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Cayley-Hamilton: every finite-dimensional endomorphism satisfies its characteristic polynomial, $χ_{T} (T) = 0$

Statement

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

$χ_{T} (T) = 0.$

Facts & Assumptions

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

[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); polynomial evaluation is $p (T) = \sum_{k} a_{k} T^{k}$ (Polynomial evaluation at an endomorphism: $p (T) = \sum_{k} a_{k} T^{k}$ ).

[L2]

The adjugate consists of signed $(n - 1) \times (n - 1)$ minors (Deleted-row-and-column minors, cofactors, the cofactor matrix and the adjugate over a commutative ring), and over a commutative ring $(x I - A) adj (x I - A) = \det (x I - A) I$ (For every positive-sized square matrix over a commutative ring, $A adj (A) = adj (A) A = \det (A) I$ ).

[L3]

Equality of polynomials is equality of all coefficients (Finitely supported coefficient sequences and trimmed finite coefficient lists define the same formal polynomials).

[L4]

Matrix representation is injective and sends sums and scalar multiples to matrix operations ( $T \mapsto [T]_{B}^{C}$ is a vector-space isomorphism $L (V, W) ≅ M_{m \times n} (F)$ ) and composition to matrix multiplication ( $[S \circ T]_{B}^{D} = [S]_{C}^{D} [T]_{B}^{C}$ ).

Proof

technique · direct coefficient comparison

1.1

If $\dim V = 0$ , [L1] gives $χ_{T} = 1$ , while $I_{V} = 0_{V}$ is the unique endomorphism of the zero space; hence $χ_{T} (T) = I_{V} = 0_{V}$ .

L1algebra

1.2

Suppose $n = \dim V \geq 1$ , choose a basis, and put $A = [T]$ . Each entry of $adj (x I - A)$ has degree at most $n - 1$ by [L2], so write $adj (x I - A) = \sum_{j = 0}^{n - 1} B_{j} x^{j}$ ; also write $χ_{A} (x) = \sum_{k = 0}^{n} c_{k} x^{k}$ , where $c_{n} = 1$ .

L1L2choosealgebra

2.1

Expanding the adjugate identity [L2] and comparing coefficients using [L3] gives $- A B_{0} = c_{0} I$ , $B_{k - 1} - A B_{k} = c_{k} I$ for $1 \leq k < n$ , and $B_{n - 1} = I$ .

step 1.2L2L3algebra

3.1

Multiply the equation indexed by $k$ on the left by $A^{k}$ , include the first equation at $k = 0$ and the last identity at $k = n$ , and add. The $A^{k} B_{k - 1}$ terms telescope against the $A^{k} B_{k - 1}$ terms of the preceding recurrence, leaving $\sum_{k = 0}^{n} c_{k} A^{k} = 0$ .

step 2.1algebra

4.1

By [L1] and [L4], the left side of step 3.1 is the representing matrix of $χ_{T} (T)$ ; injectivity of matrix representation therefore gives $χ_{T} (T) = 0$ .

step 3.1L1L4

5.1

Steps 1.1 and 4.1 prove Cayley-Hamilton in every finite dimension without treating substitution into a matrix-coefficient polynomial as a ring homomorphism.

step 1.1step 4.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 66 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

M. Khovanov, Linear Algebra II notes, §6 (standard reference, not scraped)
The Stacks Project, Lemma 10.16.1 (05G6) (standard reference, not scraped)

Cayley-Hamilton: every finite-dimensional endomorphism satisfies its characteristic polynomial, χT(T)=0χT​(T)=0