5. Adjoints and special operators

All spaces are finite-dimensional over $\mathbb R$ or $\mathbb C$.

In this unit $V$ is finite-dimensional over $\mathbb R$ or $\mathbb C$, with the inner product linear in the first argument. All operators map $V$ to itself.

The shared proofs write $T^\dagger$ for the Hermitian adjoint; in this chapter $T^*$ denotes exactly the same operator. We specialize their maps to endomorphisms of $V$. This is not the algebraic dual map, whose coordinate matrix is an ordinary transpose.

Every linear functional $\varphi$ can be written uniquely as $\varphi(v)=\langle v,w\rangle$ for a vector $w$. In an orthonormal basis $q_j$, take $w=\sum_j\overline{\varphi(q_j)}q_j$ and check on basis vectors. This representation depends on the inner product. Over $\mathbb C$ the correspondence $w\mapsto\langle\cdot,w\rangle$ is conjugate-linear, not a complex-linear identification with the algebraic dual.

For $T:V\to W$, its adjoint $T^\dagger:W\to V$ is uniquely defined by

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

Functional representation proves existence and uniqueness for each $w$; conjugations in the convention show $T^\dagger$ is linear. In orthonormal bases its matrix is $A^*=\bar A^{\mathsf T}$. By contrast, the algebraic dual map of C01 has ordinary transpose in dual bases and goes $W^*\to V^*$. These are related constructions, not identical definitions.

Evaluation in the defining identity gives

$$(ST)^\dagger=T^\dagger S^\dagger,\quad (T^\dagger)^\dagger=T,\quad (\alpha T)^\dagger=\bar\alpha T^\dagger.$$

Moreover

$$(\operatorname{im}T)^\perp=\ker T^\dagger,\qquad \operatorname{im}T^\dagger=(\ker T)^\perp.$$

The first follows because $w$ is perpendicular to every $Tv$ exactly when $T^\dagger w=0$. Apply the first to $T^\dagger$ and take orthogonal complements to obtain the second in finite dimensions.

An endomorphism is self-adjoint if $T=T^\dagger$, unitary (orthogonal over $\mathbb R$) if $T^\dagger T=TT^\dagger=I$, and normal if $T^\dagger T=TT^\dagger$. Self-adjoint and unitary maps are normal. A normal map need be neither.

Choose an orthonormal basis. If $A$ is the matrix of $T$, the matrix of $T^*$ is the conjugate transpose $A^*=\overline{A}^{\mathsf T}$. This identity also proves existence and uniqueness of the adjoint: direct substitution verifies it and the orthonormal coordinates determine it. Moreover $(ST)^*=T^*S^*$ and $(T^*)^*=T$. A unitary operator preserves inner products and norms; in finite dimension norm preservation also implies unitarity (by polarization and injectivity).

In orthonormal column coordinates, the first-linear convention is $\langle x,y\rangle=y^*x$. Thus

$$\langle Ax,y\rangle=y^*Ax=(A^*y)^*x=\langle x,A^*y\rangle.$$

If two proposed adjoints satisfy the defining identity, their difference $B$ has $\langle v,Bw\rangle=0$ for every $v,w$; choosing $v=Bw$ forces $Bw=0$. This proves uniqueness. Also

$$\langle STv,w\rangle=\langle Tv,S^*w\rangle=\langle v,T^*S^*w\rangle,$$

which proves $(ST)^*=T^*S^*$. Conjugate symmetry similarly gives $(T^*)^*=T$.

Here is the norm-to-inner-product step explicitly. Over $\mathbb C$ with the present convention,

$$4\langle v,w\rangle=\|v+w\|^2-\|v-w\|^2 +i\bigl(\|v+iw\|^2-\|v-iw\|^2\bigr).$$

Over $\mathbb R$ omit the imaginary part. A linear norm-preserving map preserves these four norms and hence the inner product. Therefore $\langle v,T^*Tw\rangle=\langle Tv,Tw\rangle=\langle v,w\rangle$ for every $v,w$, so $T^*T=I$. Norm preservation makes $T$ injective; finite-dimensional rank–nullity then makes it invertible, giving $T^*=T^{-1}$ and also $TT^*=I$. No infinite-dimensional surjectivity assertion is being used.

If $S$ is normal, then $\|Sv\|=\|S^\dagger v\|$, because the squares are determined by $S^\dagger S=SS^\dagger$. For normal $T$, the operator $S=T-\lambda I$ is also normal. Thus

$$Tv=\lambda v\quad\Longrightarrow\quad T^\dagger v=\bar\lambda v.$$

If $v$ is an eigenvector, its orthogonal complement is invariant under both $T$ and $T^\dagger$ by the adjoint identity. Their restrictions to that complement remain adjoints and commute, hence are normal.

A self-adjoint eigenvalue is real: $\langle Tv,v\rangle=\langle v,Tv\rangle$ gives $\lambda\|v\|^2=\bar\lambda\|v\|^2$. A unitary eigenvalue has modulus one because norms are preserved. These are necessary spectral properties, not sufficient conditions for an arbitrary matrix to be self-adjoint or unitary.

Proof

Norms and eigenvectors

The norm identity follows from $\|Tv\|^2=\langle T^*Tv,v\rangle$ and $\|T^*v\|^2=\langle TT^*v,v\rangle$. $T-\lambda I$ is normal, so its norm identity shows that $(T-\lambda I)v=0$ implies $(T^*-\overline\lambda I)v=0$. If $T=T^*$ and $Tv=\lambda v$ with $v\ne0$, the preceding result gives $\lambda=\overline\lambda$. If $Tu=\lambda u$, $Tv=\mu v$ and $T$ is self-adjoint, then $\lambda\langle u,v\rangle=\langle Tu,v\rangle=\langle u,Tv\rangle=\overline\mu\langle u,v\rangle$. Real distinct eigenvalues force orthogonality.

Worked example

A normal operator need not be self-adjoint

On $\mathbb C^2$, $A=\operatorname{diag}(i,1)$ has $A^*=\operatorname{diag}(-i,1)$ and $AA^*=A^*A=I$. It is unitary and normal, but not self-adjoint. Thus normality is weaker than self-adjointness.

Choosing a method and avoiding mistakes

For a matrix in an orthonormal basis compute $A^*$ and compare products; for a conceptual argument use the defining inner-product identity. The transpose alone is wrong over $\mathbb C$. In a non-orthonormal basis conjugate transpose is generally not the adjoint matrix. Orthogonal eigenvectors for distinct eigenvalues alone do not prove normality unless an orthonormal eigenbasis spans the whole space.

Exercises

1. Determine whether $A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ is self-adjoint, normal, or orthogonal over $\mathbb R$.
2. Let $P$ be orthogonal projection onto $U$. Prove $P=P^*$ and $P^2=P$.
3. Show that an eigenvalue of a unitary operator has absolute value $1$.

Full solutions

1. $A^*=A^{\mathsf T}=-A\ne A$. Since $A^{\mathsf T}A=AA^{\mathsf T}=I$, it is orthogonal and hence normal, but not self-adjoint.
2. Write $v=u+u_\perp$ and $w=z+z_\perp$ in $U\oplus U^\perp$. Then $Pv=u$, $Pw=z$ and $\langle Pv,w\rangle=\langle u,z\rangle=\langle v,Pw\rangle$. Uniqueness of the adjoint gives $P=P^*$; projecting $u$ again gives $P^2v=Pv$.
3. If $Tv=\lambda v$ for nonzero $v$, norm preservation gives $\|v\|=\|Tv\|=|\lambda|\|v\|$. Divide by the positive norm.

Connection

The spectral theorem identifies exactly which complex operators possess orthonormal eigenbases: the normal ones. Over the reals, symmetric operators have such bases.