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For every finite-dimensional space, is exactly the set of roots in of
Statement
For an endomorphism of a finite-dimensional -vector space,
Facts & Assumptions
Given: A finite-dimensional -vector space and .
In any basis, is the characteristic polynomial of the representing matrix; in dimension zero it is (The basis-independent characteristic polynomial of an endomorphism of a finite-dimensional space, including in dimension zero).
A scalar is an eigenvalue exactly when is not invertible (For a finite-dimensional space, is an eigenvalue of if and only if is not invertible).
A finite-dimensional endomorphism is invertible exactly when its determinant is nonzero (A finite-dimensional linear operator over a field is invertible if and only if its determinant is nonzero).
A root of a polynomial is a scalar at which its evaluation is zero (Evaluation and roots of a polynomial in a commutative target ring).
Proof
If , the spectrum is empty because there is no nonzero eigenvector, while [L1] gives , which has no root.
Suppose and choose a basis with matrix . Then .
The scalar is nonzero. Thus step 1.2, [L3], and [L2] give if and only if is not invertible if and only if .
Steps 1.1 and 2.1 prove the set equality in every finite dimension and prove both directions of the equivalence.
Depends on
- The basis-independent characteristic polynomial $\chi_T$ of an endomorphism of a finite-dimensional space, including $\chi_T=1$ in dimension zero
- For a finite-dimensional space, $\lambda$ is an eigenvalue of $T$ if and only if $T-\lambda I$ is not invertible
- A finite-dimensional linear operator over a field is invertible if and only if its determinant is nonzero
- Evaluation and roots of a polynomial in a commutative target ring
Used by
- Every endomorphism of a nonzero finite-dimensional vector space over an algebraically closed field has an eigenvalue Corollary
- If χ_T splits over F, every eigenvalue of χ_T(T) is 0 Corollary
- Adjacency spectrum, spectral radius, and cospectral graphs Definition
- A quarter-turn of ℝ² has characteristic polynomial x²+1 and no real eigenvalue Example
- beginpmatrix0&11&1 endpmatrix over F₂ has characteristic polynomial x²+x+1 and no eigenvalue in its base field Example
- A stationary splitting converges for every start if and only if its iteration matrix has spectral radius below 1 Theorem
- An endomorphism is diagonalisable if and only if its characteristic polynomial splits and every eigenvalue's geometric multiplicity equals its algebraic multiplicity Theorem
- If χ_T(x)=∏_i<n(x-λᵢ) in F[x], then det(T)=∏_i<nλᵢ: determinant is the product of the eigenvalues counted with algebraic multiplicity Theorem
- If χ_T(x)=∏_i<n(x-λᵢ) in F[x], then tr(T)=∑_i<nλᵢ: trace is the sum of the eigenvalues counted with algebraic multiplicity Theorem
- If χ_T(x)=∏_i<n(x-λᵢ) in F[x], then χ_p(T)(y)=∏_i<n(y-p(λᵢ)) for every p∈ F[x]: the eigenvalues of p(T) are p(λᵢ), counted with algebraic multiplicity Theorem
- Over every extension field, a scalar is an eigenvalue of the extended matrix exactly when it is a root of the minimal polynomial Theorem
- T is triangularisable iff its minimal polynomial splits iff its characteristic polynomial splits Theorem
- The minimal and characteristic polynomials have exactly the same monic irreducible factors Theorem
Dependency tree · two levels
22 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
- M. Khovanov, Linear Algebra II notes, §6 (standard reference, not scraped)