basis-free definition of determinant

The definition of determinant as a multilinear mapping on rows can bemodified to provide a basis-free definition of determinant. In orderto make it clear that we are not using bases. we shall speak in termsof an endomorphism of a vector space over $k$ rather than speaking ofa matrix whose entries belong to $k$. We start by recalling somepreliminary facts.

Suppose $V$ is a finite-dimensional vector space of dimension $n$ overa field $k$. Recall that a multilinear map $f:V^n\to k$ isalternating if $f(x)=0$ whenever there exist distinct indices$i,j\in [n]=\{1,\mathrm{\dots },n\}$ such that $x_i=x_j$. Every alternatingmap $f:V^n\to k$ is skew-symmetric, that is, for eachpermutation $\pi \in {𝔖}_n$, we have that $f(x)=\mathrm{sgn}(\pi )f(x^{\pi })$, where $x^{\pi }$ denotes${(x_{\pi (i)})}_{i\in [n]}$, the result of $\pi$ permuting the entries of$x$.

Since the trivial map $0:V^n\to k$ is alternating and any linearcombination of alternating maps is alternating, it follows thatalternating maps form a subspace of the space of multilinear maps. Inthe following proposition we show that this subspace isone-dimensional.

Theorem.

Suppose $V$ is a finite-dimensional vector space of dimension $n$ overa field $k$. Then the space of alternating maps from $V^n$ to $k$ isone-dimensional.

Proof.

We use a basis here, but we will throw it away later. We need thebasis here because each map we will consider has exactly as manyelements as a basis of $V$. So let $B=\{b_i:i\in [n]\}$ be abasis of $V$.

Suppose $f$ and $g$ are nontrivial alternating maps from $V^n$ to $k$.We claim that $f$ and $g$ are linearly dependent. Let $x\in V^n$. Wemay assume that the entries of $x$ are basis vectors, that is, that $X=\{x_i:i\in [n]\}\subset \{b_i:i\in [n]\}$.If $X⊊B$, then there exist distinct indices $i,j\in [n]$ suchthat $x_i=x_j$. Since $f$ and $g$ are alternating, it follows that$f(x)=g(x)=0$, which implies that $f(b)g(x)=g(b)f(x)$.On the other hand, if $X=B$, then there is a permutation$\pi \in {𝔖}_n$ such that $x=b^{\pi }$. Since $f$ and $g$ areskew-symmetric, it follows that

$$f(b)g(x)=\mathrm{sgn}(\pi )f(b)g(b)=g(b)f(x).$$

In either case we find that $f(b)g(x)=g(b)f(x)$. Since $f(b)$ and$g(b)$ are fixed scalars, it follows that $f$ and $g$ are linearlydependent.

So far we have shown only that the dimension of the space ofalternating maps is less than or equal to one. In order to show thatthe space is one-dimensional we simply need to find a nontrivialalternating form. To do this, let $\{b_i^{*}:i\in [n]\}$ be thenatural basis of $V^{*}$, so that $b_i^{*}(b_j)$ is the Kronecker delta of$i$ and $j$ for any $i,j\in [n]$. Define a map $f:V^n\to k$ by

$$f(x)=\sum _{\pi \in {𝔖}_n}\mathrm{sgn}(\pi )\prod _{i\in [n]}{b}_i^{*}(x_{\pi (i)}).$$

One can check that $f$ is multilinear and alternating. Moreover,$f(b)=1$, so it is nontrivial. Hence the space of alternating mapsis one-dimensional.∎

For an alternate view of the above results, we could look instead atlinear maps from the exterior product ${\bigwedge }^{n}V$ into $k$. Theproposition above can be viewed as saying that the dimension of${\bigwedge }^{n}V$ is $\left({\displaystyle \genfrac{}{}{0pt}{}{n}{n}}\right)=1$.

We define the determinant of an endomorphism in terms of the action ofthe endomorphism on alternating maps. Recall that if $M:V\to V$is an endomorphism, its pullback $M^{*}$ is the unique operator suchthat

$$(M^{*}f){(x_i)}_{i\in [n]}=f{(M(x_i))}_{i\in [n]}.$$

Since the space of alternating maps is one-dimensional andendomorphisms of a one-dimensional space reduce to scalarmultiplication, it follows that $M^{*}f$ is a scalar multiple of $f$.We call this scalar the determinant. It is well-defined because thescalar depends on $M$ but not on $f$.

Definition.

Suppose $V$ is a finite-dimensional vector space of dimension $n$ overa field $k$, and let $M\mathrm{:}V\mathrm{\to }V$ be an endomorphism. Then thedeterminant of $M$ is the unique scalar $\det \mathit{}\mathrm{(}M\mathrm{)}$ such that

$$M^{*}f=det(M)f$$

for all alternating maps $f\mathrm{:}{V}^n\mathrm{\to }k$.