Hermitian Functions

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Hermitian Functions

09 Oct 2021

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In this post we’ll study even and odd functions, the even-odd function decomposition, Hermitian functions and prove some results for Fourier transforms.

Hermitian functions are named after the 19th century French mathematician Charles Hermite. Hermitian matrices are also named after him. Charles was the advisor of Henri Poincaré.

Even and Odd functions

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function with a real domain and image, where $F$ is either the real numbers $\mathbb{R}$ or the complex numbers $\mathbb{C}$.

An even function is defined as:

\[f(x) = f(-x)\]

An example of even function is $f(x) = x^2$, since $x^2 = (-x)^2$.

Conversely, an odd function is defined as:

\[f(x) = - \overline{f(-x)}\]

An example of odd function is $f(x) = x^3$, since $x^3 = - (-x)^3$.

Odd Naming

The names odd and even comes from the fact that if $f(x)$ is of the form $f(x) = x^n$, then $f(x)$ is even when $n$ is even, and odd when $n$ is odd.

I find this terminology confusing, since $f(x) = \abs{x}$ is also even. Perhaps a better name could be symmetric and anti-symmetric, but unfortunately symmetric functions mean something else.

By calling functions this way also gives the idea that all functions are either even and odd, which is not true. Some functions are neither.

Even-Odd Decomposition

The even-odd decomposition states that for any given function $f: \mathbb{R} \rightarrow \mathbb{R}$ there exists:

\[f_e(x) = \frac{f(x) + \overline{f(-x)}}{2}\]


\[f_o(x) = \frac{f(x) - \overline{f(-x)}}{2}\]

and that $f_e$ is an even function $f_o$ and an odd function. Since,

\[f(x) = f_e(x) + f_o(x)\]

we’re saying that any function can be decomposed into an even and odd function. Furthermore such decomposition is unique. We’ll prove a more general version of this decomposition in the next sections.

Hermitian Functions

Hermitian functions are in a way a generalization of Even and Odd functions, in which we allow complex numbers as the image. The analogous of even function is the (symmetric) Hermitian function as:

\[f(x) = \overline{f(-x)}\]

An example of Hermitian function is $f(x) = x^2+ ix$, since $f(-x)= x^2 - ix = \overline{f(x)}$.

Conversely, the analogous to odd function is the antisymmetric Hermitian function:

\[f(x) = - \overline{f(-x)}\]

An example of antisymmetric Hermitian function is $f(x) = x + ix^2$, since $f(-x)= -x + ix^2 = - (x - ix^2) = -\overline{f(x)}$.

Note that a function $f$ is a Hermitian function if and only if $Re\curly{f}$ is even and $Im\curly{f}$ is odd.

Generalizing to $\mathbb{R}^n$

We can generalize the concept of Hermitian functions to the $\mathbb{R}^n$ domain, so if $f: \mathbb{R}^n \rightarrow \mathbb{C}$, then:

\[f(\vec{x}) = \overline{f(-\vec{x})}\]

Where $\vec{x} = x_0, \cdots, x_{n-1}$ and $-\vec{x} = -x_0, \cdots, -x_{n-1}$.

Hermitian Decomposition

Let $f: \mathbb{R} \rightarrow \mathbb{C}$ be a function, $f_s$ a symmetric Hermitian function and $f_a$ an antisymmetric Hermitian. If $f$ can be written as a sum of $f_s$ and $f_a$, that is:

\[(1) \qquad f(x) = f_s(x) + f_a(x)\]

then it has a Hermitian decomposition.

Proposition 1. Every function $f: \mathbb{R} \rightarrow \mathbb{C}$ has a Hermitian decomposition.


We can define

\[f_s(x) = \frac{f(x) + \overline{f(-x)}}{2}\]


\[f_a(x) = \frac{f(x) - \overline{f(-x)}}{2}\]

Which clearly satisfies the identity (1). We now claim that $f_s$ is symmetric, i.e. $f_s(x) = \overline{f_s(-x)}$. Consider some $x_p \ge 0$, and let $x_n = - x_p$. We can show that $f_s(x_p) = \overline{f_s(x_n)}$ by:

\[f_s(x_p) = \frac{f(x_p) + \overline{f(-x_p)}}{2} = \frac{f(x_p) + \overline{f(x_n)}}{2} = \overline{\frac{\overline{f(-x_n)} + f(x_n)}{2}} = \overline{f_s(x_n)}\]

which proves our claim. We can show that $f_a(x) = - \overline{f_a(-x)}$ through an analogous process. QED.

Proposition 2. The Hermitian decompositions is unique.

Proof: See Appendix.

Our definition of Hermitian decomposition and the two properties above can be trivially generalized for $f: \mathbb{R}^n \rightarrow \mathbb{C}$ since we don’t work directly with $\vec{x}$ but rather their scalar image.

Fourier Transform is Hermitian

We’ll now show that the DTFT is Hermitian if it’s applied to $\vec{x} \in \mathbb{R}^N$. Recall that the DTFT can be defined as:

\[(2) \quad \lambda(\omega) = \sum_{t = 0}^{N-1} x_t e^{-i \omega t} \quad -\pi \le \omega \le \pi\]

Let’s expand $e^{-i \omega t}$ using Euler’s formula:

\[e^{-i \omega t} = \cos(- \omega t) + i \sin( - \omega t)\]

Since $\cos(\theta) = \cos(-\theta)$ and $\sin(\theta) = -\sin(-\theta)$,

\[e^{-i \omega t} = \cos(\omega t) - i \sin(\omega t) = \overline{e^{i \omega t}}\]

If we call each term in the sum (2), $c_{t,\omega} = x_t e^{-i \omega t}$, then by the results above, $c_{t,\omega} = \overline{c_{t, -\omega}}$, and it’s easy to see this property is preserved over sums, so:

\[\lambda(\omega) = \overline{\lambda(-\omega)}\]

Which shows the Fourier transform, $\lambda: \mathbb{R} \rightarrow \mathbb{C}$ is a symmetric Hermitian function.

Polar Form

We can express a complex number $c = Re\curly{c} + i Im\curly{c}$ in polar form as $\abs{c} e^{i \varphi}$, where $\abs{c} = \sqrt{Re\curly{c}^2 + Im\curly{c}^2}$ is the modulus and $\varphi = \arctan2(Im\curly{c}, Re\curly{c})$ the phase.

In this definition, $\abs{c}$ is non-negative and $\varphi \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

Since the Fourier transform $\lambda: \mathbb{R} \rightarrow \mathbb{C}$ is a symmetric Hermitian function, we know that the real part is even $Re\curly{\lambda(\omega)} = Re\curly{\lambda(-\omega)}$ and the imaginary part is odd $Im\curly{\lambda(\omega)} = -Im\curly{\lambda(-\omega)}$.

We can compute the modulus of the Fourier transform $\abs{\lambda(\omega)}$ as

\[\abs{\lambda(\omega)}^2 = Re\curly{\lambda(\omega)}^2 + Im\curly{\lambda(\omega)}^2\]

Since $Im\curly{\lambda(\omega)}$ is squared, the negative sign is gone and thus we get:

\[\abs{\lambda(\omega)} = \abs{\lambda(-\omega)}\]

which means the modulus of a Fourier transform is an even function.

An alternative way of computing $\varphi$, assuming $\abs{c} > 0$ is [3]:

\[\begin{equation} \varphi = \left\{ \begin{array}{@{}ll@{}} \arccos (\frac{Re\curly{c}}{\abs{c}}), & \text{if}\ Im\curly{c} \ge 0 \\ - \arccos (\frac{Re\curly{c}}{\abs{c}}), & \text{if}\ Im\curly{c} < 0 \\ \end{array}\right. \end{equation}\]

Let $\varphi(\omega)$ be the phase of $\lambda(\omega)$ computed as above. We have both $Re\curly{\lambda(\omega)}$ and $\abs{\lambda(\omega)}$ as even functions and $Im\curly{\lambda(\omega)}$ as an odd function.

Assume $Im\curly{\lambda(\omega)} > 0$ for now. Then $\varphi(\omega) = \arccos (\frac{Re\curly{\lambda(\omega)}}{\abs{\lambda(\omega)}})$. Since $Im\curly{\lambda(-\omega)} = -Im\curly{\lambda(\omega)} < 0$, $\varphi(-\omega) = -\arccos (\frac{Re\curly{\lambda(-\omega)}}{\abs{\lambda(-\omega)}}) = -\arccos (\frac{Re\curly{\lambda(\omega)}}{\abs{\lambda(\omega)}}) = -\varphi(-\omega)$.

We can obtain the same result for $Im\curly{\lambda(\omega)} < 0$ and $Im\curly{\lambda(\omega)} = 0$, concluding that the phase of a Fourier transform is an odd function.


I found the even-odd decomposition counter-intuitive at first, but making analogies to how a complex number can be uniquely decomposed into a sum of a purely real and purely imaginary parts makes it a bit intuitive.

I initially intended to study even and odd functions but was interested in the results of for Fourier transforms, which led me to the more general Hermitian functions.

Interestingly, the Wikipedia entry for even and odd functions does not mention Hermitian functions even though it presents the generalized version for complex images.

In regards of decomposition, I was reminded of the Shor’s Prime Factoring Algorithm post, where we saw that $\ket{1}$ can be decomposed into some a linear combination of eigenvectors:

\[\frac{1}{\sqrt{r}} \sum_{s=0}^{r-1} \ket{u_s} = \ket{1}\]

in and Discrete Fourier Transforms we also see that a signal can be decomposed into a sum of sinusoids.

Decomposing values into “basic” parts, is prevalent in linear algebra via bases of vector spaces.


Proof of Proposition 2:

We’ll show by contradiction and some tedious algebra that there’s only one way the do a Hermitian decomposition.

Suppose there is some symmetric function $f’_s$ that is different than $f_s$. For this to be true, there must exist at least one $x’$ such that $f’_s(x’) \ne f_s(x’)$.

Since $f’_s(x’)$ is a complex number, either its real part or imaginary part, or both, must be different from that of $f_s(x’)$. First let’s assume at least the real part is different and that, without loss of generality, $Re\curly{f’_s(x’)} > Re\curly{f_s(x’)}$.

Because it has to satisfy (1) and we know that $f_a(x’) \ne f(x’) - f’_s(x’)$, so there must also be an antisymmetric function $f’_a \ne f_a$ such that $f’_a(x’) = f(x’) - f’_s(x’)$. If we only consider the real part, $Re\curly{f’_a(x’)} = Re\curly{f(x’)} - Re\curly{f’_s(x’)} < Re\curly{f(x’)} - Re\curly{f_s(x’)} = Re\curly{f_a(x’)}$, that is

\[(3) \quad Re\curly{f'_a(x')} < Re\curly{f_a(x')}\]

Since $f’_s$ is symmetric, we have $f’_s(x’) = \overline{f’_s(-x’)}$ or $Re\curly{f’_s(x’)} = Re\curly{f’_s(-x’)}$. Since $f’_s(x’)$ is also symmetric, we have

\[(4) \quad Re\curly{f'_s(-x')} > Re\curly{f_s(-x')}\]

Applying (1) for $-x’$ and considering only the real part we have $Re\curly{f’_a(-x’)} = Re\curly{f(-x’)} - Re\curly{f’_s(-x’)}$ By (4), $Re\curly{f’_a(-x’)} < Re\curly{f(-x’)} - Re\curly{f_s(-x’)} = Re\curly{f_a(-x’)}$, that is

\[Re\curly{f'_a(-x')} < Re\curly{f_a(-x')}\]


\[(5) \quad -Re\curly{f'_a(-x')} > -Re\curly{f_a(-x')}\]

Since $f’_a$ is antisymmetric, $Re\curly{f’_a(x’)} = -Re\curly{f’_a(-x’)}$, then by (5) $Re\curly{f’_a(x’)} > -Re\curly{f_a(-x’)} = Re\curly{f_a(x’)}$, which contradicts (3).

For the second part, let’s assume the imaginary part is different and that $Im\curly{f’_s(x’)} > Im\curly{f_s(x’)}$. It will be pretty similar to the first part, but we have to be careful with the signs. Because it has to satisfy (1), by an analogous reasoning we’ll get

\[(6) \quad Im\curly{f'_a(x')} < Im\curly{f_a(x')}\]

Since $f’_s$ and $f_s$ are symmetric, we have $Im\curly{f’_s(x’)} = -Im\curly{f’_s(-x’)}$ and $Im\curly{f_s(x’)} = -Im\curly{f_s(-x’)}$. We started with $Im\curly{f’_s(x’)} > Im\curly{f_s(x’)}$ which implies $-Im\curly{f’_s(x’)} < -Im\curly{f_s(x’)}$ and thus

\[(7) \quad Im\curly{f'_s(-x')} < Im\curly{f_s(-x')}\]

Applying (1) for $-x’$ we have $Im\curly{f'_a(-x')} = Im\curly{f(-x')} - Im\curly{f'_s(-x')}$. From (7) $Im\curly{f'_a(-x')} > Im\curly{f(-x')} - Im\curly{f_s(-x')} = Im\curly{f_a(-x')}$, that is

\[(8) \quad Im\curly{f'_a(-x')} > Im\curly{f_a(-x')}\]

Since $f’_a$ is antisymmetric, $Im\curly{f’_a(x’)} = Im\curly{f’_a(-x’)}$ (negation + conjugate), then by (8) $Im\curly{f’_a(x’)} > Im\curly{f_a(-x’)} = Im\curly{f_a(x’)}$, which contradicts (6).

In either case, we reach a contradiction if we assume there exists a different way of decomposing $f$ into symmetric and antisymmetric Hermitian functions, so it must be it’s unique.



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