B. Special matrices

  
B.1. Unitary matrices

A $2 \times 2\$matrix $\vec{Y}$ is unitary if $\vec{Y}\vec{Y}^{\dagger}= \mathbf{I}$. To derive its quaternion form I cast the most general quaternion $[\, a+\vec{a}\,]$ in the form

 

$$\vec{Y}= {\rm e}^{i\xi}\, [\, y + {\vec{x}}+ i\,\vec{y}\,] \,, \quad y, {\vec{x}}, \vec{y}\mbox{ real} .$$ (31)

I now expand $\vec{Y}\vec{Y}^{\dagger}$ in quaternion form and require that it equal I:

$$\begin{eqnarray*}\vec{Y}\,\vec{Y}^{\dagger}&=& [\, y + {\vec{x}}+ i\,\vec{y}\,] ... ...^2 + 2 (y{\vec{x}}+ \,{\vec{x}}\times\vec{y}) \,] = [\,1\,] . \end{eqnarray*}$$

Since ${\vec{x}}\times \vec{y}$ is perpendicular to ${\vec{x}}$, the vector part in the product can vanish only if ${\vec{x}}=\vec{0}$. It follows that

$$\begin{eqnarray*}y^2 + \vec{y}^2 = 1 . \end{eqnarray*}$$

Now, if $\vec{Y}$ is to be unimodular, $\xi$ must be 0. Hence we may rewrite Eq. (31) as

 

$$\vec{Y} [\, \cos\eta + i \vec{1}_{\vec{y}} \sin\eta \,]$$ (32)

$\vec{Y}$ is completely defined by the three real components of its Gibbs vector ([Korn & Korn 1961]) $\vec{1}_{\vec{y}} \sin\eta$. Since $\vec{1}_{\vec{y}}^2 = 1$, it can be shown from the Taylor-series expansions of the cosine and sine functions that

$$\begin{eqnarray*}\cos [\, \vec{1}_{\vec{y}} \,] \eta = \cos \eta, \quad \sin [\, \vec{1}_{\vec{y}}\,] \eta = [\, \vec{1}_{\vec{y}} \,] \sin\eta \end{eqnarray*}$$

and hence

$$\begin{eqnarray*}\vec{Y}= \exp i \,[\, \vec{1}_{\vec{y}} \eta \,] . \end{eqnarray*}$$

For small $\eta$, we may replace the exponential by its first-order approximation. The value of $\eta$ then provides a direct measure for the deviation of $\vec{Y}$ from I.

  
B.2. Complex linear polarized brightness

In the quasi-linear treatment of polarization, the linearly polarized visibility and brightness frequently appear in the form of the complex variable Q+iU, cf. (Eq. 19). This form is directly related to the quaternion exponential above. In particular, for a unit vector in the q,u plane of quaternion-vector space

$$\begin{eqnarray*}[\, \vec{1}_{\mathbf{q}}\cos\phi +\vec{1}_{\mathbf{u}}\sin\phi ... ...\mathbf{q}}\,] \,\exp -i\,[\, \vec{1}_{\mathbf{v}} \,] \,\phi . \end{eqnarray*}$$

  
B.3. Unitary Jones matrices and perfect feeds

In Sect. 8 unitary Jones matrices were postulated. Since $\vec{Y}\mathbf{I}\vec{Y}^{\dagger}= \mathbf{I}$, a feed with such a matrix transfers all incident radiation to its output: it must be loss-free and matched at its in- and outputs. Matching implies that either receptor must absorb all the radiation that the other one does not: the receptors must be of opposite polarizations ([Born & Wolf 1964]; [Cornbleet 1976]; [Thompson et al. 1986]).

This is the way feeds for radio telescopes are normally designed. Note, however, that e.g. a stationary pair of crossed dipole receptors is not matched to radiation from an arbitrary direction. Designs for arrays of phased dipoles will have to take the problems ensuring into account.

  
B.4. Positive hermitian matrices

$\vec{H}$ is hermitian or self-adjoint if $\vec{H}= \vec{H}^{\dagger}$; since the Pauli matrices are hermitian, the quaternion form $\vec{H}= [\, h + \vec{h}\,]$ of a hermitian $2 \times 2\$matrix is real, cf. Eq. (23). A $2 \times 2\$matrix is positive if its eigenvalues are both positive. An equivalent condition is that both its trace and its determinant are positive.

For the matrix to be positive hermitian and unimodular

$$\begin{eqnarray*}h > 0; \qquad h^2 - \vec{h}^2 = 1 . \end{eqnarray*}$$

We may then write it as

 

$$\vec{H}= \cosh\gamma + [\, \vec{1}_{\vec{h}} \,] \sinh\gamma \equiv \exp \,[\, \vec{1}_{\vec{h}} \,] \,\gamma .$$ (33)

It is completely defined by the three real components of $\vec{1}_{\vec{h}}\sinh\gamma$ which I will also call a Gibbs vector. It is readily shown from Eq. (33) that

 

$$\vec{H}^2 = \exp \,[\, \vec{1}_{\vec{h}} \,] \,2\gamma .$$ (34)

  
B.5. Matrix square root

We will need the positive hermitian square root $\vec{H}$ of the product $\vec{M}\vec{M}^{\dagger}$ for an arbitrary $2 \times 2\$matrix $\vec{M}$. Let

$$\begin{eqnarray*}\vec{M}= [\, m + \vec{m}\,] . \end{eqnarray*}$$

Then

$$\begin{eqnarray*}\vec{A}= \vec{M}\vec{M}^{\dagger}= [\, mm^* + \vec{m}\cdot \vec{m}^* + 2{\rm Re\,}m\vec{m}\,] \end{eqnarray*}$$

is readily shown to be positive hermitian. We now seek to find a positive hermitian matrix $\vec{H}$ such that $\vec{H}\vec{H}= \vec{A}$, that is

 

$$h^2 + \vec{h}^2 = a; \quad 2h\vec{h}= \vec{a}.$$ (35)

Conceptually the simplest way to find the root is through Eqs. (33) and (34). Computationally it is more efficient to solve the quadratic equation that Eq. (35) represents. Out of four possible solutions, the positive definite one is

$$\begin{eqnarray*}\vec{H}&=& \sqrt{(\, a + \sqrt{ a^2 - \vec{a}^2 } \;) /2 } \\ &... ...vec{a}} \,] \,\sqrt{ (\, a - \sqrt{ a^2 - \vec{a}^2 }\;) /2 } . \end{eqnarray*}$$

  
B.6. Polar decomposition

An arbitrary matrix $\vec{X}$ can be represented ([Lancaster & Tismenetsky 1985]) as the product of a unitary and a positive hermitian matrix:

$$\begin{eqnarray*}\vec{X}= \vec{H}\vec{Y}\quad\mbox{ with }\quad \vec{Y}\vec{Y}^{... ...r}=\mathbf{I}\quad\mbox{ and }\quad \vec{H}^{\dagger}= \vec{H}. \end{eqnarray*}$$

This is the matrix/quaternion analogue of the polar form of a complex scalar. Defining

$$\begin{eqnarray*}x = \sqrt{ \det \vec{X}} \end{eqnarray*}$$

we may rewrite the decomposition as

$$\begin{eqnarray*}\vec{X}= x \vec{H}\vec{Y} \end{eqnarray*}$$

where $\vec{H}$ and $\vec{Y}$ are now unimodular. To find $\vec{H}$ we form the product

$$\begin{eqnarray*}(xx*)^{-1} \vec{X}\vec{X}^{\dagger}= \vec{H}\vec{Y}\,\vec{Y}^{\dagger}\vec{H}^{\dagger} = \vec{H}\vec{H} \end{eqnarray*}$$

and find $\vec{H}$ by taking the positive hermitian square root.