Determinants#

Definition: Determinant Form

A determinant form is a non-trivial alternating multilinear form.

Theorem: Existence of a Determinant Form

Every \(n\)-dimensional (\(n \ge 1\)) vector space \((V, F)\) has at least one determinant form \(f: V^n \to F\).

Proof

The proof is by induction.

Base case: \(n = 1\)

This is trivial because every linear form \(f: V \to F\) is alternating by definition.

Induction hypothesis: TODO

Theorem: Determinant Form Scaling

Let \((V, F)\) be an \(n\)-dimensional \((n \ge 1)\) vector space.

If \(\omega_1: V^n \to F\) is a non-zero determinant form and \(\omega_2: V^n \to F\) is any other determinant form, then there exists a unique \(\lambda \in F\) such that

\[ \omega_2 = \lambda \omega_1. \]

Proof

TODO

Theorem: Determinant

Let \((V, F)\) be an \(n\)-dimensional \((n \ge 1)\) vector space.

If \(f: V \to V\) is an endomorphism, then there exists a unique scalar \(\mu \in F\) such that

\[ \mu \cdot \omega (\mathbf{v}_1, \dotsc, \mathbf{v}_n) = \omega (f(\mathbf{v}_1), \dotsc, f(\mathbf{v}_n)) \]

for all \(\mathbf{v}_1, \dotsc, \mathbf{v}_n \in V\) and all determinant forms \(\omega\).

Definition: Determinant

We call \(\mu\) the determinant of \(f\).

Notation

We denote the determinant of \(f\) by \(\det f\).

Proof

TODO

Theorem: Determinant of Composition

Let \((V, F)\) be an \(n\)-dimensional \((n \ge 1)\) vector space and let \(f, g: V \to V\) be endomorphisms.

The determinant of the composition \(f \circ g\) is equal to the product of the determinants of \(f\) and \(g\):

\[\det f \circ g = \det f \cdot \det g\]

Proof

By definition we have

\[\begin{aligned}(\det f \circ g) \cdot \omega & = \omega(f(g(\cdot)), \dotsc, f(g(\cdot))) \\ & = (\det f)\cdot \omega(g(\cdot), \dotsc, g(\cdot)) \\ & = (\det f) \cdot ((\det g)\cdot \omega) \\ & = (\det f \cdot \det g) \cdot \omega\end{aligned}\]

for all determinant forms \(\omega\). Therefore:

\[\det f \circ g = \det f \cdot \det g\]

Theorem: Bijectivity via Determinants

Let \((V, F)\) be a finite dimensional vector space with \(\dim V \ge 1\).

An endomorphism \(f: V \to V\) is bijective if and only if its determinant \(\det f\) is not zero.

Proof

TODO

Theorem: Eigenvalues and Determinant

Let \((V, F)\) be a finite-dimensional vector space and let \(f: V \to V\) be an endomorphism.

If \(f\) has \(l\) distinct eigenvalues \(\lambda_1, \dotsc, \lambda_l\) and the sum of their algebraic multiplicities is equal to \(\dim V\), then the determinant of \(f\) is given as follows:

\[\det f = \prod_{k=1}^l \lambda_k ^{\operatorname{alg} (\lambda_k)}\]

Proof

TODO