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For every finite-dimensional space, $σ_{F} (T)$ is exactly the set of roots in $F$ of $χ_{T}$

Statement

For an endomorphism $T$ of a finite-dimensional $F$ -vector space,

$σ_{F} (T) = {λ \in F : χ_{T} (λ) = 0} .$

Facts & Assumptions

Given: A finite-dimensional $F$ -vector space $V$ and $T \in L (V)$ .

[L1]

In any basis, $χ_{T}$ is the characteristic polynomial of the representing matrix; in dimension zero it is $1$ (The basis-independent characteristic polynomial $χ_{T}$ of an endomorphism of a finite-dimensional space, including $χ_{T} = 1$ in dimension zero).

[L2]

A scalar $λ$ is an eigenvalue exactly when $T - λ I_{V}$ is not invertible (For a finite-dimensional space, $λ$ is an eigenvalue of $T$ if and only if $T - λ I$ is not invertible).

[L3]

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).

[L4]

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

technique · direct

1.1L1L4

If $\dim V = 0$ , the spectrum is empty because there is no nonzero eigenvector, while [L1] gives $χ_{T} = 1$ , which has no root.

1.2L1algebra

Suppose $\dim V = n \geq 1$ and choose a basis with matrix $A = [T]$ . Then $χ_{T} (λ) = \det (λ I_{n} - A) = (- 1)^{n} \det (A - λ I_{n})$ .

2.1step 1.2L2L3L4

The scalar $(- 1)^{n}$ is nonzero. Thus step 1.2, [L3], and [L2] give $χ_{T} (λ) = 0$ if and only if $T - λ I_{V}$ is not invertible if and only if $λ \in σ_{F} (T)$ .

3.1step 1.1step 2.1∎

Steps 1.1 and 2.1 prove the set equality in every finite dimension and prove both directions of the equivalence.

Depends on

Dependency tree · 2 levels

Sources

M. Khovanov, Linear Algebra II notes, §6

For every finite-dimensional space, σF(T)σF​(T) is exactly the set of roots in FF of χTχT​