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Symmetric plus skew-symmetric decomposition

Short answer

The transpose Aᵀ is obtained by interchanging rows and columns. Any square matrix A can be uniquely expressed as P + Q, where P = ½(A + Aᵀ) is symmetric and Q = ½(A − Aᵀ) is skew-symmetric. Uniqueness follows by solving A = R + S with Rᵀ = R and Sᵀ = −S.

Define the transpose of a matrix. Prove that every square matrix can be expressed as a sum of a symmetric matrix and a skew-symmetric matrix uniquely.

The transpose of a matrix A = [aᵢⱼ] of order m × n is the matrix Aᵀ of order n × m obtained by interchanging its rows and columns, so (Aᵀ)ᵢⱼ = aⱼᵢ. A matrix is symmetric if Aᵀ = A and skew-symmetric if Aᵀ = −A.

A square matrix A can always be written as A = P + Q, where P = ½(A + Aᵀ) is symmetric and Q = ½(A − Aᵀ) is skew-symmetric, and this representation is unique.

Existence

Let A be any square matrix. Define

P = ½(A + Aᵀ),    Q = ½(A − Aᵀ)
  • Pᵀ = ½(Aᵀ + (Aᵀ)ᵀ) = ½(Aᵀ + A) = P, so P is symmetric.
  • Qᵀ = ½(Aᵀ − (Aᵀ)ᵀ) = ½(Aᵀ − A) = −Q, so Q is skew-symmetric.
  • P + Q = ½(A + Aᵀ) + ½(A − Aᵀ) = A.

Hence A is the sum of a symmetric and a skew-symmetric matrix.

Uniqueness

Suppose A = R + S is another such decomposition, where R is symmetric (Rᵀ = R) and S is skew-symmetric (Sᵀ = −S). Take the transpose of A:

Aᵀ = Rᵀ + Sᵀ = R − S

Now add and subtract A = R + S and Aᵀ = R − S:

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

So R = P and S = Q. The decomposition is unique.

Example. For A = [1 2; 4 3]: Aᵀ = [1 4; 2 3], P = [1 3; 3 3], Q = [0 −1; 1 0], and P + Q = A.

Therefore every square matrix is uniquely the sum of a symmetric and a skew-symmetric matrix. ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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