Recall that a complex matrix is called Hermitian if $A^*=A$, where $A^*=\bar{A}^{\trans}$.
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.

Since $A$ is Hermitian, we have
\[\bar{A}^{\trans}=A.\]
Taking the conjugate of this identity, we also have
\[A^{\trans}=\bar{A}. \tag{*}\]

Let
\[B=\frac{1}{2}(A+\bar{A})\]
and
\[C=\frac{1}{2i}(A-\bar{A}).\]
We claim that $B$ is a real symmetric matrix and $C$ is a real skew-symmetric matrix.

We have
\begin{align*}
\bar{B}=\frac{1}{2}\overline{(A+\bar{A})}=\frac{1}{2}(\bar{A}+\bar{\bar{A}})=\frac{1}{2}(\bar{A}+A)=B.
\end{align*}
Thus, the matrix $B$ is real. To prove $B$ is symmetric, we compute
\begin{align*}
&B^{\trans}=\frac{1}{2}(A+\bar{A})^{\trans}\\
&=\frac{1}{2}(A^{\trans}+\bar{A}^{\trans})\\
&=\frac{1}{2}(\bar{A}+A) && \text{by (*) and $A$ is Hermitian}\\
&=B.
\end{align*}
This proves that $B$ is symmetric.

The matrix $C$ is real because we have
\begin{align*}
\bar{C}=\frac{1}{-2i}(\bar{A}-\bar{\bar{A}})=\frac{1}{2i}(A-\bar{A})=C.
\end{align*}
We also have
\begin{align*}
&C^{\trans}=\frac{1}{2i}(A^{\trans}-\bar{A}^{\trans})\\
&=\frac{1}{2i}(\bar{A}-A)&& \text{by (*) and $A$ is Hermitian}\\
&=-\frac{1}{2i}(A-\bar{A})=-C.
\end{align*}
Hence $C$ is a skew-symmetric matrix.

Finally, we compute
\begin{align*}
B+iC&=\frac{1}{2}(A+\bar{A})+i\cdot \frac{1}{2i}(A-\bar{A})\\
&=\frac{1}{2}(A+\bar{A})+\frac{1}{2}(A-\bar{A})\\
&=A.
\end{align*}
Therefore, we have obtained the sum as described in the problem.

Related Question.

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

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