I am reading "Linear Algebra" by Hoffman and Kunze.
A determinant is defined as multi-linear alternating function which maps sq. matrices on a ring $K^{n\times n}$ to the ring $K$ itself.
[ch 5, pg 144]
It is unambiguous that determinants are functions after all.
There is another notion of a (linear) function space defined as (a linear subspace of) a set of all maps among a set of linear spaces. [ch 3, pg 75]
Is it fair to speak of this (linear) function space as a linear space in itself?
Further, we are met with this notion of linear functionals (forms), that are maps from a linear space to the field over the space. [ch 3, pg 97]
I am thinking if determinants could be thought of as linear forms? I understand the definition holds for fields and not rings. But, conventionally, determinants are over real and complex field.
If so, would it unlock other concepts of forms for our determinants?
- Determinants being the dual space of the linear function space among vector spaces.
- Determinants of maps collapsing to nullspace possessing properties of annihilators?
Best Answer
"The" determinant (denoted $\det$) is a function from $K^{n\times n}$ to $K$, where $K$ is a commutative ring (often $\mathbb R$ or $\mathbb C$, but there are many other choices) and where $n$ is a nonnegative integer. So its input is an $n \times n$-matrix with entries in $K$, and its output is an element of $K$. There is one such function for each choice of $K$ and $n$.
The determinant of a matrix $A$ (denoted $\det A$) is the value of this function at $A$.
Functions from a given set to a vector space do themselves form a vector space (with pointwise addition and scaling). So, yes, you can view the function $\det : K^{n \times n} \to K$ as a vector in the vector space of functions from $K^{n \times n}$ to $K$. This viewpoint is occasionally useful but not, by itself, insightful. In particular, there are lots of functions from $K^{n \times n}$ to $K$. If $\det$ was a linear function, then we could get some more out of this, but alas it is not (unless $n = 1$). Thus, in particular, nullspaces and dual spaces are not relevant here.
What is true is that $\det$ is a multilinear function, at least if we rewrite its input (originally a matrix) as an $n$-tuple of column vectors (or row vectors). This viewpoint is quite fruitful and can be used to give slick proofs for the basic properties of determinants. See, for example, Section 9C of Sheldon Axler, Linear Algebra Done Right, 4th edition 2024 (I think this is not in the earlier editions).