The Adjoint of an Endomorphism

Introduction

A non-degenerate pairing on a linear space turns the endomorphism $A$ into a second endomorphism $A^{*}$, the adjoint, by moving $A$ from one side of the pairing to the other, and the article is about that operator: its existence and uniqueness, the description of its kernel and image as the paired complements of the image and kernel of $A$, the preservation of rank, invertibility and spectrum, the rule for the adjoint of a composite, and the identification of the adjoint with the transpose once a basis is fixed. The involution that the assignment $A \mapsto A^{*}$ defines on the endomorphism algebra is Involutions of the Endomorphism Algebra; the article here is its operator-level counterpart in the *-operator group, and it treats the adjoint as an operator rather than the involution as a structure.

The pairing, its non-degeneracy and the notion of a reflexive pairing are as in Involutions of the Endomorphism Algebra; the transpose of a linear map, its contravariance and the annihilator description of its kernel and image are The Transpose of a Linear Map; the dual involution is Involutions of the Dual Space. The passage from the pairing duality to the dual-space duality is The Involution on the Dual Operator; the group of endomorphisms with $A^{*}=A^{-1}$ is Unitary Endomorphisms; the forms themselves, their norms and the analysis they support are Hilbert Algebras, in Part II.

Throughout, $F$ is a field, $V$ is a finite-dimensional $F$-linear space, $E = \operatorname{End}_F(V)$, and $B$ is a non-degenerate reflexive pairing on $V$: bilinear with $B(y,x) = \varepsilon B(x,y)$ for a sign $\varepsilon$, or sesquilinear with respect to an involution $\varsigma$ of $F$. The adjoint of $A$ is $A^{*}$, defined by $B(Ax,y) = B(x,A^{*}y)$. No norm and no topology is used.

The Adjoint Operator

Theorem (existence, uniqueness, linearity). For every $A \in E$ there is a unique endomorphism $A^{*} \in E$ with

$$ B(Ax,y) = B(x,A^{*}y) \qquad \text{for all } x,y \in V , $$

and the assignment $A \mapsto A^{*}$ is additive, of order two and anti-multiplicative: $(A^{*})^{*} = A$, $(A+B)^{*} = A^{*}+B^{*}$, $(AB)^{*} = B^{*}A^{*}$, and $(\lambda A)^{*} = \varsigma(\lambda)A^{*}$.

Proof. For fixed $y$ the map $x \mapsto B(Ax,y)$ is linear; since $z \mapsto B(\cdot,z)$ is a bijection $V \to V^{*}$ by non-degeneracy, there is a unique $A^{*}y$ representing it, and uniqueness in $y$ makes $A^{*}$ a map. Linearity of $A^{*}$ and the four laws follow by uniqueness exactly as in Involutions of the Endomorphism Algebra, which owns them.

Definition. An endomorphism is self-adjoint when $A^{*}=A$, skew-adjoint when $A^{*}=-A$, and unitary when $A^{*}A = AA^{*} = \mathrm{id}$.

Kernel, Image and the Paired Complement

Definition. For a subspace $U \subseteq V$ the paired complement is

$$ U^{\mathrm{c}} = \{y \in V : B(x,y) = 0 \text{ for all } x \in U\} . $$

Proposition. $U^{\mathrm{c}}$ is a subspace and $\dim_F U^{\mathrm{c}} = \dim_F V - \dim_F U$; moreover $(U^{\mathrm{c}})^{\mathrm{c}} = U$ when $B$ is reflexive, so the assignment $U \mapsto U^{\mathrm{c}}$ is an order-reversing bijection of the lattice of subspaces onto itself.

Proof. $U^{\mathrm{c}}$ is the kernel of the linear map $V \to U^{*}$, $y \mapsto B(\cdot,y)$, which is surjective because $B$ is non-degenerate; hence the dimension formula. Reflexivity gives $(U^{\mathrm{c}})^{\mathrm{c}} = U$, and both statements are the standard pairings of a form.

Theorem (kernel and image of the adjoint). For every $A \in E$,

$$ \ker A^{*} = (\operatorname{im}A)^{\mathrm{c}}, \qquad \operatorname{im}A^{*} = (\ker A)^{\mathrm{c}} , \qquad \operatorname{rk}A^{*} = \operatorname{rk}A . $$

Proof. $A^{*}y = 0$ means $B(x,A^{*}y) = 0$ for all $x$, that is $B(Ax,y) = 0$ for all $x$, which says $y \in (\operatorname{im}A)^{\mathrm{c}}$; this is the first identity. For the second, $A^{*}y$ ranges over the paired complement of $\ker A$: indeed $B(x,A^{*}y) = B(Ax,y)$ vanishes for all $y$ exactly when $x \in \ker A$, so $(\operatorname{im}A^{*})^{\mathrm{c}} = \ker A$, and taking paired complements gives the identity. The rank is unchanged because $\dim\ker A^{*} = \dim(\operatorname{im}A)^{\mathrm{c}} = \dim\ker A$.

Corollary (invertibility and inverse). $A$ is invertible if and only if $A^{*}$ is, and then $(A^{-1})^{*} = (A^{*})^{-1}$; the adjoint of the identity is the identity, and the adjoint of a scalar $\lambda$ is $\varsigma(\lambda)$.

Proof. $\operatorname{rk}A^{*} = \operatorname{rk}A$ makes invertibility correspond; from $AA^{-1}=\mathrm{id}$ and anti-multiplicativity, $(A^{-1})^{*}A^{*} = \mathrm{id}$, which is the inverse statement.

Spectrum and the Matrix Form

Proposition (spectrum). $A^{*}$ has the same trace, the same determinant and the same characteristic polynomial as $A$; consequently the spectrum of $A^{*}$, with multiplicities, is the spectrum of $A$. In the bilinear case, if $\Phi$ is the Gram matrix of $B$ in a basis, then $[A^{*}] = \Phi^{-1}[A]^{\mathsf{T}}\Phi$, so $A^{*}$ is similar to the transpose of $A$.

Proof. $[A^{*}] = \Phi^{-1}[A]^{\mathsf{T}}\Phi$ is the matrix form of Involutions of the Endomorphism Algebra; a matrix and its transpose are similar over the semigroup generated by transposition, and they have the same characteristic polynomial, $\det(xI-A) = \det((xI-A)^{\mathsf{T}})$; conjugation by $\Phi$ preserves the characteristic polynomial. The trace and determinant are the coefficients of the characteristic polynomial in degrees $n-1$ and $0$.

Example. With the standard pairing $B(x,y) = \sum x_iy_i$ on $F^n$ the Gram matrix is the identity and $A^{*} = A^{\mathsf{T}}$; with $B(x,y) = x_1y_1 - x_2y_2$ on $F^2$ one has $A^{*} = \operatorname{diag}(1,-1)A^{\mathsf{T}}\operatorname{diag}(1,-1)$, which for $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $\begin{pmatrix} a & -c \\ -b & d \end{pmatrix}$: the diagonal entries are unchanged and the two off-diagonal entries change sign.

The Adjoints of a Composite and the Involution

Proposition (the adjoint of a composite). For $A_1,\dots,A_k \in E$,

$$ (A_1A_2\cdots A_k)^{*} = A_k^{*}\cdots A_2^{*}A_1^{*} , \qquad \Bigl(\sum_i A_i\Bigr)^{*} = \sum_i A_i^{*} . $$

Proof. Induction on $k$ from $(AB)^{*}=B^{*}A^{*}$ and additivity.

Remark (the involution on the endomorphism algebra). The assignment $A \mapsto A^{*}$ is an involution of the algebra $E$ in the sense of Involutive Rings, of the first kind for a bilinear pairing and of the second kind for a sesquilinear one; its fixed and skew parts, its matrix description, the classification of the involutions it produces and the unitary elements are the subject of Involutions of the Endomorphism Algebra, and only the operator facts about a single adjoint are established here. The two articles divide the subject: that one treats the involution as a structure on the algebra, this one the adjoint as an operator on $V$.

Remark (the involution and the operator adjoint). There are two maps that the word "adjoint" names, and the article's symbol $\ast$ is the element involution on $E$ while the operator adjoint of a map on $E$ — the adjoint of $L_A$ for the natural pairing of the endomorphism algebra — is written ${}^{\dagger}$ and is The Adjoint of the Left Multiplication on a Linear Space.

Summary

A non-degenerate reflexive pairing $B$ on a finite-dimensional space $V$ attaches to each endomorphism $A$ its adjoint $A^{*}$, uniquely determined by $B(Ax,y) = B(x,A^{*}y)$; the assignment is additive, of order two, anti-multiplicative and $\varsigma$-semilinear, so it is an involution of the endomorphism algebra, treated as a structure in Involutions of the Endomorphism Algebra. As an operator the adjoint is described by the paired complement: $\ker A^{*} = (\operatorname{im}A)^{\mathrm{c}}$, $\operatorname{im}A^{*} = (\ker A)^{\mathrm{c}}$, the paired complement being the order-reversing bijection $U \mapsto U^{\mathrm{c}}$ of the subspace lattice with $\dim U^{\mathrm{c}} = n - \dim U$. Consequently the rank is preserved, invertibility corresponds, and the inverse passes to the adjoint, $(A^{-1})^{*} = (A^{*})^{-1}$; the trace, determinant and characteristic polynomial are preserved, so the spectrum of $A^{*}$ is that of $A$, and in a basis $A^{*}$ is $\Phi^{-1}A^{\mathsf{T}}\Phi$. The adjoint of a product is the product of the adjoints in the reverse order. The dual-space duality and its compatibility with the pairing duality are The Involution on the Dual Operator.

Summary of Notation

Symbol	Meaning
$F$, $\varsigma$	the field and the involution of the sesquilinear case
$V$, $n$	the space and its dimension
$E=\operatorname{End}_F(V)$	the endomorphism algebra
$B$	a non-degenerate reflexive pairing
$A^{*}$	the adjoint, $B(Ax,y)=B(x,A^{*}y)$
$U^{\mathrm{c}}$	the paired complement, $\{y : B(x,y)=0 \ \forall x\in U\}$
$\ker A^{*}=(\operatorname{im}A)^{\mathrm{c}}$	the kernel of the adjoint
$\operatorname{im}A^{*}=(\ker A)^{\mathrm{c}}$	the image of the adjoint
$\Phi$	the Gram matrix; $[A^{*}]=\Phi^{-1}[A]^{\mathsf{T}}\Phi$
$A^{*}=\pm A$	self-adjoint and skew-adjoint
$A^{}A=AA^{}=\mathrm{id}$	unitary