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

Linear Algebra Problems and Solutions

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.

 
LoadingAdd 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.


LoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

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“. […]

Leave a Reply

Your email address will not be published. Required fields are marked *

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

More in Linear Algebra
Problems and Solutions of Eigenvalue, Eigenvector 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