# Every Complex Matrix Can Be Written as $A=B+iC$, where $B, C$ are Hermitian Matrices ## Problem 425

(a) Prove that each complex $n\times n$ matrix $A$ can be written as
$A=B+iC,$ where $B$ and $C$ are Hermitian matrices.

(b) Write the complex matrix
$A=\begin{bmatrix} i & 6\\ 2-i& 1+i \end{bmatrix}$ as a sum $A=B+iC$, where $B$ and $C$ are Hermitian matrices. Add to solve later

## Definition (Hermitian matrix).

Recall that a complex matrix $M$ is said to be Hermitian if $M^*=M$.
Here $A^*$ is the conjugate transpose matrix $M^*=\bar{M}^*$.

## Proof.

Let
$B=\frac{A+A^*}{2} \text{ and } C=\frac{A-A^*}{2i}.$ We claim that $B$ and $C$ are Hermitian matrices.
Using the fact that $(A^*)^*=A$, we compute
\begin{align*}
B^*&=\left(\, \frac{A+A^*}{2} \,\right)^*\\
&=\frac{A^*+(A^*)^*}{2}\\
&=\frac{A^*+A}{2}=B.
\end{align*}
It yields that the matrix $B$ is Hermitian.

We also have
\begin{align*}
C^*&=\left(\, \frac{A-A^*}{2i} \,\right)^*\\
&=\frac{A^*-(A^*)^*}{-2i}\\
&=\frac{A^*-A}{-2i}\\
&=\frac{A-A^*}{2i}=C.
\end{align*}
Thus, the matrix $C$ is also Hermitian.

Finally, note that we have
\begin{align*}
B+iC&=\frac{A+A^*}{2}+i\frac{A-A^*}{2i}\\
&=\frac{A+A^*}{2}+\frac{A-A^*}{2}\\
&=A.
\end{align*}
Therefore, each complex matrix $A$ can be written as $A=B+iC$, where $B$ and $C$ are Hermitian matrices.

\item By the proof of part (a), it suffices to compute
$B=\frac{A+A^*}{2} \text{ and } C=\frac{A-A^*}{2i}.$

We have
$A^*=\begin{bmatrix} -i & 2+i\\ 6& 1-i \end{bmatrix}.$

A direct computation yields that
$B=\begin{bmatrix} 0 & 4+\frac{i}{2}\\[6pt] 4-\frac{i}{2}& 1 \end{bmatrix} \text{ and } C=\begin{bmatrix} 1 & -\frac{1}{2}-2i\\[6pt] -\frac{1}{2}+2i& 1 \end{bmatrix}.$

By the result of part (a), these matrices are Hermitian and satisfy $A=B+iC$, as required.

## Related Question.

Problem. Prove that every Hermitian matrix $A$ can be written as the sum
$A=B+iC,$ where $B$ is a real symmetric matrix and $C$ is a real skew-symmetric matrix.

See the post “Express a Hermitian matrix as a sum of real symmetric matrix and a real skew-symmetric matrix” for a proof. Add to solve later

### 1 Response

1. 05/22/2017

[…] For a proof of this problem, see the post “Every complex matrix can be written as $A=B+iC$, where $B, C$ are Hermitian matrices“. […]

This site uses Akismet to reduce spam. Learn how your comment data is processed.

###### More in Linear Algebra ##### If Two Matrices Have the Same Eigenvalues with Linearly Independent Eigenvectors, then They Are Equal

Let $A$ and $B$ be $n\times n$ matrices. Suppose that $A$ and $B$ have the same eigenvalues $\lambda_1, \dots, \lambda_n$...

Close