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If $χ_{T} (x) = \prod_{i < n} (x - λ_{i})$ in $F [x]$ , then $\det (T) = \prod_{i < n} λ_{i}$ : determinant is the product of the eigenvalues counted with algebraic multiplicity

Statement

Let $T$ be an endomorphism of an $n$ -dimensional vector space over $F$ . If

$χ_{T} (x) = \prod_{i < n} (x - λ_{i})$

in $F [x]$ , then $\det (T) = \prod_{i < n} λ_{i}$ . Thus the determinant is the product of the eigenvalues counted with algebraic multiplicity.

Facts & Assumptions

Given: $T$ as stated and a displayed factorization $χ_{T} (x) = \prod_{i < n} (x - λ_{i})$ .

[L1]

The operator characteristic polynomial is computed from any representing matrix and equals $1$ in dimension zero (The basis-independent characteristic polynomial $χ_{T}$ of an endomorphism of a finite-dimensional space, including $χ_{T} = 1$ in dimension zero).

[L2]

In positive size, the constant coefficient of a characteristic polynomial is $(- 1)^{n}$ times the determinant ( $χ_{A} (x)$ is monic of degree $n$ ; for $n \geq 1$ its $x^{n - 1}$ coefficient is $- tr (A)$ and its constant coefficient is $(- 1)^{n} \det (A)$ , while $χ_{0 \times 0} = 1$ ).

[L3]

The determinant of an endomorphism is the determinant of any representing matrix and equals $1$ in dimension zero (The determinant of an endomorphism of a finite-dimensional vector space: its matrix determinant in an ordered basis in positive dimension, and $1$ on the zero space).

[L4]

Roots of $χ_{T}$ are precisely eigenvalues (For every finite-dimensional space, $σ_{F} (T)$ is exactly the set of roots in $F$ of $χ_{T}$ ), and algebraic multiplicity is the exponent of the corresponding linear factor (Algebraic multiplicity as the exponent of $x - λ$ in $χ_{T}$ , and geometric multiplicity as $\dim E_{λ} (T)$ ).

Proof

technique · direct

1.1

If $n = 0$ , [L3] gives $\det (T) = 1$ , while the product indexed by the empty set is $1$ .

L1L3algebra

1.2

Suppose $n \geq 1$ . By [L1]–[L3], the constant coefficient of $χ_{T}$ is $(- 1)^{n} \det (T)$ . The constant coefficient of the given product is $\prod_{i < n} (- λ_{i}) = (- 1)^{n} \prod_{i < n} λ_{i}$ .

L1L2L3givenalgebra

2.1

Equality of coefficients and cancellation of the nonzero scalar $(- 1)^{n}$ give $\det (T) = \prod_{i < n} λ_{i}$ . By [L4], the factors list exactly the eigenvalues with their algebraic multiplicities.

step 1.2L4algebra

3.1

Steps 1.1 and 2.1 prove the formula in every finite dimension.

step 1.1step 2.1∎

Depends on

Used by

The matrix-tree theorem becomes an eigenvalue product formula Corollary

Dependency tree · two levels

19 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

H. Pinkham, Linear Algebra, §12.3.2 (standard reference, not scraped)

If χT(x)=∏i<n(x−λi)χT​(x)=∏i<n​(x−λi​) in F[x]F[x], then det⁡(T)=∏i<nλidet(T)=∏i<n​λi​: determinant is the product of the eigenvalues counted with algebraic multiplicity