Expressing Matrix as Sum of Hermitian and Skew-Hermitian Matrices

Introduction

A complex square matrix $A$ can be analyzed using its conjugate transpose, denoted as $A^\theta$ (or $A^*$), which is obtained by taking the transpose of the matrix and replacing each entry with its complex conjugate. By definition, a matrix is Hermitian if $A^\theta = A$, and skew-Hermitian if $A^\theta = -A$. Every square matrix with complex entries can be uniquely expressed as the sum of a Hermitian matrix and a skew-Hermitian matrix.

1. Theoretical Foundation and Existence

Let $A$ be any square complex matrix. We can algebraically decompose $A$ as:

$$A = \frac{1}{2}(A + A^\theta) + \frac{1}{2}(A - A^\theta)$$

Let us define:

• Hermitian component: $P = \frac{1}{2}(A + A^\theta)$
• Skew-Hermitian component: $Q = \frac{1}{2}(A - A^\theta)$

Thus, $A = P + Q$.

2. Verification of Matrix Properties

Applying the conjugate transpose operator $()^\theta$, and using the properties $(X + Y)^\theta = X^\theta + Y^\theta$, $(kA)^\theta = \bar{k}A^\theta$, and $(A^\theta)^\theta = A$:

• Verification for $P$ (Hermitian):

  $$P^\theta = \left[\frac{1}{2}(A + A^\theta)\right]^\theta = \frac{1}{2}(A^\theta + (A^\theta)^\theta) = \frac{1}{2}(A^\theta + A) = P$$

  Since $P^\theta = P$, the matrix $P$ is Hermitian.

• Verification for $Q$ (Skew-Hermitian):

  $$Q^\theta = \left[\frac{1}{2}(A - A^\theta)\right]^\theta = \frac{1}{2}(A^\theta - (A^\theta)^\theta) = \frac{1}{2}(A^\theta - A) = -\frac{1}{2}(A - A^\theta) = -Q$$

  Since $Q^\theta = -Q$, the matrix $Q$ is skew-Hermitian.

3. Uniqueness of Decomposition

Assume $A = R + S$, where $R^\theta = R$ and $S^\theta = -S$. Taking the conjugate transpose yields:

$$A^\theta = (R + S)^\theta = R^\theta + S^\theta = R - S$$

• Solving for $R$: Adding $A$ and $A^\theta$ gives $A + A^\theta = 2R \implies R = \frac{1}{2}(A + A^\theta) = P$.
• Solving for $S$: Subtracting $A^\theta$ from $A$ gives $A - A^\theta = 2S \implies S = \frac{1}{2}(A - A^\theta) = Q$.

This establishes that the decomposition is unique.

4. Application to the Given Matrix

Given the matrix:

$$A = \begin{bmatrix} 1+i & 2 & 5-5i \\ 2i & 2+i & 4+2i \\ -1+i & -4 & 7 \end{bmatrix}$$

The conjugate transpose $A^\theta = (\bar{A})^T$ is:

$$A^\theta = \begin{bmatrix} 1-i & -2i & -1-i \\ 2 & 2-i & -4 \\ 5+5i & 4-2i & 7 \end{bmatrix}$$

Calculating the Hermitian component $P = \frac{1}{2}(A + A^\theta)$:

$$P = \begin{bmatrix} 1 & 1-i & 2-3i \\ 1+i & 2 & i \\ 2+3i & -i & 7 \end{bmatrix}$$

Calculating the skew-Hermitian component $Q = \frac{1}{2}(A - A^\theta)$:

$$Q = \begin{bmatrix} i & 1+i & 3-2i \\ -1+i & i & 4+i \\ -3-2i & -4+i & 0 \end{bmatrix}$$

Hence, $A = P + Q$, where $P$ is Hermitian ($P^\theta = P$) and $Q$ is skew-Hermitian ($Q^\theta = -Q$).

Conclusion

Any square matrix with complex entries can be uniquely decomposed into symmetric and anti-symmetric conjugate parts. The computed matrices $P$ and $Q$ satisfy $A = P + Q$ with real diagonal elements for the Hermitian component and purely imaginary or zero diagonal elements for the skew-Hermitian component, demonstrating the fundamental decomposition theorem of linear algebra.