Definition.
Complex matrices
Example.
Some complex matrices
Here are some examples of complex matrices:
$$\boldsymbol{A}_1=
\begin{pmatrix}
2+3i&4i\\5&1+2i\\6i&8+9i
\end{pmatrix},\;\;\;\;\;
\boldsymbol{A}_2=
\begin{pmatrix}
1&2\\3&4
\end{pmatrix}$$
Note that $\boldsymbol{A}_2$ can be treated as a complex matrix because real numbers are a subset of complex numbers, which means that the entries of $\boldsymbol{A}_2$ are also complex numbers.
Definition.
Complex conjugate of a complex matrix
The complex conjugate of a complex matrix $\boldsymbol{A}$ is a matrix whose entries are the complex conjugatelink of the entries of $\boldsymbol{A}$.
Example.
Finding the complex conjugate of a complex matrix
Find the complex conjugate of the following matrix:
$$\boldsymbol{A}=
\begin{pmatrix}
2+3i&5i\\
4&1+6i
\end{pmatrix}$$
Solution. The complex conjugate of $\boldsymbol{A}$ is:
$$\overline{\boldsymbol{A}}=
\begin{pmatrix}
2-3i&-5i\\
4&1-6i
\end{pmatrix}$$
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Properties of matrix conjugates
Theorem.
Complex conjugate of the product between a real matrix and a complex vector
If $\boldsymbol{A}$ is a real matrix and $\boldsymbol{z}$ is a complex vectorlink, then:
$$\overline{\boldsymbol{Az}}=
\boldsymbol{A}\overline{\boldsymbol{z}}$$
Proof. The product $\boldsymbol{Az}$ is:
$$\begin{align*}
\boldsymbol{Az}&=
\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\smash\ddots&\vdots\\
a_{m1}&a_{m2}&\cdots&a_{mn}
\end{pmatrix}\begin{pmatrix}z_1\\z_2\\\vdots\\z_n\end{pmatrix}\\&=
\begin{pmatrix}a_{11}z_1+a_{12}z_2+\cdots+a_{1n}z_n\\
a_{21}z_1+a_{22}z_2+\cdots+a_{2n}z_n\\\vdots\\
a_{m1}z_1+a_{m2}z_2+\cdots+a_{mn}z_n\end{pmatrix}
\end{align*}$$
The complex conjugate of $\boldsymbol{Az}$ is:
$$\begin{align*}
\overline{\boldsymbol{Az}}&=
\begin{pmatrix}
\overline{a_{11}z_1+a_{12}z_2+\cdots+a_{1n}z_n}\\
\overline{a_{21}z_1+a_{22}z_2+\cdots+a_{2n}z_n}\\\vdots\\
\overline{a_{m1}z_1+a_{m2}z_2+\cdots+a_{mn}z_n}
\end{pmatrix}\\&=\begin{pmatrix}
a_{11}\overline{z_1}+a_{12}\overline{z_2}+\cdots+a_{1n}\overline{z_n}\\
a_{21}\overline{z_1}+a_{22}\overline{z_2}+\cdots+a_{2n}\overline{z_n}\\\vdots\\
a_{m1}\overline{z_1}+a_{m2}\overline{z_2}+\cdots+a_{mn}\overline{z_n}
\end{pmatrix}\\&=
\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\smash\ddots&\vdots\\
a_{m1}&a_{m2}&\cdots&a_{mn}
\end{pmatrix}
\begin{pmatrix}\overline{z_1}\\\overline{z_2}\\\vdots\\\overline{z_n}\end{pmatrix}\\
&=\boldsymbol{A}\overline{\boldsymbol{z}}
\end{align*}$$
Here, the 2nd equality holds by theoremlink. This completes the proof.
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Theorem.
Complex conjugate of a sum of two complex matrices
If $\boldsymbol{A}$ and $\boldsymbol{B}$ are $m\times{n}$ complex matrices, then:
$$\overline{\boldsymbol{A}+\boldsymbol{B}}=
\overline{\boldsymbol{A}}+
\overline{\boldsymbol{B}}$$
Proof. The sum $\boldsymbol{A}+\boldsymbol{B}$ is:
$$\begin{align*}
\boldsymbol{A}+\boldsymbol{B}&=
\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\smash\ddots&\vdots\\
a_{m1}&a_{m2}&\cdots&a_{mn}
\end{pmatrix}+\begin{pmatrix}
b_{11}&b_{12}&\cdots&b_{1n}\\
b_{21}&b_{22}&\cdots&b_{2n}\\
\vdots&\vdots&\smash\ddots&\vdots\\
b_{m1}&b_{m2}&\cdots&b_{mn}
\end{pmatrix}\\
&=\begin{pmatrix}
a_{11}+b_{11}&a_{12}+b_{12}&\cdots&a_{1n}+b_{1n}\\
a_{21}+b_{21}&a_{22}+b_{22}&\cdots&a_{2n}+b_{2n}\\
\vdots&\vdots&\smash\ddots&\vdots\\
a_{m1}+b_{m1}&a_{m2}+b_{m2}&\cdots&a_{mn}+b_{mn}
\end{pmatrix}
\end{align*}$$
We now take the complex conjugate of both sides and apply theoremlink to get:
$$\begin{align*}
\overline{\boldsymbol{A}+\boldsymbol{B}}
&=\begin{pmatrix}
\overline{a_{11}+b_{11}}&\overline{a_{12}+b_{12}}&\cdots&\overline{a_{1n}+b_{1n}}\\
\overline{a_{21}+b_{21}}&\overline{a_{22}+b_{22}}&\cdots&\overline{a_{2n}+b_{2n}}\\
\vdots&\vdots&\smash\ddots&\vdots\\
\overline{a_{m1}+b_{m1}}&\overline{a_{m2}+b_{m2}}&\cdots&\overline{a_{mn}+b_{mn}}
\end{pmatrix}\\
&=\begin{pmatrix}
\overline{a_{11}}+\overline{b_{11}}&\overline{a_{12}}+\overline{b_{12}}&\cdots&\overline{a_{1n}}+\overline{b_{1n}}\\
\overline{a_{21}}+\overline{b_{21}}&\overline{a_{22}}+\overline{b_{22}}&\cdots&\overline{a_{2n}}+\overline{b_{2n}}\\
\vdots&\vdots&\smash\ddots&\vdots\\
\overline{a_{m1}}+\overline{b_{m1}}&\overline{a_{m2}}+\overline{b_{m2}}&\cdots&\overline{a_{mn}}+\overline{b_{mn}}
\end{pmatrix}\\
&=\begin{pmatrix}
\overline{a_{11}}&\overline{a_{12}}&\cdots&\overline{a_{1n}}\\
\overline{a_{21}}&\overline{a_{22}}&\cdots&\overline{a_{2n}}\\
\vdots&\vdots&\smash\ddots&\vdots\\
\overline{a_{m1}}&\overline{a_{m2}}&\cdots&\overline{a_{mn}}
\end{pmatrix}+
\begin{pmatrix}
\overline{b_{11}}&\overline{b_{12}}&\cdots&\overline{b_{1n}}\\
\overline{b_{21}}&\overline{b_{22}}&\cdots&\overline{b_{2n}}\\
\vdots&\vdots&\smash\ddots&\vdots\\
\overline{b_{m1}}&\overline{b_{m2}}&\cdots&\overline{b_{mn}}
\end{pmatrix}\\
&=\overline{\boldsymbol{A}}+\overline{\boldsymbol{B}}
\end{align*}$$
This completes the proof.
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Theorem.
Complex conjugation of a product between a complex number and a complex matrix
If $\boldsymbol{A}$ is a complex matrix and $z\in\mathbb{C}$, then:
$$\overline{z\boldsymbol{A}}=
\overline{z}\overline{\boldsymbol{A}}$$
Proof. We start with the left-hand side:
$$\begin{align*}
z\boldsymbol{A}&=
z\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\smash\ddots&\vdots\\
a_{m1}&a_{m2}&\cdots&a_{mn}
\end{pmatrix}\\&=
\begin{pmatrix}
za_{11}&za_{12}&\cdots&za_{1n}\\
za_{21}&za_{22}&\cdots&za_{2n}\\
\vdots&\vdots&\smash\ddots&\vdots\\
za_{m1}&za_{m2}&\cdots&za_{mn}
\end{pmatrix}\\
\end{align*}$$
We now take the complex conjugate of both sides and apply theoremlink to get:
$$\begin{align*}
\overline{z\boldsymbol{A}}&=
\begin{pmatrix}
\overline{za_{11}}&\overline{za_{12}}&\cdots&\overline{za_{1n}}\\
\overline{za_{21}}&\overline{za_{22}}&\cdots&\overline{za_{2n}}\\
\vdots&\vdots&\smash\ddots&\vdots\\
\overline{za_{m1}}&\overline{za_{m2}}&\cdots&\overline{za_{mn}}
\end{pmatrix}\\
&=\begin{pmatrix}
\overline{z}\;\overline{a_{11}}&\overline{z}\;\overline{a_{12}}&\cdots&\overline{z}\;\overline{a_{1n}}\\
\overline{z}\;\overline{a_{21}}&\overline{z}\;\overline{a_{22}}&\cdots&\overline{z}\;\overline{a_{2n}}\\
\vdots&\vdots&\smash\ddots&\vdots\\
\overline{z}\;\overline{a_{m1}}&\overline{z}\;\overline{a_{m2}}&\cdots&\overline{z}\;\overline{a_{mn}}
\end{pmatrix}\\
&=\overline{z}\begin{pmatrix}
\overline{a_{11}}&\overline{a_{12}}&\cdots&\overline{a_{1n}}\\
\overline{a_{21}}&\overline{a_{22}}&\cdots&\overline{a_{2n}}\\
\vdots&\vdots&\smash\ddots&\vdots\\
\overline{a_{m1}}&\overline{a_{m2}}&\cdots&\overline{a_{mn}}
\end{pmatrix}\\
&=\overline{z}\overline{\boldsymbol{A}}
\end{align*}$$
This completes the proof.
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Theorem.
Complex conjugate of a conjugate of a complex matrix
If $\boldsymbol{A}$ is a complex matrix, then:
$$\overline{\overline{\boldsymbol{A}}}=\boldsymbol{A}$$
Proof. The complex conjugate of matrix $\boldsymbol{A}$ is:
$$\begin{align*}
\overline{\boldsymbol{A}}&=
\begin{pmatrix}
\overline{a_{11}}&\overline{a_{12}}&\cdots&\overline{a_{1n}}\\
\overline{a_{21}}&\overline{a_{22}}&\cdots&\overline{a_{2n}}\\
\vdots&\vdots&\smash\ddots&\vdots\\
\overline{a_{m1}}&\overline{a_{m2}}&\cdots&\overline{a_{mn}}
\end{pmatrix}
\end{align*}$$
Taking the complex conjugate of $\overline{\boldsymbol{A}}$ and applying theoremlink gives:
$$\begin{align*}
\overline{\overline{\boldsymbol{A}}}&=
\begin{pmatrix}
\overline{\overline{a_{11}}}&\overline{\overline{a_{12}}}&\cdots&\overline{\overline{a_{1n}}}\\
\overline{\overline{a_{21}}}&\overline{\overline{a_{22}}}&\cdots&\overline{\overline{a_{2n}}}\\
\vdots&\vdots&\smash\ddots&\vdots\\
\overline{\overline{a_{m1}}}&\overline{\overline{a_{m2}}}&\cdots&\overline{\overline{a_{mn}}}
\end{pmatrix}\\
&=\begin{pmatrix}
a_{11}&a_{12}&\cdots&a_{1n}\\
a_{21}&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\smash\ddots&\vdots\\
a_{m1}&a_{m2}&\cdots&a_{mn}
\end{pmatrix}\\
&=\boldsymbol{A}
\end{align*}$$
This completes the proof.
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Theorem.
Complex conjugate of matrix transpose
If $\boldsymbol{A}$ is a complex matrix, then:
$$\overline{\boldsymbol{A}^T}=
(\overline{\boldsymbol{A}})^T$$
Proof. Let $\boldsymbol{A}$ be an $m\times{n}$ matrix. The transpose $\boldsymbol{A}^T$ is:
$$\begin{align*}
\boldsymbol{A}^T&=
\begin{pmatrix}
a_{11}&a_{21}&\cdots&a_{m1}\\
a_{12}&a_{22}&\cdots&a_{m2}\\
\vdots&\vdots&\smash\ddots&\vdots\\
a_{1n}&a_{2n}&\cdots&a_{mn}
\end{pmatrix}
\end{align*}$$
The complex conjugate of $\boldsymbol{A}^T$ is:
$$\begin{equation}\label{eq:HPJkaK2epYiLllFGes1}
\overline{\boldsymbol{A}^T}=
\begin{pmatrix}
\overline{a_{11}}&\overline{a_{21}}&\cdots&\overline{a_{m1}}\\
\overline{a_{12}}&\overline{a_{22}}&\cdots&\overline{a_{m2}}\\
\vdots&\vdots&\smash\ddots&\vdots\\
\overline{a_{1n}}&\overline{a_{2n}}&\cdots&\overline{a_{mn}}
\end{pmatrix}
\end{equation}$$
Now, $\overline{\boldsymbol{A}}$ is:
$$\begin{align*}
\overline{\boldsymbol{A}}&=
\begin{pmatrix}
\overline{a_{11}}&\overline{a_{12}}&\cdots&\overline{a_{1n}}\\
\overline{a_{21}}&\overline{a_{22}}&\cdots&\overline{a_{2n}}\\
\vdots&\vdots&\smash\ddots&\vdots\\
\overline{a_{m1}}&\overline{a_{m2}}&\cdots&\overline{a_{mn}}
\end{pmatrix}
\end{align*}$$
Taking the transpose gives:
$$\begin{equation}\label{eq:qZCrB6lwgO4YxtCJxtL}
(\overline{\boldsymbol{A}})^T=
\begin{pmatrix}
\overline{a_{11}}&\overline{a_{21}}&\cdots&\overline{a_{m1}}\\
\overline{a_{12}}&\overline{a_{22}}&\cdots&\overline{a_{m2}}\\
\vdots&\vdots&\smash\ddots&\vdots\\
\overline{a_{1n}}&\overline{a_{2n}}&\cdots&\overline{a_{mn}}
\end{pmatrix}
\end{equation}$$
Notice how \eqref{eq:HPJkaK2epYiLllFGes1} and \eqref{eq:qZCrB6lwgO4YxtCJxtL} are identical, which means:
$$\overline{\boldsymbol{A}^T}=
(\overline{\boldsymbol{A}})^T$$
This completes the proof.
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Properties of complex column vectors
Theorem.
Product between a transpose of a column vector and the vector
If $\boldsymbol{v}$ is a complex vector in $\mathbb{C}^n$, then:
$$\overline{\boldsymbol{v}}^T
\boldsymbol{v}=
\vert{v_1}\vert^2+
\vert{v_2}\vert^2+\cdots+
\vert{v_n}\vert^2$$
Where $\vert{v_1}\vert$ is the moduluslink of complex number $v_1$.
Proof. Let $\boldsymbol{v}$ be a complex vector in $\mathbb{C}^n$ with complex entries $v_1$, $v_2$, $\cdots$, $v_n$. Using theoremlink, we get:
$$\begin{align*}
\overline{\boldsymbol{v}}^T\boldsymbol{v}&=
\overline{v_1}v_1+
\overline{v_2}v_2
+\cdots+\overline{v_n}v_n\\
&=\vert{v_1}\vert^2+
\vert{v_2}\vert^2+\cdots+
\vert{v_n}\vert^2
\end{align*}$$
This completes the proof.
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Theorem.
Product between a transpose of a non-zero column vector and the vector is greater than zero
If $\boldsymbol{v}$ is a non-zero complex vector in $\mathbb{C}^n$, then:
$$\overline{\boldsymbol{v}}^T\boldsymbol{v}\gt0$$
Proof. By theoremlink, we have that:
$$\overline{\boldsymbol{v}}^T
\boldsymbol{v}=
\vert{v_1}\vert^2+
\vert{v_2}\vert^2+\cdots+
\vert{v_n}\vert^2$$
Since the modulus of a complex number is non-negative by definitionlink, it suffices to show that $\overline{\boldsymbol{v}}^T\boldsymbol{v}\ne0$ when $\boldsymbol{v}\ne\boldsymbol{0}$. If $\boldsymbol{v}\ne\boldsymbol{0}$, then at least one of the entries $v_1$, $v_2$, $\cdots$, $v_n$ is non-zero - let's call this $v_i$. This implies that $\vert{v_i}\vert\gt0$, which makes $\overline{\boldsymbol{v}}^T\boldsymbol{v}\ne0$. This completes the proof.
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