Self-Adjoint Operators and Real Eigenvalues

Self-Adjointness in a Complex Inner-Product Space

On a complex vector space, the adjoint of an operator captures how that operator interacts with the inner product. An operator that equals its adjoint has a strong constraint on its eigenvalues: any eigenvalue it has must be real. Its eigenvectors belonging to different eigenvalues also have a precise geometric relation: they are orthogonal. These facts follow directly from the adjoint identity and the convention that the inner product is linear in its first slot and conjugate-linear in its second.

Let \(V\) be a complex inner-product space, with inner product \(\langle\mathbf{v},\mathbf{w}\rangle\) linear in \(\mathbf{v}\) and conjugate-linear in \(\mathbf{w}\). For a linear operator \(T:V\to V\), its adjoint \(T^*\) satisfies \(\langle T\mathbf{v},\mathbf{w}\rangle=\langle\mathbf{v},T^*\mathbf{w}\rangle\) for all \(\mathbf{v},\mathbf{w}\in V\). An eigenvalue \(\lambda\in\mathbb{C}\) of \(T\) is a scalar for which \(T\mathbf{v}=\lambda\mathbf{v}\) for some nonzero \(\mathbf{v}\in V\); such a \(\mathbf{v}\) is an eigenvector. These definitions do not require \(V\) to be finite-dimensional.

For a matrix, the adjoint depends on the inner product used for coordinates. In the standard inner product on \(\mathbb{C}^n\), \(\langle\mathbf{x},\mathbf{y}\rangle=\sum_{j=1}^n x_j\overline{y_j}\), the matrix of the adjoint of \(\mathbf{A}\) is \(\mathbf{A}^*\). More generally, if coordinates are taken in an orthonormal basis, the matrix of \(T^*\) is the conjugate transpose of the matrix of \(T\). This is why a self-adjoint operator has a Hermitian matrix in an orthonormal basis; a matrix in an arbitrary basis need not be Hermitian.

Proof. Suppose \(T\mathbf{v}=\lambda\mathbf{v}\) for an eigenvector \(\mathbf{v}\ne\mathbf{0}\). Self-adjointness and the adjoint identity give \(\langle T\mathbf{v},\mathbf{v}\rangle=\langle\mathbf{v},T\mathbf{v}\rangle\). The left side is \(\lambda\langle\mathbf{v},\mathbf{v}\rangle\), because the first slot is linear. The right side is \(\overline{\lambda}\langle\mathbf{v},\mathbf{v}\rangle\), because the second slot is conjugate-linear. Thus \((\lambda-\overline{\lambda})\langle\mathbf{v},\mathbf{v}\rangle=0\). Since \(\mathbf{v}\ne\mathbf{0}\), positive definiteness gives \(\langle\mathbf{v},\mathbf{v}\rangle>0\). It follows that \(\lambda=\overline{\lambda}\), so \(\lambda\in\mathbb{R}\).

Now let \(T\mathbf{v}=\lambda\mathbf{v}\) and \(T\mathbf{w}=\mu\mathbf{w}\), where both vectors are nonzero and \(\lambda\ne\mu\). By the first part, \(\lambda,\mu\in\mathbb{R}\). Self-adjointness gives \(\langle T\mathbf{v},\mathbf{w}\rangle=\langle\mathbf{v},T\mathbf{w}\rangle\). Substituting the eigenvalue equations yields \(\lambda\langle\mathbf{v},\mathbf{w}\rangle=\overline{\mu}\langle\mathbf{v},\mathbf{w}\rangle=\mu\langle\mathbf{v},\mathbf{w}\rangle\). Since \(\lambda-\mu\ne0\), this forces \(\langle\mathbf{v},\mathbf{w}\rangle=0\). Therefore \(\mathbf{v}\perp\mathbf{w}\). \(\square\)

Recognizing a Hermitian Matrix

For a matrix, taking the conjugate transpose means first conjugating every entry and then transposing. Consequently, \(\mathbf{A}=\mathbf{A}^*\) requires real diagonal entries and entries on opposite sides of the diagonal that are complex conjugates of one another. For a real matrix, this condition reduces to symmetry: \(\mathbf{A}=\mathbf{A}^{\top}\).

Proof. Write \(\mathbf{A}=(a_{ij})\) and use \(\langle\mathbf{x},\mathbf{y}\rangle=\sum_i x_i\overline{y_i}\). For any \(\mathbf{x},\mathbf{y}\in\mathbb{C}^n\), expanding the left side of the adjoint identity gives \(\langle\mathbf{A}\mathbf{x},\mathbf{y}\rangle=\sum_{i,j}a_{ij}x_j\overline{y_i}\). Expanding the other side with \(\mathbf{A}^*\) gives \(\langle\mathbf{x},\mathbf{A}^*\mathbf{y}\rangle=\sum_{i,j}x_j\overline{(\mathbf{A}^*)_{ji}}=\sum_{i,j}x_j\overline{a_{ij}}\) only after matching indices carefully; equivalently, \((\mathbf{A}^*)_{ji}=\overline{a_{ij}}\), so conjugating that entry gives \(a_{ij}\), and the expansion is \(\sum_{i,j}x_j a_{ij}\overline{y_i}\). These expressions agree for every \(\mathbf{x},\mathbf{y}\), so \(\mathbf{A}^*\) is the adjoint matrix. Thus the operator is self-adjoint exactly when its matrix equals its adjoint matrix. \(\square\)

The coordinate calculation also explains why complex conjugation cannot be omitted. A Hermitian matrix need not be symmetric as an ordinary transpose, even though it represents a self-adjoint operator. The examples below use the standard inner product on \(\mathbb{C}^2\), so the inner product is \(\langle(x_1,x_2),(y_1,y_2)\rangle=x_1\overline{y_1}+x_2\overline{y_2}\).

Worked Hermitian-Matrix Examples

What the Conclusions Do—and Do Not—Say

The theorem is conditional: if an eigenvalue exists, it is real. It does not say that every operator on an infinite-dimensional space has an eigenvalue. It also says that eigenvectors are orthogonal when their eigenvalues are distinct; it does not say that every pair of eigenvectors is orthogonal. In particular, it gives no orthogonality conclusion for two vectors in the same eigenspace. A repeated eigenvalue may have many eigenvectors, and the theorem alone does not require every such pair to be orthogonal.

Common pitfall: forgetting conjugate-linearity. A mistaken eigenvalue proof may write \(\langle\mathbf{v},\lambda\mathbf{v}\rangle=\lambda\langle\mathbf{v},\mathbf{v}\rangle\). That is false under the stated convention. For the concrete vector \(\mathbf{e}_1=(1,0)\) and scalar \(i\), the correct calculation is \(\langle\mathbf{e}_1,i\mathbf{e}_1\rangle=1\cdot\overline{i}=-i\), not \(i\). The correct factor is \(\overline{\lambda}\) because the scalar multiplies the second slot. Using that factor gives \(\lambda\langle\mathbf{v},\mathbf{v}\rangle=\overline{\lambda}\langle\mathbf{v},\mathbf{v}\rangle\), which is the step that forces an eigenvalue of a self-adjoint operator to be real.

A related matrix pitfall is to test symmetry using \(\mathbf{A}^{\top}\) instead of Hermitian symmetry using \(\mathbf{A}^*\). For example, \(\begin{bmatrix}0&i\\-i&0\end{bmatrix}\) is not equal to its transpose, but it is equal to its conjugate transpose. It is therefore Hermitian, and the self-adjoint results apply. In a real inner-product space, by contrast, the adjoint matrix is the transpose, so self-adjoint matrices are symmetric.

These results let us check important structural claims without solving every eigenvector problem in advance: self-adjointness rules out nonreal eigenvalues, and distinct eigenspaces are mutually orthogonal. They do not by themselves establish that an operator has enough eigenvectors to form a basis, or that it can be diagonalized. Those are stronger questions requiring additional results.