How to understand the dual of determinants

Question

We consider the determinant of a $n\times n$ square matrix as a linear transformation. $$\det : M_{n\times n}\to \mathbb{C}.$$ How to discribe its dual $\det ^*$?

Answer 1

As noted in the comments, $\det$ is not linear because $$\det(\alpha A+B)\ne \alpha\det A+\det B$$ in general. For example if $I$ is the $2\times 2$ identity matrix over $\mathbb{C}$, then $\det(2I)=4\ne 2$. However, $\det$ is multilinear in the rows and columns of a matrix. If we view it this way, there's a meaningful notion of dual determinant function. I'll present this in a slightly more general setting than $\mathbb{C}^n$.

Let $\mathbb{F}$ be a field and let $V,V^*$ be a pair of dual $n$-dimensional vector spaces over $\mathbb{F}$, so there's a non-degenerate pairing $\langle-,-\rangle:V^*\times V\to\mathbb{F}$. Consider the function $\Omega:(V^*)^n\times V^n\to\mathbb{F}$ defined by $$\Omega(x^*_1,\ldots,x^*_n\,;x_1,\ldots,x_n)=\det\bigl(\langle x^*_j,x_i\rangle\bigr)\tag{1}$$ By bilinearity of the pairing, and alternating multilinearity of $\det$ in the rows and columns of the matrix in (1), it follows that

• $\Omega(x^*_1,\ldots,x^*_n\,;-):V^n\to\mathbb{F}$ is a determinant function on $V$ for every selection of $x^*_1,\ldots,x^*_n\in V^*$
• $\Omega(-\,;x_1,\ldots,x_n):(V^*)^n\to\mathbb{F}$ is a determinant function on $V^*$ for every selection of $x_1,\ldots,x_n\in V$

Let $\Delta$ be an arbitrary determinant function on $V$ with $\Delta(v_1,\ldots,v_n)=1$ and $\Delta^*$ an arbitrary determinant function on $V^*$ with $\Delta^*(v^*_1,\ldots,v^*_n)=1$. By (1) and the universal properties of $\Delta$ and $\Delta^*$, it follows that $$\begin{align*} \det\bigl(\langle x^*_j,x_i\rangle\bigr)&=\Omega(x^*_1,\ldots,x^*_n\,;x_1,\ldots,x_n)\\ &=\Omega(x^*_1,\ldots,x^*_n\,;v_1,\ldots,v_n)\cdot\Delta(x_1,\ldots,x_n)\\ &=\Omega(v^*_1,\ldots,v^*_n\,;v_1,\ldots,v_n)\cdot\Delta^*(x^*_1,\ldots,x^*_n)\cdot\Delta(x_1,\ldots,x_n)\\ &=\det\bigl(\langle v^*_j,v_i\rangle\bigr)\cdot\Delta^*(x^*_1,\ldots,x^*_n)\cdot\Delta(x_1,\ldots,x_n) \end{align*}$$ Letting $\delta=\det\bigl(\langle v^*_j,v_i\rangle\bigr)^{-1}$, we have $$\Delta^*(x^*_1,\ldots,x^*_n)\cdot\Delta(x_1,\ldots,x_n)=\delta\cdot\det\bigl(\langle x^*_j,x_i\rangle\bigr)\tag{2}$$ The determinant functions $\Delta,\Delta^*$ are called dual when $\delta=1$ in (2). Every determinant function has a unique dual determinant function, and the relationship can be used to establish important duality properties.