## Unit1.2.1Absolute value

###### Remark1.2.1.1.

Don't Panic!

In this course, we mostly allow scalars, vectors, and matrices to be complex-valued. This means we will use terms like "conjugate" and "Hermitian" quite liberally. You will think this is a big deal, but actually, if you just focus on the real case, you will notice that the complex case is just a natural extension of the real case.

Recall that $\vert \cdot \vert: \mathbb C \rightarrow \mathbb R$ is the function that returns the absolute value of the input. In other words, if $\alpha = \alpha_r + \alpha_c i \text{,}$ where $\alpha_r$ and $\alpha_c$ are the real and imaginary parts of $\alpha \text{,}$ respectively, then

\begin{equation*} \vert \alpha \vert = \sqrt{ \alpha_r^2 + \alpha_c^2 }. \end{equation*}

The absolute value (magnitude) of a complex number can also be thought of as the (Euclidean) distance from the point in the complex plane to the origin of that plane, as illustrated below for the number $3 + 2i \text{.}$

Alternatively, we can compute the absolute value as

\begin{equation*} \begin{array}{l} \vert \alpha \vert \\ ~~~=~~~ \\ \sqrt{ \alpha_r^2 + \alpha_c^2 } \\ ~~~=~~~ \\ \sqrt{ \alpha_r^2 - \alpha_c \alpha_r i + \alpha_r \alpha_c i + \alpha_c^2 } \\ ~~~=~~~ \\ \sqrt{ ( \alpha_r - \alpha_c i ) ( \alpha_r + \alpha_c i )} \\ ~~~=~~~ \\ \sqrt{ \overline \alpha \alpha }~~ , \end{array} \end{equation*}

where $\overline \alpha$ denotes the complex conjugate of $\alpha \text{:}$

\begin{equation*} \overline \alpha = \overline{\alpha_r + \alpha_c i} = \alpha_r - \alpha_c i\text{.} \end{equation*}

The absolute value function has the following properties:

• $\alpha \neq 0 \Rightarrow \vert \alpha \vert > 0$ ($\vert \cdot \vert$ is positive definite),

• $\vert \alpha \beta \vert = \vert \alpha \vert \vert \beta \vert$ ($\vert \cdot \vert$ is homogeneous), and

• $\vert \alpha + \beta \vert \leq \vert \alpha \vert + \vert \beta \vert$ ($\vert \cdot \vert$ obeys the triangle inequality).

Norms are functions from a domain to the real numbers that are positive definite, homogeneous, and obey the triangle inequality. This makes the absolute value function an example of a norm.

The below exercises help refresh your fluency with complex arithmetic.

###### Homework1.2.1.1.
1. $\displaystyle ( 1 + i )( 2 - i ) =$

2. $\displaystyle ( 2 - i )( 1 + i ) =$

3. $\displaystyle \overline{( 1 - i )}( 2 - i ) =$

4. $\displaystyle \overline{ \overline{( 1 - i )} ( 2 - i )} =$

5. $\displaystyle \overline{( 2 - i )} ( 1 - i ) =$

6. $\displaystyle (1-i) \overline{( 2 - i )} =$

Solution
1. $\displaystyle ( 1 + i )( 2 - i ) = 2 + 2i - i - i^2 = 2 + i + 1 = 3 + i$

2. $\displaystyle ( 2 - i )( 1 + i ) = 2 -i + 2i - i^2 = 2 + i + 1 = 3 + i$

3. $\displaystyle \overline{( 1 - i )}( 2 - i ) = ( 1 + i )( 2 - i ) = 2 -i + 2i - i^2 = 3 + i$

4. $\displaystyle \overline{ \overline{( 1 - i )} ( 2 - i )} = \overline{ {( 1 + i )} ( 2 - i )} = \overline{ 2 - i + 2 i - i^2 } = \overline{ 2 + i +1} = \overline{ 3 + i } = 3 - i$

5. $\displaystyle \overline{( 2 - i )} ( 1 - i ) = ( 2 + i ) ( 1 - i ) = 2 -2 i + i - i^2 = 2 - i + 1 = 3 - i$

6. $\displaystyle (1-i) \overline{( 2 - i )} = ( 1 - i ) ( 2 + i ) = 2 + i - 2 i - i^2 = 2 - i + 1 = 3 - i$

###### Homework1.2.1.2.

Let $\alpha, \beta \in \mathbb C \text{.}$

1. ALWAYS/SOMETIMES/NEVER: $\alpha \beta = \beta \alpha \text{.}$

2. ALWAYS/SOMETIMES/NEVER: ${\overline \alpha \beta} = \overline \beta \alpha \text{.}$

Hint

Let $\alpha = \alpha_r + \alpha_c i$ and $\beta = \beta_r + \beta_c i \text{,}$ where $\alpha_r , \alpha_c, \beta_r, \beta_c \in \mathbb R \text{.}$

1. ALWAYS: $\alpha \beta = \beta \alpha \text{.}$

2. SOMETIMES: ${\overline \alpha \beta} = \overline \beta \alpha \text{.}$

Solution
1. ALWAYS: $\alpha \beta = \beta \alpha \text{.}$

Proof:

\begin{equation*} \begin{array}{l} \alpha \beta \\ ~~~~=~~~~~ \lt \mbox{substitute} \gt \\ (\alpha_r + \alpha_c i) (\beta_r + \beta_c i) \\ ~~~~=~~~~~ \lt \mbox{multiply out} \gt \\ \alpha_r \beta_r + \alpha_r \beta_c i + \alpha_c \beta_r i - \alpha_c \beta_c \\ ~~~~=~~~~~ \lt \mbox{commutativity of real multiplication} \gt \\ \beta_r \alpha_r + \beta_r \alpha_c i + \beta_c \alpha_r i - \beta_c \alpha_c \\ ~~~~=~~~~~ \lt \mbox{factor} \gt \\ (\beta_r + \beta_c i) (\alpha_r + \alpha_c i) \\ ~~~~=~~~~~ \lt \mbox{substitute} \gt \\ \beta \alpha. \end{array} \end{equation*}
2. SOMETIMES: ${\overline \alpha \beta} = \overline \beta \alpha \text{.}$

An example where it is true: $\alpha = \beta = 0 \text{.}$

An example where it is false: $\alpha = 1$ and $\beta = i \text{.}$ Then $\overline \alpha \beta = 1 \times i = i$ and $\overline \beta \alpha = -i \times 1 = -i \text{.}$

###### Homework1.2.1.3.

Let $\alpha, \beta \in \mathbb C \text{.}$

ALWAYS/SOMETIMES/NEVER: $\overline{\overline \alpha \beta} = \overline \beta \alpha \text{.}$

Hint

Let $\alpha = \alpha_r + \alpha_c i$ and $\beta = \beta_r + \beta_c i \text{,}$ where $\alpha_r , \alpha_c, \beta_r, \beta_c \in \mathbb R \text{.}$

ALWAYS

Now prove it!

Solution 1
\begin{equation*} \begin{array}{l} \overline{ \overline{ \alpha } \beta } \\ ~~~~ = ~~~ \lt \alpha = \alpha_r + \alpha_c i ; \beta = \beta_r + \beta_c i \gt \\ \overline{ \overline{ (\alpha_r + \alpha_c i ) } (\beta_r + \beta_c i) } \\ ~~~~ = ~~~ \lt \mbox{ conjugate } \alpha \gt \\ \overline{ (\alpha_r - \alpha_c i ) (\beta_r + \beta_c i) } \\ ~~~~ = ~~~ \lt \mbox{ multiply out } \gt \\ \overline{ (\alpha_r \beta_r - \alpha_c \beta_r i + \alpha_r \beta_c i + \alpha_c \beta_c ) } \\ ~~~~ = ~~~ \lt \mbox{ conjugate } \gt \\ \alpha_r \beta_r + \alpha_c \beta_r i - \alpha_r \beta_c i + \alpha_c \beta_c \\ ~~~~ = ~~~ \lt \mbox{ rearrange } \gt \\ \beta_r \alpha_r + \beta_r \alpha_c i - \beta_c \alpha_r i + \beta_c \alpha_c \\ ~~~~ = ~~~ \lt \mbox{ factor } \gt \\ ( \beta_r - \beta_c i ) ( \alpha_r + \alpha_c i ) \\ ~~~~ = ~~~ \lt \mbox{ definition of conjugation } \gt \\ \overline{( \beta_r + \beta_c i) } ( \alpha_r + \alpha_c i ) \\ ~~~~ = ~~~ \lt \alpha = \alpha_r + \alpha_c i ; \beta = \beta_r + \beta_c i \gt \\ \overline{\beta} \alpha \end{array} \end{equation*}
Solution 2

Proofs in mathematical textbooks seem to always be wonderfully smooth arguments that lead from the left-hand side of an equivalence to the right-hand side. In practice, you may want to start on the left-hand side, and apply a few rules:

\begin{equation*} \begin{array}{l} \overline{ \overline{ \alpha } \beta } \\ ~~~~ = ~~~ \lt \alpha = \alpha_r + \alpha_c i ; \beta = \beta_r + \beta_c i \gt \\ \overline{ \overline{ (\alpha_r + \alpha_c i ) } (\beta_r + \beta_c i) } \\ ~~~~ = ~~~ \lt \mbox{ conjugate } \alpha \gt \\ \overline{ (\alpha_r - \alpha_c i ) (\beta_r + \beta_c i) } \\ ~~~~ = ~~~ \lt \mbox{ multiply out } \gt \\ \overline{ (\alpha_r \beta_r - \alpha_c \beta_r i + \alpha_r \beta_c i + \alpha_c \beta_c ) } \\ ~~~~ = ~~~ \lt \mbox{ conjugate } \gt \\ \alpha_r \beta_r + \alpha_c \beta_r i - \alpha_r \beta_c i + \alpha_c \beta_c \end{array} \end{equation*}

and then move on to the right-hand side, applying a few rules:

\begin{equation*} \begin{array}{l} \overline{ \beta } \alpha \\ ~~~~ = ~~~ \lt \alpha = \alpha_r + \alpha_c i ; \beta = \beta_r + \beta_c i \gt \\ \overline{(\beta_r + \beta_c i) } (\alpha_r + \alpha_c i ) \\ ~~~~ = ~~~ \lt \mbox{ conjugate } \beta \gt \\ (\beta_r - \beta_c i) (\alpha_r + \alpha_c i ) \\ ~~~~ = ~~~ \lt \mbox{ multiply out } \gt \\ \beta_r \alpha_r + \beta_r \alpha_c i - \beta_c \alpha_r i + \beta_c \alpha_c. \end{array} \end{equation*}

At that point, you recognize that

\begin{equation*} \alpha_r \beta_r + \alpha_c \beta_r i - \alpha_r \beta_c i + \alpha_c \beta_c \\ ~~~= ~~~~ \\ \beta_r \alpha_r + \beta_r \alpha_c i - \beta_c \alpha_r i + \beta_c \alpha_c \end{equation*}

since the second is a rearrangement of the terms of the first. Optionally, you then go back and presents these insights as a smooth argument that leads from the expression on the left-hand side to the one on the right-hand side:

\begin{equation*} \begin{array}{l} \overline{ \overline{ \alpha } \beta } \\ ~~~~ = ~~~ \lt \alpha = \alpha_r + \alpha_c i ; \beta = \beta_r + \beta_c i \gt \\ \overline{ \overline{ (\alpha_r + \alpha_c i ) } (\beta_r + \beta_c i) } \\ ~~~~ = ~~~ \lt \mbox{ conjugate } \alpha \gt \\ \overline{ (\alpha_r - \alpha_c i ) (\beta_r + \beta_c i) } \\ ~~~~ = ~~~ \lt \mbox{ multiply out } \gt \\ \overline{ (\alpha_r \beta_r - \alpha_c \beta_r i + \alpha_r \beta_c i + \alpha_c \beta_c ) } \\ ~~~~ = ~~~ \lt \mbox{ conjugate } \gt \\ \alpha_r \beta_r + \alpha_c \beta_r i - \alpha_r \beta_c i + \alpha_c \beta_c \\ ~~~~ = ~~~ \lt \mbox{ rearrange } \gt \\ \beta_r \alpha_r + \beta_r \alpha_c i - \beta_c \alpha_r i + \beta_c \alpha_c \\ ~~~~ = ~~~ \lt \mbox{ factor } \gt \\ ( \beta_r - \beta_c i ) ( \alpha_r + \alpha_c i ) \\ ~~~~ = ~~~ \lt \mbox{ definition of conjugation } \gt \\ \overline{( \beta_r + \beta_c i) } ( \alpha_r + \alpha_c i ) \\ ~~~~ = ~~~ \lt \alpha = \alpha_r + \alpha_c i ; \beta = \beta_r + \beta_c i \gt \\ \overline{\beta} \alpha . \end{array} \end{equation*}
Solution 3

Yet another way of presenting the proof uses an "equivalence style proof." The idea is to start with the equivalence you wish to prove correct:

\begin{equation*} \overline{\overline \alpha \beta} = \overline \beta \alpha \end{equation*}

and through a sequence of equivalent statements argue that this evaluates to TRUE:

\begin{equation*} \begin{array}{l} \overline{\overline \alpha \beta} = \overline \beta \alpha \\ ~~~ \Leftrightarrow ~~~~ \lt \alpha = \alpha_r + \alpha_c i ; \beta = \beta_r + \beta_c i \gt \\ \overline{\overline{ ( \alpha_r + \alpha_c i)} ( \beta_r + \beta_c i )} = \overline {( \beta_r + \beta_c i )}( \alpha_r + \alpha_c i ) \\ ~~~ \Leftrightarrow ~~~~ \lt \mbox{ conjugate } \times 2 \gt \\ \overline{{ ( \alpha_r - \alpha_c i)} ( \beta_r + \beta_c i )} = {( \beta_r - \beta_c i )}( \alpha_r + \alpha_c i ) \\ ~~~ \Leftrightarrow ~~~~ \lt \mbox{ multiply out } \times 2 \gt \\ \overline{\alpha_r \beta_r + \alpha_r \beta_c i - \alpha_c \beta_r i + \alpha_c \beta_c } = \beta_r \alpha_r + \beta_r \alpha_c i - \beta_c \alpha_r i + \beta_c \alpha_c \\ ~~~ \Leftrightarrow ~~~~ \lt \mbox{ conjugate } \gt \\ {\alpha_r \beta_r - \alpha_r \beta_c i + \alpha_c \beta_r i + \alpha_c \beta_c } = \beta_r \alpha_r + \beta_r \alpha_c i - \beta_c \alpha_r i + \beta_c \alpha_c \\ ~~~ \Leftrightarrow ~~~~ \lt \mbox{ subtract equivalent terms from left-hand side and right-hand side } \gt \\ 0 = 0 \\ ~~~ \Leftrightarrow ~~~~ \lt \mbox{ algebra } \gt \\ \mbox{TRUE}. \end{array} \end{equation*}

By transitivity of equivalence, we conclude that $\overline{\overline \alpha \beta} = \overline \beta \alpha$ is TRUE.

###### Homework1.2.1.4.

Let $\alpha \in \mathbb C \text{.}$

ALWAYS/SOMETIMES/NEVER: $\overline{\alpha} \alpha \in \mathbb R$

ALWAYS.

Now prove it!

Solution

Let $\alpha = \alpha_r + \alpha_c i \text{.}$ Then

\begin{equation*} \begin{array}{l} \overline{\alpha} \alpha \\ ~~~ = ~~~~ \lt \mbox{ instantiate } \gt \\ \overline{( \alpha_r + \alpha_c i )}(\alpha_r + \alpha_c i ) \\ ~~~ = ~~~~ \lt \mbox{ conjugate } \gt \\ {( \alpha_r - \alpha_c i )}(\alpha_r + \alpha_c i ) \\ ~~~ = ~~~~ \lt \mbox{ multiply out } \gt \\ \alpha_r^2 + \alpha_c^2, \end{array} \end{equation*}

which is a real number.

###### Homework1.2.1.5.

Prove that the absolute value function is homogeneous: $\vert \alpha \beta \vert = \vert \alpha \vert \vert \beta \vert$ for all $\alpha, \beta \in \mathbb C \text{.}$

Solution
\begin{equation*} \begin{array}{l} \vert \alpha \beta \vert = \vert \alpha \vert \vert \beta \vert \\ ~~~ \Leftrightarrow ~~~~ \lt \mbox{ squaring both sides simplifies }\gt \\ \vert \alpha \beta \vert^2 = \vert \alpha \vert^2 \vert \beta \vert^2 \\ ~~~ \Leftrightarrow ~~~~ \lt \mbox{ instantiate }\gt \\ \vert (\alpha_r + \alpha_c i ) (\beta_r + \beta_c i ) \vert^2 = \vert \alpha_r + \alpha_c i \vert^2 \vert \beta_r + \beta_c i \vert^2 \\ ~~~ \Leftrightarrow ~~~~ \lt \mbox{ algebra }\gt \\ \vert (\alpha_r \beta_r - \alpha_c \beta_c ) + ( \alpha_r \beta_c + \alpha_c \beta_r ) i \vert^2 = (\alpha_r^2 + \alpha_c^2) (\beta_r^2 + \beta_c^2) \\ ~~~ \Leftrightarrow ~~~~ \lt \mbox{ algebra }\gt \\ (\alpha_r \beta_r - \alpha_c \beta_c )^2 + ( \alpha_r \beta_c + \alpha_c \beta_r )^2 = (\alpha_r^2 + \alpha_c^2) (\beta_r^2 + \beta_c^2) \\ ~~~ \Leftrightarrow ~~~~ \lt \mbox{ algebra }\gt \\ \alpha_r^2 \beta_r^2 - 2 \alpha_r \alpha_c \beta_r \beta_c + \alpha_c^2 \beta_c^2 + \alpha_r^2 \beta_c^2 + 2 \alpha_r \alpha_c \beta_r \beta_c + \alpha_c^2 \beta_r^2 \\ ~~~~~~~~~~~~~~~~ = \alpha_r^2 \beta_r^2 + \alpha_r^2 \beta_c^ + \alpha_c^2 \beta_r^2 + \alpha_c^2 \beta_c^2 \\ ~~~ \Leftrightarrow ~~~~ \lt \mbox{ subtract equivalent terms from both sides }\gt \\ 0 = 0 \\ ~~~ \Leftrightarrow ~~~~ \lt \mbox{ algebra }\gt \\ T \end{array} \end{equation*}
###### Homework1.2.1.6.

Let $\alpha \in \mathbb C \text{.}$

ALWAYS/SOMETIMES/NEVER: $\vert \overline \alpha \vert = \vert \alpha \vert \text{.}$

Let $\alpha = \alpha_r + \alpha_c i \text{.}$