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Determinants from the Top Power, and the Symmetric Algebra

Because $\Lambda^n(V)$ is one-dimensional, any linear map $T: V \to V$ acts on it by a single scalar — and that scalar is the determinant, defined with no formula and no choice of basis. We prove $\det(ST) = \det S \det T$ in one line, then meet the symmetric algebra, the commutative twin of $\Lambda(V)$, and watch it turn into a polynomial ring.

The determinant, with no formula

Let $\dim V = n$, so $\Lambda^n(V)$ is one-dimensional. A linear map $T: V \to V$ induces $\Lambda^n(T): \Lambda^n(V) \to \Lambda^n(V)$ by $v_1 \wedge \cdots \wedge v_n \mapsto Tv_1 \wedge \cdots \wedge Tv_n$. Any linear endomorphism of a one-dimensional space is multiplication by a scalar — and that scalar is the [[determinant-exterior-power|determinant]] of $T$. No permutation sum, no cofactor expansion, no basis: $\det T$ is simply the factor by which $T$ scales top forms.

Multiplicativity in one line.  Lambda^n is a functor on linear maps:
    Lambda^n(S . T) = Lambda^n(S) . Lambda^n(T).
Reading off scalars on the 1-dim space Lambda^n(V):
    det(S.T) = det(S) . det(T).             Done. No index gymnastics.

Recovering the classical formula.  Fix basis e_1,...,e_n and let T have
matrix A = (a_{ij}), i.e. T e_j = sum_i a_{ij} e_i.  Then
   T e_1 ^ ... ^ T e_n
     = ( sum_i a_{i1} e_i ) ^ ... ^ ( sum_i a_{in} e_i ).
Expand; any term with a repeated e_i dies (e_i ^ e_i = 0), so only
permutations survive, each contributing a sign = sgn(sigma):
   = ( sum_{sigma in S_n} sgn(sigma) a_{sigma(1),1} ... a_{sigma(n),n} )
       . (e_1 ^ ... ^ e_n).
The scalar in parentheses is exactly the Leibniz formula for det(A).

Check on 2x2,  A = [a, b; c, d]:
   (a e_1 + c e_2) ^ (b e_1 + d e_2)
     = ad (e_1^e_2) + cb (e_2^e_1) = (ad - bc)(e_1^e_2).   det = ad - bc.
Functoriality of $\Lambda^n$ gives $\det(ST)=\det S\det T$ instantly; expanding in a basis regenerates the Leibniz permutation sum and the familiar $ad-bc$.

The symmetric algebra: commutativity instead of sign

Quotient $T(V)$ by the other natural relation — $v \otimes w - w \otimes v$ — and you get the symmetric algebra $S(V) = T(V) / (v \otimes w - w \otimes v)$, in which multiplication is commutative. Where the exterior algebra demanded a sign flip, the symmetric algebra demands that order not matter at all. Its degree-$k$ piece $S^k(V)$ has basis the unordered monomials in a basis of $V$, so $\dim S^k(V) = \binom{n+k-1}{k}$.

The punchline: if $V$ has basis $x_1, \ldots, x_n$, then $S(V)$ is canonically the polynomial ring $R[x_1, \ldots, x_n]$, graded by total degree. So the symmetric algebra is the coordinate-free, basis-independent meaning of ‘polynomial ring’ — and it satisfies the universal property: linear maps $V \to A$ into a commutative algebra extend uniquely to algebra maps $S(V) \to A$. This is the bridge from multilinear algebra to commutative algebra and the geometry built on it.