Bilinear and Hermitian Forms, Adjoints, and Isometries

A form turns a bare vector space into a space with geometric comparisons. A nondegenerate bilinear form identifies vectors with covectors linearly; a positive-definite Hermitian form supplies norms and orthogonality and gives an antilinear identification with the complex dual. Once nondegenerate forms of the same kind are fixed on the domain and codomain, every linear map has a corresponding adjoint. An isometry has its adjoint as a left inverse; only a bijective isometry has its adjoint equal to its inverse. The details depend on whether the form is bilinear or sesquilinear and whether it is positive or indefinite.

Required background. Vector Spaces, Duals, Linear Maps, and Bases supplies the dual-space and change-of-basis language used to distinguish a form from its matrix.

This page is finite dimensional. Questions about domains of unbounded adjoints and self-adjoint extensions belong to Functional and Spectral Analysis.

Bilinear forms and the map to the dual

A bilinear form on a real or complex vector space $V$ is a map

$$B:V\times V\longrightarrow\mathbb F$$

that is linear in each argument. It is symmetric when $B(v,w)=B(w,v)$ and alternating when $B(v,v)=0$ for every $v$. Over $\mathbb R$ or $\mathbb C$, alternating implies $B(v,w)=-B(w,v)$.

Each bilinear form defines linear maps to the algebraic dual. Choosing the second-slot version,

$$R_B:V\longrightarrow V^*, \qquad R_B(w)=B(\,\cdot\,,w),$$

one calls $B$ nondegenerate when $R_B$ is an isomorphism. Equivalently,

$$B(v,w)=0\ \text{for every }v \quad\Longrightarrow\quad w=0.$$

In finite dimension, the analogous condition in the first slot is equivalent. A nondegenerate form therefore supplies the extra structure needed to identify $V$ with $V^*$. A bare vector space has no such preferred identification.

For a basis $\{e_a\}$, define the Gram matrix

$$G_{ab}=B(e_a,e_b).$$

Then

$$B(v,w)=v^aG_{ab}w^b.$$

If $e'_a=e_bM^b{}_a$, the same form has matrix

$$G'=M^{\mathsf T}GM.$$

Nondegeneracy is equivalent to $\det G\neq0$. This congruence transformation differs from the similarity transformation of a linear operator because both arguments of the form change basis in the same direction.

Real symmetric forms and signature

For a real symmetric bilinear form, a suitable basis diagonalizes $G$ by congruence. In the nondegenerate case, rescaling the basis vectors gives

$$G \sim \operatorname{diag}( \underbrace{1,\ldots,1}_{p}, \underbrace{-1,\ldots,-1}_{q}).$$

Sylvester’s law of inertia says that the pair $(p,q)$ is independent of the diagonalizing basis. It is the signature data of the form. A positive-definite real inner product has $(p,q)=(n,0)$; the Minkowski metric in the site’s $(+---)$ convention has $(p,q)=(1,3)$.

An indefinite nondegenerate form still defines orthogonality,

$$v\perp_B w \quad\Longleftrightarrow\quad B(v,w)=0,$$

but it does not define a norm. Nonzero null vectors can satisfy $B(v,v)=0$, and $B(v,v)$ can be negative. Thus the expression $\sqrt{B(v,v)}$ is not a real length on all of $V$.

For the Lorentz form

$$\eta=\operatorname{diag}(1,-1,-1,-1), \qquad v\cdot w=\eta_{\mu\nu}v^\mu w^\nu,$$

the form implements the explicit identification

$$v_\mu=\eta_{\mu\nu}v^\nu.$$

Raising or lowering an index is therefore an application of $\eta$ or its inverse, not a change in typography.

Hermitian forms and the physics convention

On a complex vector space, a Hermitian form satisfies

$$\langle v|w\rangle = \langle w|v\rangle^*,$$

with conjugate-linearity in the bra and linearity in the ket:

$$\begin{aligned} \langle av+bv'|w\rangle &= a^*\langle v|w\rangle +b^*\langle v'|w\rangle,\\ \langle v|aw+bw'\rangle &= a\langle v|w\rangle +b\langle v|w'\rangle. \end{aligned}$$

Sources that take the first slot to be linear have every displayed conjugation reversed. Translating both slots together preserves the norm, orthogonality, and adjoint identities below.

The form is nondegenerate when

$$\langle v|w\rangle=0\ \text{for every }v \quad\Longrightarrow\quad w=0.$$

In finite dimension this is equivalent to the analogous first-slot condition and to invertibility of every Gram matrix.

It is positive definite when

$$\langle v|v\rangle>0 \quad\text{for every }v\neq0.$$

A positive Hermitian form is an inner product and defines

$$\|v\|=\sqrt{\langle v|v\rangle}.$$

The Cauchy–Schwarz inequality,

$$|\langle v|w\rangle| \leq \|v\|\,\|w\|,$$

then gives the triangle inequality. Positivity is essential; a nondegenerate indefinite Hermitian form is not a Hilbert inner product.

In a basis $\{e_a\}$,

$$G_{ab}=\langle e_a|e_b\rangle, \qquad \langle v|w\rangle =(v^a)^*G_{ab}w^b.$$

The matrix $G$ is Hermitian, and it is positive definite exactly when the form is. Under $e'=eM$,

$$G'=M^\dagger GM,$$

where the dagger on a coordinate matrix denotes conjugate transpose.

The finite-dimensional Riesz map is

$$\mathcal R:V\longrightarrow V^*, \qquad \mathcal R(v)=\langle v|.$$

With the convention used here, $\mathcal R$ is conjugate-linear:

$$\mathcal R(av)=a^*\mathcal R(v).$$

This is the precise sense in which an inner product turns a ket into a bra. It is not the linear identification produced by a bilinear form.

Orthogonality and orthonormal bases

For a subspace $S\subseteq V$ of a positive inner-product space, define

$$S^\perp = \{v\in V:\langle s|v\rangle=0 \text{ for every }s\in S\}.$$

Finite-dimensional positivity gives

$$V=S\oplus S^\perp.$$

If $\{|e_a\rangle\}_{a=1}^k$ is an orthonormal basis of $S$, the orthogonal projector is

$$P_S = \sum_{a=1}^k|e_a\rangle\langle e_a|.$$

Indeed,

$$P_S|v\rangle = \sum_a|e_a\rangle\langle e_a|v\rangle, \qquad P_S^2=P_S, \qquad P_S^\dagger=P_S.$$

Starting from an independent list $v_1,\ldots,v_k$, Gram–Schmidt constructs such a basis recursively:

$$|u_j\rangle = |v_j\rangle - \sum_{a<j}|e_a\rangle\langle e_a|v_j\rangle, \qquad |e_j\rangle = \frac{|u_j\rangle}{\|u_j\|}.$$

Independence ensures $u_j\neq0$. In an indefinite space, a nonzero $u_j$ can be null, so this positive-definite algorithm cannot be imported without modification.

Adjoints depend on the chosen forms

Let $T:V\to W$, where $V$ and $W$ carry nondegenerate Hermitian forms. The adjoint is the unique map

$$T^\dagger:W\longrightarrow V$$

such that

$$\langle Tv|w\rangle_W = \langle v|T^\dagger w\rangle_V$$

for every $v\in V$ and $w\in W$. Nondegeneracy gives existence and uniqueness in finite dimension. For positive forms, this is the familiar Hilbert-space adjoint.

If the chosen bases have Gram matrices $G_V$ and $G_W$, then

$$[T^\dagger] = G_V^{-1}[T]^{\dagger_{\mathrm{mat}}}G_W.$$

Only in orthonormal bases, where both Gram matrices are identities, is the adjoint matrix simply the conjugate transpose of $[T]$. Thus “take the dagger” is not a basis-free matrix recipe unless the basis and form are known. Conrad, n.d., §§ 1–3, PDF develops form-dependent adjoints and congruence, while Axler 2024, Chapters 6–7 gives the positive-Hermitian specialization and its spectral consequences.

Adjoints reverse composition and conjugate scalars:

$$(ST)^\dagger=T^\dagger S^\dagger, \qquad (aT+bR)^\dagger = a^*T^\dagger+b^*R^\dagger.$$

An endomorphism $A:V\to V$ is Hermitian or self-adjoint in this finite-dimensional setting when $A^\dagger=A$, and anti-Hermitian when $A^\dagger=-A$.

For nondegenerate bilinear forms $B_V$ and $B_W$, the bilinear adjoint $T^\sharp:W\to V$ is instead defined by

$$B_W(Tv,w)=B_V(v,T^\sharp w).$$

Its matrix is

$$[T^\sharp] = G_V^{-1}[T]^{\mathsf T}G_W,$$

with no complex conjugation. If a form is degenerate, an adjoint may fail to exist or fail to be unique; the inverse Gram matrices in these formulas do not exist.

Isometries, orthogonal maps, and unitary maps

A map $U:V\to W$ is an isometry of Hermitian spaces when

$$\langle Uv|Uw\rangle_W = \langle v|w\rangle_V.$$

Equivalently,

$$U^\dagger U=\mathbb 1_V.$$

An isometry is injective. When $V$ and $W$ have the same finite dimension, it is invertible and

$$U^{-1}=U^\dagger.$$

An isometric endomorphism of a complex positive inner-product space is unitary. In a real positive inner-product space it is orthogonal. For a real form of signature $(p,q)$, choose a basis in which

$$G=\operatorname{diag}(\mathbb 1_p,-\mathbb 1_q).$$

The group preserving this standard form is

$$O(p,q) = \{A\in GL_{p+q}(\mathbb R):A^{\mathsf T}GA=G\}.$$

Lorentz transformations satisfy

$$\Lambda^{\mathsf T}\eta\Lambda=\eta.$$

They preserve an indefinite spacetime form and should not be confused with unitary maps on a positive state space.

QFT-facing checks

Finite-dimensional unitary evolution

The bra–ket and unitary conventions in this check agree with MIT OpenCourseWare 2017, Lectures 2–3.

Let $H^\dagger=H$ on a finite-dimensional positive inner-product space and

$$U(t)=e^{-itH}.$$

Then

$$U(t)^\dagger=e^{+itH}, \qquad U(t)^\dagger U(t)=\mathbb 1.$$

Equivalently, if a state obeys

$$i\frac{d}{dt}|\psi(t)\rangle = H|\psi(t)\rangle,$$

then

$$\begin{aligned} \frac{d}{dt}\langle\psi(t)|\psi(t)\rangle &= \langle\dot\psi|\psi\rangle +\langle\psi|\dot\psi\rangle\\ &= i\langle\psi|H^\dagger|\psi\rangle -i\langle\psi|H|\psi\rangle =0. \end{aligned}$$

This calculation isolates the algebraic role of the adjoint. For an unbounded Hamiltonian, self-adjointness, its domain, and the existence of the unitary group require separate theorems.

Polarization completeness

The Lorentz-form and polarization conventions translated below are those of Tong 2006–2007, §§ 1.2 and 6.2.

For nonzero spatial momentum $\mathbf p$, choose two orthonormal transverse polarizations $\boldsymbol\epsilon_r(\mathbf p)$, $r=1,2$, satisfying

$$\boldsymbol\epsilon_r^*\cdot \boldsymbol\epsilon_s=\delta_{rs}, \qquad \mathbf p\cdot\boldsymbol\epsilon_r=0.$$

Their completeness relation is the orthogonal projector onto the plane perpendicular to $\mathbf p$:

$$\sum_{r=1}^2 \epsilon_r^i(\mathbf p) \epsilon_r^{j*}(\mathbf p) = \delta^{ij} - \frac{p^ip^j}{|\mathbf p|^2}.$$

This is the same basis-independent projector formula $P_S=\sum_r|e_r\rangle\langle e_r|$ written in components.

A Lorentz-covariant four-polarization basis instead uses the indefinite metric $\eta_{\lambda\lambda'}$ in its completeness relation. Timelike and longitudinal gauge modes are not thereby positive-norm physical states. Separating the auxiliary covariant space from the positive physical state space is part of the developed treatment on Hilbert Positivity and Unitary Evolution.

Common pitfalls

Equating transpose with adjoint. A transpose represents the dual map in dual bases. A Hermitian adjoint also uses chosen forms and complex conjugation; in non-orthonormal bases the Gram matrices are essential.

Calling every nondegenerate form an inner product. In physics one often speaks informally of an “indefinite inner product.” It defines a nondegenerate Hermitian pairing, but not a positive norm and not a Hilbert space.

Forgetting which slot is linear. Here the bra is conjugate-linear and the ket is linear. A mathematics source using the opposite convention must have its scalar conjugations translated before adjoint formulas are copied.

Raising an index without naming the form. The map $V\to V^*$ is supplied by $G$. Different nondegenerate forms give different lowered covectors and different adjoints.

Extending matrix self-adjointness to unbounded operators. In finite dimension, Hermitian and self-adjoint coincide. For an unbounded operator, symmetry on a dense domain is weaker than self-adjointness, and the domain is part of the operator.

Exercises

1. On $\mathbb R^2$, let

   $$G= \begin{pmatrix} 1&0\\ 0&2 \end{pmatrix}, \qquad T= \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}.$$

   Compute the adjoint of $T$ relative to $B(v,w)=v^{\mathsf T}Gw$.

   Solution

   The bilinear-adjoint formula gives

   $$T^\sharp = G^{-1}T^{\mathsf T}G = \begin{pmatrix} 0&2\\ \tfrac12&0 \end{pmatrix}.$$

   It differs from $T^{\mathsf T}=T$ because the standard basis is not orthonormal for this form.

2. In $1+1$ dimensions, verify that

   $$\Lambda(\chi) = \begin{pmatrix} \cosh\chi&\sinh\chi\\ \sinh\chi&\cosh\chi \end{pmatrix}$$

   preserves $\eta=\operatorname{diag}(1,-1)$.

   Solution

   Direct multiplication and $\cosh^2\chi-\sinh^2\chi=1$ give

   $$\Lambda(\chi)^{\mathsf T} \eta \Lambda(\chi) = \eta.$$

   Thus the boost is an isometry of the Lorentz form. It is not orthogonal for the Euclidean matrix $\mathbb 1$ unless $\chi=0$.

3. Let $P=P^\dagger=P^2$. Show that $\operatorname{im}P$ and $\ker P$ are orthogonal and that $V=\operatorname{im}P\oplus\ker P$.

   Solution

   If $x=Pu$ and $y\in\ker P$, then

   $$\langle x|y\rangle = \langle Pu|y\rangle = \langle u|P^\dagger y\rangle = 0.$$

   Also $v=Pv+(\mathbb 1-P)v$, with the first term in $\operatorname{im}P$ and the second in $\ker P$. If $z\in\operatorname{im}P\cap\ker P$, write $z=Px$; then $z=P^2x=Pz=0$. Hence the sum is direct.

References

• Sheldon Axler, Linear Algebra Done Right, 4th ed., Springer, 2024, Chapters 6 and 7, for inner-product spaces, orthogonality, adjoints, and isometries. Axler’s inner-product slot convention is translated to the bra–ket convention used here.
• Keith Conrad, Bilinear Forms, PDF, University of Connecticut lecture notes, n.d., §§ 1–3, for forms as maps to the dual, nondegeneracy, congruence transformations, and bilinear adjoints; Theorem 6.19 and Corollary 6.20 for invariance of real signature.
• MIT OpenCourseWare, Quantum Theory I: Lecture Notes, 8.321, Fall 2017, Lectures 2 and 3, Massachusetts Institute of Technology, for bra–ket inner products, adjoint operators, and unitary transformations.
• David Tong, Lectures on Quantum Field Theory, Cambridge Part III lecture notes, University of Cambridge, 2006–2007, §§ 1.2 and 6.2, for Lorentz-form preservation and polarization completeness.