I need please revision of my reasoning for this problem:
"Let $S$ be the set of $n \times n$ matrices over a field $F$. Let $V$ be the vector space of all functions from $S$ into $F$. Let $W$ be the set of alternating $n$-linear functions on $S$. What is the dimension of $W$?"
Now, I considered the annihilator of $W$, that is, the set of all $n \times n$ matrices such that 2 rows in them are equal. Let $W^\circ$ be that set. The dimension values would be then related by $$ \dim \,W + \dim \,W^\circ = \dim \,V $$ and then I would be able to find the dimension of $W$. But I found very complicated to find the dimension of $W^\circ$. I am not sure if I am thinking in the right direction. Any hint would be appreciated!
Let $D$ and $D'$ be two arbitrary alternating $n$-linear functions from $S$ to $F$.
Now, $D(A)=(\det A)D(I)$ and $D'(A)=(\det A)D'(I)$ where $I$ is the $n\times n$ identity matrix.
Then dividing them we can see $D'(A)=C\times D(A)$ where $C$ is a constant. So $W$ has dimension $1$.