(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.

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.

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

The post Every Complex Matrix Can Be Written as $A=B+iC$, where $B, C$ are Hermitian Matrices first appeared on Problems in Mathematics.

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.

Proof.

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.

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“.

The post Express a Hermitian Matrix as a Sum of Real Symmetric Matrix and a Real Skew-Symmetric Matrix first appeared on Problems in Mathematics.