well-known property of the determinant from the two properties. Consider any $v_1,\ldots, v_n$. Suppose the dimension of $\operatorname{span}\{v_i\}$ is less than $n$. Then we claim $D(v_1,\ldots v_n)=0$.

These $n$ vectors are linearly dependent. By applying property 2 repeatedly, you can show that $D(v_1,\ldots, v_n)=D(0,v_2,\ldots, v_n)$, and then apply property 1 to show that this quantity is zero.

We now return to your question. If the dimension of $\operatorname{span}\{v_i\}_{i \ne k}$ is less than $n-1$, then the above shows that the claim you want to show becomes $0=0+0$. So from here on we assume $\operatorname{span}\{v_i\}_{i \ne k}$ has dimension $n-1$.

If $u_k$ is in the span of $\{v_i\}_{i\ne k}$, then $D(v_1,\ldots, u_k,\ldots, v_n)=0$, and you can show that $D(v_1,\ldots, u_k+v_k,\ldots, v_n)=D(v_1,\ldots, v_k, \ldots, v_n)$ using the second property.

If $v_k$ is in the span of $\{v_i\}_{i\ne k}$, a similar argument works.

Finally, if neither $u_k$ nor $v_k$ lie in the span of $\{v_i\}_{i\ne k}$, then $u_k=\alpha v_k$ for some $\alpha$ [edit: see correction in comments] (because $\operatorname{span}\{v_i\}_{i \ne k}$ has dimension $n-1$), and you have shown this case.