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Linear Algebra
52 guides
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1. Vectors
2. Matrices
3. Linear equations
4. Matrix determinant
5. Vector space
6. Special matrices
7. Eigenvalues and Eigenvectors
8. Orthogonality
9. Matrix decomposition
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# Guide on Complex Vectors

schedule Mar 5, 2023
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Definition.

# Complex vectors

A complex vector is a vector whose entries are complex numbers. Just like how real vectors belong to the real vector space $\mathbb{R}^n$, complex vectors belong to the complex vector space $\mathbb{C}^n$.

Example.

## Some complex vectors

Below are some examples of complex vectors:

$$\boldsymbol{z}_1= \begin{pmatrix} 2+3i\\5\\6i \end{pmatrix},\;\;\;\; \boldsymbol{z}_2= \begin{pmatrix} 2\\5 \end{pmatrix}$$

Note that $\boldsymbol{z}_2$ can be regarded as a complex vector even though its entries are all real because real numbers are a subset of complex numbers.

Definition.

# Conjugate of a complex vector

The conjugate of a complex vector $\boldsymbol{z}$ can be obtained by taking the complex conjugatelink of every entry of $\boldsymbol{z}$.

Example.

## Finding the conjugate of a complex vector:

Find the complex conjugate of the following complex vector:

$$\boldsymbol{z}= \begin{pmatrix} 2+3i\\5\\6i \end{pmatrix}$$

Solution. To obtain the complex conjugate of $\boldsymbol{z}$, we take the complex conjugate of each entry of $\boldsymbol{z}$ like so:

$$\overline{\boldsymbol{z}}= \begin{pmatrix} \overline{2+3i}\\\overline{5}\\\overline{6i} \end{pmatrix}= \begin{pmatrix} 2-3i\\5\\-6i \end{pmatrix}$$
Definition.

# Euclidean inner product in a complex vector space

If $\boldsymbol{v}$ and $\boldsymbol{w}$ are vectors in $\mathbb{C}^n$, then the Euclidean inner product of $\boldsymbol{v}$ and $\boldsymbol{w}$ is defined as:

$$\boldsymbol{v}\cdot \boldsymbol{w}= v_1\overline{w_1}+ v_2\overline{w_2}+ \cdots+ v_n\overline{w_n}$$
Example.

## Computing the Euclidean inner product between two complex vectors

Consider the vectors in $\mathbb{C}^3$ below:

$$\boldsymbol{v}= \begin{pmatrix} 2+i\\3\\4+2i \end{pmatrix},\;\;\;\;\; \boldsymbol{w}= \begin{pmatrix} 1+4i\\2i\\1 \end{pmatrix}$$

Compute $\boldsymbol{v}\cdot\boldsymbol{w}$.

Solution. The dot product of $\boldsymbol{v}$ and $\boldsymbol{w}$ is:

\begin{align*} \boldsymbol{v}\cdot\boldsymbol{w}&= \begin{pmatrix}2+i\\3\\4+2i\end{pmatrix}\cdot \begin{pmatrix}1+4i\\2i\\1\end{pmatrix}\\ &=(2+i)(1-4i)+(3)(-2i)+(4+2i)(1)\\ &=2-7i-4i^2-6i+4+2i\\ &=10-11i \end{align*}

# Properties of the Euclidean Inner Product

In this section, we will state and prove some basic properties of the Euclidean inner product. For the proofs, we will assume that the complex vectors $\boldsymbol{u}, \boldsymbol{v}, \boldsymbol{w}\in\mathbb{C}^n$ are:

$$\boldsymbol{u}= \begin{pmatrix} u_1\\u_2\\\vdots\\u_n \end{pmatrix},\;\;\;\;\; \boldsymbol{v}= \begin{pmatrix} v_1\\v_2\\\vdots\\v_n \end{pmatrix},\;\;\;\;\; \boldsymbol{w}= \begin{pmatrix} w_1\\w_2\\\vdots\\w_n \end{pmatrix}$$

We will use the term "dot product" and "Euclidean inner product" interchangeably. Let's get started!

Theorem.

## Complex conjugate of the dot product between two complex vectors

If $\boldsymbol{v}$ and $\boldsymbol{w}$ are vectors in $\mathbb{C}^n$ and $z\in\mathbb{C}$, then:

$$\overline{\boldsymbol{v}\cdot\boldsymbol{w}}= \overline{\boldsymbol{w}\cdot\boldsymbol{v}}$$

Proof. We start from the left-hand side. By definitionlink of complex dot product, we have that:

\begin{align*} \overline{\boldsymbol{v}\cdot\boldsymbol{w}}&= \overline{v_1\overline{w_1}+ v_2\overline{w_2}+ \cdots+ v_n\overline{w_n}} \end{align*}

Now, by propertylink, we have that:

\begin{align*} \overline{\boldsymbol{v}\cdot\boldsymbol{w}}&= \overline{v_1\overline{w_1}}+ \overline{v_2\overline{w_2}}+ \cdots+\overline{v_n\overline{w_n}}\\ &=\overline{v_1}w_1+\overline{v_2}w_2+\cdots+\overline{v_n}w_n\\ &=w_1\overline{v_1}+w_2\overline{v_2} +\cdots+w_n\overline{v_n}\\ &=\overline{\boldsymbol{w}\cdot\boldsymbol{v}} \end{align*}

Here, the 2nd equality holds by theoremlink. This completes the proof.

Theorem.

# Another form of the dot product between two complex vectors (1)

If $\boldsymbol{v}$ and $\boldsymbol{w}$ are vectors in $\mathbb{C}^n$ and $z\in\mathbb{C}$, then:

$$(z\boldsymbol{v})\cdot\boldsymbol{w}= z(\boldsymbol{v}\cdot\boldsymbol{w})$$

Proof. We start from the left-hand side:

\begin{align*} (z\boldsymbol{v})\cdot\boldsymbol{w} &=z\begin{pmatrix} v_1\\v_2\\\vdots\\v_n \end{pmatrix}\cdot\boldsymbol{w}\\ &=\begin{pmatrix} zv_1\\zv_2\\\vdots\\zv_n \end{pmatrix}\cdot\boldsymbol{w}\\ &=zv_1\overline{w_1}+ zv_2\overline{w_2}+\cdots+zv_n\overline{w_n}\\ &=z(v_1\overline{w_1}+ v_2\overline{w_2}+\cdots+v_n\overline{w_n})\\ &=z(\boldsymbol{v}\cdot\boldsymbol{w}) \end{align*}

This completes the proof.

Theorem.

# Another form of the dot product between two complex vectors (2)

If $\boldsymbol{v}$ and $\boldsymbol{w}$ are vectors in $\mathbb{C}^n$ and $z\in\mathbb{C}$, then:

$$\boldsymbol{v}\cdot(z\boldsymbol{w})= \overline{z}(\boldsymbol{v}\cdot\boldsymbol{w})$$

Proof. We start from the left-hand side:

\begin{align*} \boldsymbol{v}\cdot(z\boldsymbol{w}) &=\boldsymbol{v}\cdot {z\begin{pmatrix}w_1\\w_2\\\vdots\\w_n\end{pmatrix}}\\ &=\boldsymbol{v}\cdot \begin{pmatrix}zw_1\\zw_2\\\vdots\\zw_n\end{pmatrix}\\ &=v_1\overline{zw_1}+v_2\overline{zw_2}+\cdots+ v_n\overline{zw_n}\\ &=v_1\overline{z}\;\overline{w_1}+ v_2\overline{z}\;\overline{w_2}+\cdots+ v_n\overline{z}\;\overline{w_n}\\ &=\overline{z}(v_1\overline{w_1}+ v_2\overline{w_2}+\cdots+ v_n\overline{w_n})\\ &=\overline{z}(\boldsymbol{v}\cdot\boldsymbol{w}) \end{align*}

Here, we used propertylink for the 4th equality. This completes the proof.

Theorem.

# Distributive property of complex dot products

If $\boldsymbol{u}$, $\boldsymbol{v}$ and $\boldsymbol{w}$ are vectors in $\mathbb{C}^n$, then:

$$(\boldsymbol{u}+\boldsymbol{v})\cdot\boldsymbol{w}= \boldsymbol{u}\cdot\boldsymbol{w}+ \boldsymbol{v}\cdot\boldsymbol{w}$$

Proof. We start from the left-hand side:

\begin{align*} (\boldsymbol{u}+\boldsymbol{v})\cdot\boldsymbol{w}&= \left(\begin{pmatrix}u_1\\u_2\\\vdots\\u_n \end{pmatrix}+\begin{pmatrix} v_1\\v_2\\\vdots\\v_n\end{pmatrix}\right)\cdot \begin{pmatrix}w_1\\w_2\\\vdots\\w_n\end{pmatrix}\\ &=\begin{pmatrix} u_1+v_1\\u_2+v_2\\\vdots\\u_n+v_n \end{pmatrix}\cdot \begin{pmatrix}w_1\\w_2\\\vdots\\w_n \end{pmatrix}\\ &=(u_1+v_1)\overline{w_1}+(u_2+v_2)\overline{w_2}+\cdots+ (u_n+v_n)\overline{w_n}\\ &=u_1\overline{w_1}+v_1\overline{w_1}+ u_2\overline{w_2}+v_2\overline{w_2}+\cdots+ u_n\overline{w_n}+v_n\overline{w_n}\\ &=(u_1\overline{w_1}+u_2\overline{w_2}+\cdots+ u_n\overline{w_n}) +(v_1\overline{w_1}+v_2\overline{w_2}+\cdots+ +v_n\overline{w_n})\\ &=\boldsymbol{u}\cdot\boldsymbol{w}+ \boldsymbol{v}\cdot\boldsymbol{w} \end{align*}

This completes the proof.

Theorem.

# Dot product of a complex vector with itself is equal to the sum of the squared modulus of each component

If $\boldsymbol{v}$ is a complex vector in $\mathbb{C}^n$, then the Euclidean inner product between $\boldsymbol{v}$ and itself is:

$$\boldsymbol{v}\cdot\boldsymbol{v} =\vert{v_1}\vert^2+ \vert{v_2}\vert^2+ \cdots+\vert{v_n}\vert^2$$

Proof. By definitionlink of complex dot product and propertylink of complex conjugates, we get:

\begin{align*} \boldsymbol{v}\cdot\boldsymbol{v} &=v_1\overline{v_1} +v_2\overline{v_2}+\cdots+ v_n\overline{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.

# Euclidean inner product of a complex vector with itself is greater than or equal to zero

If $\boldsymbol{v}$ is a complex vector, then the Euclidean inner product between $\boldsymbol{v}$ and itself is:

$$\boldsymbol{v}\cdot\boldsymbol{v} \ge0$$

Proof. By theoremlink, we have that:

$$\boldsymbol{v}\cdot\boldsymbol{v} =\vert{v_1}\vert^2+ \vert{v_2}\vert^2+ \cdots+\vert{v_n}\vert^2$$

Since the modulus is non-negative by definitionlink, we have that $\boldsymbol{v}\cdot\boldsymbol{v}\ge0$. This completes the proof.

Definition.

# Euclidean norm of a complex vector

The Euclidean norm (or length) of a complex vector $\boldsymbol{v}$ in $\mathbb{C}^n$ is defined as:

$$\Vert\boldsymbol{v}\Vert= (\boldsymbol{v}\cdot\boldsymbol{v})^{1/2}$$
Theorem.

# Euclidean norm of a complex vector expressed using the modulus

The Euclidean norm of $\boldsymbol{v}\in\mathbb{C}^n$ can also be expressed as:

$$\Vert\boldsymbol{v}\Vert=(\vert{v_1}\vert^2+\vert{v_2}\vert^2+\cdots+\vert{v_n}\vert^2)^{1/2}$$

Where $\vert{v_1}\vert$ is the moduluslink of the first entry of $\boldsymbol{v}$.

Proof. We apply propertylink of complex dot product to get:

\begin{align*} \Vert\boldsymbol{v}\Vert&= (\boldsymbol{v}\cdot\boldsymbol{v})^{1/2}\\ &=(\vert{v_1}\vert^2+\vert{v_2}\vert^2+\cdots+\vert{v_n}\vert^2)^{1/2} \end{align*}

This completes the proof.

Example.

## Finding the Euclidean norm of a complex vector

Find the Euclidean norm of the following complex vector:

$$\boldsymbol{v}= \begin{pmatrix} 2+3i\\4i\\ \end{pmatrix}$$

Solution. Let's find the squared modulus of each component of $\boldsymbol{v}$ first:

\begin{align*} \vert{v_1}\vert^2=(2)^2+(3)^2&=13\\ \vert{v_2}\vert^2=(0)^2+(4)^2&=16\\ \end{align*}

By theoremlink, the Euclidean norm of $\boldsymbol{v}$ is:

\begin{align*} \Vert\boldsymbol{v}\Vert&= (\vert{v_1}\vert^2+\vert{v_2}\vert^2)^{1/2}\\ &=(13+16)^{1/2}\\ &=\sqrt{29} \end{align*}
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