Skip to main content

Hermitian and skew-Hermitian matrices

Short answer

A matrix is Hermitian if A* = A (conjugate transpose) and skew-Hermitian if A* = −A. Every square matrix A can be uniquely expressed as A = P + Q with P = ½(A+A*) Hermitian and Q = ½(A−A*) skew-Hermitian.

Define Hermitian and skew-Hermitian matrices. Show that every square matrix can be uniquely expressed as the sum of a Hermitian and a skew-Hermitian matrix.

Definitions

Let A* = (Ā)ᵀ be the conjugate transpose of a complex square matrix A.

  • A is Hermitian if A* = A (aᵢⱼ = conj(aⱼᵢ); the diagonal entries are real). Example: [3 2−i; 2+i 5].
  • A is skew-Hermitian if A* = −A (aᵢⱼ = −conj(aⱼᵢ); the diagonal entries are zero or purely imaginary). Example: [2i 1+i; −1+i 0].

Existence

Define P = ½(A + A*) and Q = ½(A − A*). Using (A*)* = A and (kA)* = k̄A* with k = ½ real:

  • P* = ½(A* + A) = P, so P is Hermitian.
  • Q* = ½(A* − A) = −Q, so Q is skew-Hermitian.
  • P + Q = A.

Uniqueness

Suppose A = R + S with R Hermitian and S skew-Hermitian. Taking the conjugate transpose:

A* = R* + S* = R − S

Adding and subtracting with A = R + S:

A + A* = 2R  ⇒  R = ½(A + A*) = P
A − A* = 2S  ⇒  S = ½(A − A*) = Q

So the decomposition is unique.

Example. A = [1 2+i; 3 i]. Then A* = [1 3; 2−i −i]. P = ½(A + A*) = [1 (5+i)/2; (5−i)/2 0] (Hermitian), and Q = ½(A − A*) = [0 (−1+i)/2; (1+i)/2 i] (skew-Hermitian), with P + Q = A. ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics III chapter 1 questions →