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The inverse of an invertible finite-dimensional endomorphism is a polynomial in that endomorphism

Statement

If $T$ is an invertible endomorphism of a finite-dimensional vector space, then there is a polynomial $q \in F [x]$ such that $T^{- 1} = q (T)$ .

Facts & Assumptions

Given: An invertible endomorphism $T$ of a finite-dimensional $F$ -vector space $V$ .

[L1]

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

[L2]

The characteristic polynomial of an endomorphism is that of any representing matrix $A = [T]$ , and the determinant of an endomorphism is $\det (A)$ for any such matrix (The basis-independent characteristic polynomial $χ_{T}$ of an endomorphism of a finite-dimensional space, including $χ_{T} = 1$ 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). For $A \in M_{n} (F)$ with $n \geq 1$ , $χ_{A} (x) = x^{n} + c_{n - 1} x^{n - 1} + \dots + c_{0}$ with constant coefficient $c_{0} = (- 1)^{n} \det (A)$ ( $χ_{A} (x)$ 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 $χ_{0 \times 0} = 1$ ). Hence in positive dimension $c_{0} = (- 1)^{n} \det (T)$ .

[L3]

An invertible finite-dimensional endomorphism has nonzero determinant (A finite-dimensional linear operator over a field is invertible if and only if its determinant is nonzero).

Proof

technique · direct

1.1

If $V = {0}$ , the unique endomorphism is at once $T^{- 1}$ and the zero endomorphism, so the zero polynomial evaluates to $T^{- 1}$ .

givenalgebra

1.2

Suppose $n = \dim V \geq 1$ and use the coefficients in [L2]. By [L3], $c_{0} \neq 0$ . Cayley-Hamilton gives $T^{n} + c_{n - 1} T^{n - 1} + \dots + c_{1} T + c_{0} I = 0$ .

L1L2L3

2.1

Multiply step 1.2 by $T^{- 1}$ and solve for the inverse: $T^{- 1} = - c_{0}^{- 1} (T^{n - 1} + c_{n - 1} T^{n - 2} + \dots + c_{2} T + c_{1} I)$ .

step 1.2givenalgebra

3.1

The right side of step 2.1 is a polynomial in $T$ , and step 1.1 handles the zero space.

step 1.1step 2.1∎

Depends on

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Sources

J. Demmel, Applied Numerical Linear Algebra, Lecture 14 (standard reference, not scraped)