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33. Adv: Hermitian and unitary matrices (complex case)

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§21.0's Spectral Theorem was stated for real symmetric matrices. This lesson extends every idea in it to complex vector spaces, where transpose alone stops being the right operation — it needs one small but essential upgrade, conjugation, and everything else follows the same pattern as before.

The conjugate transpose

For a complex matrix A, the conjugate transpose (or Hermitian adjoint) is A^*=\bar A^T — transpose, then conjugate every entry (or conjugate then transpose; the two commute). For a real matrix, conjugation does nothing, so A^*=A^T exactly — every definition below reduces to its §16–20 counterpart whenever A is real, which is the guiding check for all of this lesson.

Why plain transpose stops being enough: §20.0's dot product \vec u\cdot\vec u=\sum u_i^2 can be negative for complex \vec u (e.g. \vec u=(i,0) gives u_1^2=i^2=-1), breaking positive definiteness (§20.1) outright. The fix is the complex (Hermitian) inner product \langle\vec u,\vec v\rangle=\vec u^*\vec v=\sum\bar u_iv_i, for which \langle\vec v,\vec v\rangle=\sum|v_i|^2\ge0 always — genuinely nonnegative, restoring §20.1's axiom 3, with equality only at \vec v=\vec0.

Hermitian matrices

A is Hermitian if A^*=A — the direct complex analogue of symmetric (A^T=A). The Spectral Theorem holds for Hermitian matrices exactly as stated in §21.0: real eigenvalues, and an orthonormal (now with respect to the Hermitian inner product) eigenbasis, giving A=Q\Lambda Q^* (conjugate transpose replacing plain transpose throughout). Every proof from §21.0 carries over by replacing \cdot with \langle\cdot,\cdot\rangle and ^T with ^* verbatim.

A genuinely real matrix that's symmetric is automatically Hermitian (since A^*=A^T=A when every entry is real) — so §21.0's theorem is literally the special case of this one restricted to real entries, not a separate result needing its own proof.

Unitary matrices

Q is unitary if Q^*Q=I — the complex analogue of an orthogonal matrix (Q^TQ=I, §20.4). Unitary matrices preserve the Hermitian inner product exactly as orthogonal matrices preserve the dot product: \langle Q\vec u,Q\vec v\rangle=(Q\vec u)^*(Q\vec v)=\vec u^*Q^*Q\vec v=\vec u^*\vec v=\langle\vec u,\vec v\rangle — lengths and angles both unchanged. Q^{-1}=Q^* for unitary Q, exactly matching §20.4's P^{-1}=P^T for orthogonal P.

A real orthogonal matrix is automatically unitary (same reasoning as above), so "unitary" is the correct general term whenever complex entries might appear — §18.2's rotation matrices are real orthogonal matrices, and hence also unitary, without any extra work.

Doing it in Python

import numpy as np

A = np.array([[2, 1+1j], [1-1j, 3]])   # Hermitian: A* = A

print("A* == A:", np.allclose(A.conj().T, A), " (Hermitian)")

eigvals = np.linalg.eigvalsh(A)   # eigvalsh: for Hermitian matrices
print("eigenvalues (should be real):", [round(float(v), 4) for v in eigvals])

A* == A: True  (Hermitian)
eigenvalues (should be real): [1.0, 4.0]

A unitary matrix — a complex rotation-like transformation — checked directly:

import numpy as np

theta = np.pi / 4
Q = np.array([[np.cos(theta), -np.sin(theta)],
              [np.sin(theta),  np.cos(theta)]]).astype(complex)   # real orthogonal -> unitary

print("Q* Q == I:", bool(np.allclose(Q.conj().T @ Q, np.eye(2))))

# A genuinely complex unitary matrix
U = np.array([[1, 1j], [1j, 1]]) / np.sqrt(2)
print("U* U == I:", bool(np.allclose(U.conj().T @ U, np.eye(2))))

Q* Q == I: True
U* U == I: True

Worked example

Verify A=\begin{pmatrix}1&2-i\\2+i&3\end{pmatrix} is Hermitian, and find its eigenvalues.

A^*: transpose gives \begin{pmatrix}1&2+i\\2-i&3\end{pmatrix}, then conjugate every entry: \begin{pmatrix}1&2-i\\2+i&3\end{pmatrix}=A ✓ Hermitian (the diagonal entries 1,3 are real, automatically self-conjugate, and the off-diagonal entries are conjugates of each other — the general pattern for any Hermitian matrix).

\operatorname{tr}A=4. \det A=1(3)-(2-i)(2+i)=3-(4-i^2)=3-(4+1)=3-5=-2 (using (2-i)(2+i)=4-i^2=4+1=5, the standard complex-conjugate-product identity). Characteristic equation: \lambda^2-4\lambda-2=0\Rightarrow\lambda=\frac{4\pm\sqrt{16+8}}2=2\pm\sqrt6.

\boxed{\lambda=2+\sqrt6\approx4.449,\quad\lambda=2-\sqrt6\approx-0.449}

Sanity check. Both eigenvalues are real, exactly as the extended Spectral Theorem guarantees for a Hermitian matrix, despite A itself having genuinely complex (non-real) entries — the theorem's promise holds regardless of whether the entries look real. Sum: (2+\sqrt6)+(2-\sqrt6)=4=\operatorname{tr}A ✓; product: (2+\sqrt6)(2-\sqrt6)=4-6=-2=\det A ✓.

Your turn

1. Is A=\begin{pmatrix}0&i\\-i&0\end{pmatrix} Hermitian? Compute A^* directly.

2. True or false: a real orthogonal matrix is always unitary.

3. For unitary Q, what is \det Q's magnitude, |\det Q|? (Hint: use Q^*Q=I and \det(Q^*)=\overline{\det Q}.)

Solutions

1. Yes, Hermitian. Transpose: \begin{pmatrix}0&-i\\i&0\end{pmatrix}. Conjugate every entry: \begin{pmatrix}0&i\\-i&0\end{pmatrix}=A ✓. (This is one of the Pauli matrices, foundational in quantum mechanics — Hermitian matrices represent physical observables there precisely because their eigenvalues are guaranteed real, matching measurable quantities.)

2. True. Shown directly in this lesson: for real Q, conjugation does nothing, so Q^*=Q^T, and Q^TQ=I (orthogonal, §20.4) is literally the same equation as Q^*Q=I (unitary) — no additional condition to check.

3. |\det Q|=1. \det(Q^*Q)=\det I=1. \det(Q^*)\det(Q)=1, and \det(Q^*)=\overline{\det Q} (conjugating every entry conjugates the determinant, which is a polynomial in the entries), so \overline{\det Q}\cdot\det Q=|\det Q|^2=1\Rightarrow|\det Q|=1. (This matches §18.2's real case exactly: a rotation matrix has \det=\pm1, both of magnitude 1.)

Check yourself in code

Verify A=\begin{pmatrix}4&2-2i\\2+2i&1\end{pmatrix} is Hermitian and find its eigenvalues (rounded to 4 decimal places).

Print exactly this:

Hermitian: True
eigenvalues: [-0.7016, 5.7016]

import numpy as np

A = np.array([[4, 2-2j], [2+2j, 1]])
print("Hermitian:", bool(np.allclose(A.conj().T, A)))
# print the eigenvalues (via np.linalg.eigvalsh), rounded to 4 decimal places

import numpy as np

A = np.array([[4, 2-2j], [2+2j, 1]])
print("Hermitian:", bool(np.allclose(A.conj().T, A)))

eigvals = np.linalg.eigvalsh(A)
print("eigenvalues:", [round(float(v), 4) for v in eigvals])

The conjugate transpose A^*=\bar A^T is the complex-safe replacement for plain transpose: Hermitian (A^*=A) generalizes symmetric, unitary (Q^*Q=I) generalizes orthogonal, and the Spectral Theorem carries over verbatim — real eigenvalues, orthonormal eigenbasis. Every real symmetric/orthogonal matrix is automatically Hermitian/unitary, so nothing from §21.0 is lost, only extended.

Next, back to the real case with a different structural question: which symmetric matrices have positive eigenvalues, and why that turns out to matter enormously in optimization and statistics.

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