Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22
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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Spectrum in a finite-dimensional matrix algebra

Example

Let n1 and let Mn(C) carry the Euclidean operator norm induced by identifying Mn(C) with B(Cn), Cn having the Euclidean norm, and by the operator norm on that space (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum). Then Mn(C) is a unital complex Banach algebra (Unital Banach algebra), and for every AMn(C)

σ(A)={λC:det(λIA)=0}

with the spectrum taken in that algebra (Spectrum and resolvent set in a Banach algebra) and the determinant of For n1, the determinant over a commutative ring by the Leibniz formula, and detA for a real matrix.

Facts & Assumptions

Given: An integer n1, the algebra Mn(C) of n×n complex matrices with the operator norm, and a matrix AMn(C).

[L1]

The operator norm is submultiplicative and I=1; an element is invertible in Mn(C) exactly when it has a two-sided inverse matrix, and λσ(A) exactly when λIA is not invertible (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, Spectrum and resolvent set in a Banach algebra, Unital Banach algebra).

[L3]

Determinants are multiplicative: det(BC)=det(B)det(C) (For same-sized finite square matrices over a commutative ring, det(AB)=det(A)det(B)).

[L4]

Finite-dimensional normed spaces are complete, so Mn(C)=B(Cn) is complete for the operator norm (Every finite-dimensional normed space is Banach).

Verification

technique · direct
1.1

Mn(C) is an associative complex algebra under matrix multiplication with unit I; by [L1] the operator norm is submultiplicative with I=1, and by [L4] the space is complete; hence it is a unital complex Banach algebra.

L1L4
2.1

If det(λIA)0 then B:=λIA has the two-sided inverse det(B)1adj(B) by [L2], so λσ(A) by [L1].

step 1.1L2L1
2.2

Conversely, if λIA has a two-sided inverse C in the algebra of [step 1.1], then [L3] gives det(λIA)det(C)=det(I)=1, so det(λIA)0.

step 1.1L3algebra
3.1

Combining [step 2.1] and [step 2.2] with the characterization of the spectrum in [L1]: λσ(A) precisely when λIA is not invertible, which by the two steps happens precisely when det(λIA)=0.

step 2.1step 2.2L1

Depends on

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