search
Search
Unlock 100+ guides
search toc
close
account_circle
Profile
exit_to_app
Sign out
What does this mean?
Why is this true?
Give me some examples!
search
keyboard_voice
close
Searching Tips
Search for a recipe:
"Creating a table in MySQL"
Search for an API documentation: "@append"
Search for code: "!dataframe"
Apply a tag filter: "#python"
Useful Shortcuts
/ to open search panel
Esc to close search panel
to navigate between search results
d to clear all current filters
Enter to expand content preview
Doc Search
Code Search Beta
SORRY NOTHING FOUND!
mic
Start speaking...
Voice search is only supported in Safari and Chrome.
Shrink
Navigate to
function
Other math topics
10 guides
keyboard_arrow_down
check_circle
Mark as learned
thumb_up
0
thumb_down
0
chat_bubble_outline
0
Comment
auto_stories Bi-column layout
settings

# Complex matrices and their properties with proofs

schedule Aug 12, 2023
Last updated
local_offer
Tags
mode_heat
Master the mathematics behind data science with 100+ top-tier guides
Start your free 7-days trial now!
Definition.

# Complex matrices

Complex matrices are matrices whose entries are complex numbers.

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:

$$\overline{\boldsymbol{A}}= \begin{pmatrix} 2-3i&-5i\\ 4&1-6i \end{pmatrix}$$

# 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.

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.

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:

$$\overline{z\boldsymbol{A}}= \overline{z}\overline{\boldsymbol{A}}$$

\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.

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.

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:

$$$$\label{eq:HPJkaK2epYiLllFGes1} \overline{\boldsymbol{A}^T}= \begin{pmatrix} \overline{a_{11}}&\overline{a_{21}}&\cdots&\overline{a_{m1}}\\ \overline{a_{12}}&\overline{a_{22}}&\cdots&\overline{a_{m2}}\\ \vdots&\vdots&\smash\ddots&\vdots\\ \overline{a_{1n}}&\overline{a_{2n}}&\cdots&\overline{a_{mn}} \end{pmatrix}$$$$

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:

$$$$\label{eq:qZCrB6lwgO4YxtCJxtL} (\overline{\boldsymbol{A}})^T= \begin{pmatrix} \overline{a_{11}}&\overline{a_{21}}&\cdots&\overline{a_{m1}}\\ \overline{a_{12}}&\overline{a_{22}}&\cdots&\overline{a_{m2}}\\ \vdots&\vdots&\smash\ddots&\vdots\\ \overline{a_{1n}}&\overline{a_{2n}}&\cdots&\overline{a_{mn}} \end{pmatrix}$$$$

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.

# 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.

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.

Edited by 0 others
thumb_up
thumb_down
Comment
Citation
Ask a question or leave a feedback...
thumb_up
0
thumb_down
0
chat_bubble_outline
0
settings
Enjoy our search
Hit / to insta-search docs and recipes!