The Overdetermined System $A\vec u = \vec b$

The linear algebra aspects of Maxwell's wave operator $\mathcal A$ are illustrated by the following problem from linear algebra:

Solve $A\vec u = \vec b$ for $\vec u$ , under the stipulation that

$$A: ~~R^4 \longrightarrow R^4$$

$$\vec u_r: ~~A\vec u_r=\vec 0\quad \textrm{so that }{\mathcal N}(A)=span \{\vec u_r \}$$

$$\vec u_\ell^T: ~~\vec u_\ell^T A=\vec 0\quad \textrm{so that } {\mathcal N}(A^T)=span \{\vec u_\ell \}$$

$$\vec b: ~~\vec b \in {\mathcal R}(A)\quad \textrm{so that }\vec u_\ell^T \vec b=0$$ (653)

The fact that $A$ is singular and $\vec b$ belongs to the range of $A$ makes the system over-determined but consistent. This means that there are more equations than there are unknowns.
One solves the problem in two steps.

Step I:

Let $\{ \vec v_1,\vec v_2,\vec v_3 \}$ be the set of eigenvectors having non-zero eigenvalues. Whatever $A$ is, the task of finding three vectors that satisfy

$$\left. \begin{array}{c} A\vec v_1 c_1=~\vec v_1 \lambda_1c_1\\ A\... ..._3 \lambda_3c_3 \end{array} \right\} \lambda_i\not =0,~c_i~\textrm{are scalars}$$ (654)
and

$$~A\vec u_r c_4=\vec 0~.$$ (655)

Being spanned by the three eigenvectors with non-zero eigenvalues, the range space of $A$ ,

$${\mathcal R}(A)=span \{\vec v_1,\vec v_2,\vec v_3\}~,$$

is well-determined. However, the scalars $c_i$ are at this stage as yet undetermined.

Step II:

Continuing the development, recall that quite generally

$$A\vec u =\vec b~\quad\quad\textrm{has a solution} \Leftrightarrow \vec b\in{\mathcal R}(A)$$

$$~ \Leftrightarrow \vec b =\vec v_1 b_1+\vec v_2 b_2+\vec v_3 b_3~,$$ (656)

and that if

$$\vec u =\vec v_1 c_1+\vec v_2 c_2+\vec v_3 c_3+\vec u_4 c_4~,$$
then

$$A\vec u =~\vec v_1 \lambda_1c_1+\vec v_2 \lambda_2c_2+\vec v_3 \lambda_3c_3~.$$ (657)

It is appropriate to alert the reader that in the ensuing section the vectors $\vec v_i$ and the eigenvalues $\lambda_i$ become differential operators which act on scalar fields $c_i$ and that the three subscript labels will refer to the TE, TM, and TEM eletromagnetic⁶⁶ vector potentials respectively.
Equating (6.56) and (6.57), one finds that the linear independence of $\{ \vec v_1,\vec v_2,\vec v_3 \}$ implies the following equations for $c_1$ , $c_2$ , and $c_3$ :

$$\lambda_1c_1 =b_1\longrightarrow c_1=\frac{1}{\lambda_1}\,b_1$$ (658)

$$\lambda_2c_2 =b_2\longrightarrow c_2=\frac{1}{\lambda_2}\,b_2$$ (659)

$$\lambda_3c_3 =b_3\longrightarrow c_3=\frac{1}{\lambda_3}\,b_3$$ (660)

Consequently, the solution is

$$\vec u=\vec v_1 \frac{1}{\lambda_1}\,b_1+\vec v_2 \frac{1}{\lambda_2}\,b_2+\vec v_3 \frac{1}{\lambda_3}\,b_3+\vec u_4 c_4$$

where $\vec u_4 c_4$ is an indeterminate multiple of the null space vector $\vec u_4$ .

If one represents the stated problem $A\vec u = \vec b$ ($\vec u$ determines $\vec b$ ) as an input-output process, as in Figure 6.3,

Figure 6.3: The matrix $A$ defines an input-output process.

then its solution is represented by the inverse input-output process as in Figure 6.4.

Figure: The solution to $A\vec u = \vec b$ defines an inverse input-output process.

In general, the task of finding the eigenvectors of a 4$\times$ 4 matrix can be a nontrivial task. However, given the fact that the solution to

$$\vec u_\ell^T A=\vec0$$

is already known, one finds that the associated constraints on the eigenvectors,

$$\vec u_\ell^T \vec v_i=0$$

make the task quite easy, if not trivial.

Footnotes

... eletromagnetic⁶⁶

The acronyms TE, TM, as well as TEM stand for transverse electric, transverse magnetic, and transeverse electric magnetic The justification for these apellations are given on Pages , , and , repectively.