The Quaternion Fourier Number Transform

Abstract

In this paper, we introduce the quaternion Fourier number transform (QFNT), which corresponds to a quaternionic version of the well-known number-theoretic transform. We derive several theoretical results necessary to define the QFNT and investigate its main properties. Differently from other quaternion transforms, which are defined over Hamilton’s quaternions, the QFNT requires considering a quaternion algebra over a finite field. Thus, its computation involves integer arithmetic only, avoiding truncation and rounding-off errors. We give an illustrative example regarding the application of the QFNT to digital color image processing.

Appendix

If $p\equiv 1\,(\mathrm {mod}\,\,4)$, one has $i=\sqrt{-1}\in {\mathbb {F}}_p$ [5]. The multiplicative order of $\mathbf {Q}$ can be determined according to the following cases.

• Case 1: $b^2+c^2+d^2=0$. In this case, one has $\lambda _1=\lambda _2=a\in {\mathbb {F}}_p$. The following subcases have to be considered.

  • Subcase 1.1: $b=c=d=0$. The condition determining the current subcase is identical to that considered in subcase 1.1 for $p\equiv 3\,(\mathrm {mod}\,\,4)$ and, therefore, leads to the same result.

  • Subcase 1.2: $b= 0$ and $c\ne 0$ and $d\ne 0$. In this case, one has $c^2=-d^2\Rightarrow c=\pm di$. Assuming that $c=di$, $\mathbf {Q}$ reduces to a matrix in the form

    $$\begin{aligned} \mathbf {Q}=\left[ \begin{array}{cc} a&{}\quad 2di\\ 0&{}\quad a \end{array}\right] . \end{aligned}$$

    (35)

    Denoting by $\mathbf {v} = [v(0)\,\, v(1)]$ an eigenvector of $\mathbf {Q}$, one may write $\mathbf {Q}\mathbf {v}^T=a\mathbf {v}^T$, which produces the system of equations

    $$\begin{aligned} \left\{ \begin{array}{ll} a v(0) + 2div(1)=av(0)\\ av(1)=av(1). \end{array}\right. \end{aligned}$$

    (36)

    The solution for (36) is simply $v(1)=0$, which indicates that the geometric multiplicity of $\lambda _1=\lambda _2=a$ is $m_g(a)=1$. Therefore, $\mathbf {Q}$ is not diagonalizable and admits the Jordan normal form

    $$\begin{aligned} \mathbf {J_Q}=\left[ \begin{array}{cc} a&{}\quad 1\\ 0&{}\quad a \end{array}\right] . \end{aligned}$$

    (37)

    Analogously to subcase 1.2 for $p\equiv 3\,(\mathrm {mod}\,\,4)$, one concludes that $\mathrm {ord}(\mathbf {Q})=\mathrm {ord}(\mathbf {J_Q})=\mathrm {lcm}(\mathrm {ord}(a),p)$. One obtains the same result if $c=-di$ or $c=0$ and $b\ne 0$ and $d\ne 0$ or $d=0$ and $b\ne 0$ and $c\ne 0$.

  • Subcase 1.3: $b\ne 0$ and $c\ne 0$ and $d\ne 0$. In this case, one may write $b^2=-(c^2+d^2)\Rightarrow b=\pm i\sqrt{c^2+d^2}$. Assuming that $b=i\sqrt{c^2+d^2}$, $\mathbf {Q}$ reduces to a matrix in the form

    $$\begin{aligned} \mathbf {Q}=\left[ \begin{array}{cc} a-\sqrt{c^2+d^2}&{}c+di\\ -c+di&{}a+\sqrt{c^2+d^2} \end{array}\right] . \end{aligned}$$

    (38)

    Denoting by $\mathbf {v} = [v(0)\,\, v(1)]$ an eigenvector of $\mathbf {Q}$, one may write $\mathbf {Q}\mathbf {v}^T=a\mathbf {v}^T$, which produces the system of equations

    $$\begin{aligned} \left\{ \begin{array}{ll} -\sqrt{c^2+d^2} v(0) + (c+di)v(1)=0\\ (-c+di)v(0)+\sqrt{c^2+d^2}v(1)=0 \end{array}\right. . \end{aligned}$$

    (39)

    From (39), one obtains the relationship

    $$\begin{aligned} v(0=-\frac{\sqrt{c^2+d^2}}{-c+di}v(1), \end{aligned}$$

    which indicates that the geometric multiplicity of $\lambda _1=\lambda _2=a$ is $m_g(a)=1$. Therefore, $\mathbf {Q}$ is not diagonalizable and admits the Jordan normal form (37). Analogously to subcase 1.2 for $p\equiv 3\,(\mathrm {mod}\,\,4)$, one concludes that $\mathrm {ord}(\mathbf {Q})=\mathrm {ord}(\mathbf {J_Q})=\mathrm {lcm}(\mathrm {ord}(a),p)$. One obtains the same result if $b=-i\sqrt{c^2+d^2}$. A similar development can be carried out if we use $c=\pm i\sqrt{b^2+d^2}$ or $d=\pm i\sqrt{b^2+c^2}$.

• Case 2: $b^2+c^2+d^2\ne 0$. In this case, $\mathbf {Q}$ has two distinct eigenvalues and admits the diagonal form

  $$\begin{aligned} \mathbf {\Lambda _Q}=\left[ \begin{array}{cc} \lambda _1&{}0\\ 0&{}\lambda _2 \end{array}\right] . \end{aligned}$$

  (40)

  Thus, $\mathrm {ord}(\mathbf {Q})=\mathrm {ord}(\mathbf {\Lambda _Q})=\mathrm {lcm}(\lambda _1,\lambda _2)$.