Unlike the previous two cases that involve only 3 × 3 matrices Λ_{1} and Λ_{2}, the third case that we shall study involves the 4 × 4 matrix Λ_{3}. The zero structure of Λ_{3} ensures that *u*_{2}, *u*_{4}, *v*_{1},*v*_{3} are not present in the equations, so they are free vectors in the solutions. The following lemma gives all the solutions of the last equation of (8).

#### Lemma 4.1

*The solutions* *K* *of the equation (9)* *with* Λ = Λ_{3} *are*
$$\begin{array}{}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}{K}_{11}\equiv \left[\begin{array}{cccc}0& {a}_{12}& 0& {a}_{14}\\ 0& {a}_{22}& 0& {a}_{24}\\ 0& {a}_{32}& 0& {a}_{34}\\ 0& {a}_{42}& 0& {a}_{44}\end{array}\right];\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{K}_{12}\equiv \left[\begin{array}{cccc}0& {a}_{12}& 0& {a}_{14}\\ 0& {a}_{22}& 0& {a}_{24}\\ 0& {a}_{32}& {a}_{33}& {a}_{34}\\ 0& 0& 0& 0\end{array}\right],\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{a}_{33}\ne 0;\\ \\ {K}_{13}\equiv \left[\begin{array}{cccc}0& {a}_{12}& {a}_{13}& {a}_{14}\\ 0& {a}_{22}& 0& {a}_{24}\\ 0& {a}_{32}& {a}_{33}& {a}_{34}\\ 0& 0& 0& 0\end{array}\right],{a}_{13}\ne 0;\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{K}_{14}\equiv \left[\begin{array}{cccc}0& {a}_{12}& {a}_{13}& {a}_{14}\\ 0& 0& 0& 0\\ {a}_{31}& {a}_{32}& {a}_{33}& {a}_{34}\\ 0& 0& 0& 0\end{array}\right],{a}_{13}{a}_{31}\ne 0;\\ \\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{K}_{15}\equiv \left[\begin{array}{cccc}0& {a}_{12}& 0& {a}_{14}\\ 0& -\frac{{a}_{33}{a}_{42}}{{a}_{31}}& 0& -\frac{{a}_{33}{a}_{44}}{{a}_{31}}\\ {a}_{31}& {a}_{32}& {a}_{33}& {a}_{34}\\ 0& {a}_{42}& 0& {a}_{44}\end{array}\right],{a}_{31}\ne 0;\\ \\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{K}_{16}\equiv \left[\begin{array}{cccc}{a}_{11}& {a}_{12}& {a}_{13}& {a}_{14}\\ 0& -\frac{{a}_{13}{a}_{42}}{{a}_{11}}& 0& -\frac{{a}_{13}{a}_{44}}{{a}_{11}}\\ {a}_{31}& {a}_{32}& \frac{{a}_{13}{a}_{31}}{{a}_{11}}& {a}_{34}\\ 0& {a}_{42}& 0& {a}_{44}\end{array}\right],{a}_{11}\ne 0;\\ \\ {K}_{17}\equiv \left[\begin{array}{cccc}{a}_{11}& {a}_{12}& {a}_{13}& {a}_{14}\\ 0& 0& 0& 0\\ {a}_{31}& {a}_{32}& {a}_{33}& {a}_{34}\\ 0& 0& 0& 0\end{array}\right],{a}_{11}\ne 0,\left[\begin{array}{cc}{a}_{11}& {a}_{13}\\ {a}_{31}& {a}_{33}\end{array}\right]\ne 0.\end{array}$$

#### Proof

Denote
$$\begin{array}{}K=\left[\begin{array}{llll}{a}_{11}& {a}_{12}& {a}_{13}& {a}_{14}\\ {a}_{21}& {a}_{22}& {a}_{23}& {a}_{24}\\ {a}_{31}& {a}_{32}& {a}_{33}& {a}_{34}\\ {a}_{41}& {a}_{42}& {a}_{43}& {a}_{44}\end{array}\right].\end{array}$$

Then the equation *K* Λ_{3} *K* = Λ_{3}*K*Λ_{3} is just
$$\begin{array}{}\left[\begin{array}{c}{a}_{11}{a}_{21}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{a}_{13}{a}_{41}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{a}_{11}{a}_{22}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{a}_{13}{a}_{42}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{a}_{11}{a}_{23}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{a}_{13}{a}_{43}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{a}_{11}{a}_{24}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{a}_{13}{a}_{44}\\ {a}_{21}{a}_{21}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{a}_{23}{a}_{41}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{a}_{21}{a}_{22}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{a}_{23}{a}_{42}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{a}_{21}{a}_{23}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{a}_{23}{a}_{43}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{a}_{21}{a}_{24}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{a}_{23}{a}_{44}\\ {a}_{31}{a}_{21}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{a}_{33}{a}_{41}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{a}_{31}{a}_{22}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{a}_{33}{a}_{42}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{a}_{31}{a}_{23}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{a}_{33}{a}_{43}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{a}_{31}{a}_{24}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{a}_{33}{a}_{44}\\ {a}_{41}{a}_{21}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{a}_{43}{a}_{41}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{a}_{41}{a}_{22}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{a}_{43}{a}_{42}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{a}_{41}{a}_{23}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{a}_{43}{a}_{43}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{a}_{41}{a}_{24}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}{a}_{43}{a}_{44}\end{array}\right]\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left[\begin{array}{cccc}0& {a}_{21}& 0& {a}_{23}\\ 0& 0& 0& 0\\ 0& {a}_{41}& 0& {a}_{43}\\ 0& 0& 0& 0\end{array}\right].\end{array}$$(22)

Comparing the (2,1),(4,3),(2,3),(4,1) entries above gives the four equations
$$\begin{array}{}\left\{\begin{array}{l}{a}_{21}^{2}=-{a}_{23}{a}_{41},\\ {a}_{43}^{2}=-{a}_{23}{a}_{41},\\ {a}_{23}({a}_{21}+{a}_{43})=0,\\ {a}_{41}({a}_{21}+{a}_{43})=0.\end{array}\right.\end{array}$$(23)

We discuss two cases separately. First assume *a*_{41} = 0. Then *a*_{21} = *a*_{43} = 0 from (23). If *a*_{23} ≠ 0, then *a*_{11} = *a*_{31} = *a*_{44} = 0 from equating entries (1,3),(3,3),(2,4) of both sides in (22), from which *a*_{23} = *a*_{11} *a*_{24} + *a*_{13}*a*_{44} = 0, a contradiction. Thus *a*_{23} = 0 and (22) is reduced to
$$\begin{array}{}\left\{\begin{array}{l}{a}_{11}{a}_{22}+{a}_{13}{a}_{42}=0,\\ {a}_{31}{a}_{22}+{a}_{33}{a}_{42}=0,\\ {a}_{11}{a}_{24}+{a}_{13}{a}_{44}=0,\\ {a}_{31}{a}_{24}+{a}_{33}{a}_{44}=0\end{array}\right.\end{array}$$(24)

with *a*_{21} = *a*_{23} = *a*_{41} = *a*_{43} = 0.

Now we assume *a*_{41}≠ 0. Then the last equation of (23) implies that *a*_{21} = −*a*_{43}. On the other hand, the entries (4,2) and (1,1) equations of (22) gives that *a*_{22} = −*a*_{43}*a*_{42}/*a*_{41} and *a*_{13} = −*a*_{11}*a*_{21}/*a*_{41}. Hence
$${a}_{21}={a}_{11}{a}_{22}+{a}_{13}{a}_{42}=-\frac{{a}_{11}{a}_{43}{a}_{42}}{{a}_{41}}-\frac{{a}_{11}{a}_{21}{a}_{42}}{{a}_{41}}=\frac{{a}_{11}{a}_{21}{a}_{42}}{{a}_{41}}-\frac{{a}_{11}{a}_{21}{a}_{42}}{{a}_{41}}=0,$$

and so *a*_{43} = 0. On the other hand, the entries (3,1) and (4,2) equalities give *a*_{33} = *a*_{22} = 0, so *a*_{41} = *a*_{31}*a*_{22} + *a*_{33}*a*_{42} = 0 from the entry (3,2) equality of (22), contradictory to the assumption that *a*_{41}≠ 0.

Therefore, (24) is the only equation for us to solve. Let *a*_{11} = 0. Then (24) becomes *a*_{13}*a*_{42} = 0,*a*_{31}*a*_{22} + *a*_{42} = 0,*a*_{13}*a*_{44} = 0, *a*_{31}*a*_{24} + *a*_{33}*a*_{44} = 0. If *a*_{13} = 0, then the above system is *a*_{31}*a*_{22} + *a*_{33}*a*_{42} = 0, *a*_{31}*a*_{24} + *a*_{33}*a*_{44} = 0. The first case is *a*_{31} = 0. Then *a*_{33}*a*_{42} = 0 and *a*_{33}*a*_{44} = 0. If *a*_{33} = 0, then we have solution matrix *K*_{11}; otherwise *K*_{12} occurs. For the second case *a*_{31} ≠ 0, there hold *a*_{22} = −*a*_{33}*a*_{42}/*a*_{31} and *a*_{24} = −*a*_{33}*a*_{44}/*a*_{31}, giving *K*_{15}. If *a*_{13} ≠ 0, then *a*_{42} = 0 and *a*_{44} = 0, so *a*_{31}*a*_{22} = 0, *a*_{31}*a*_{24} = 0. The case that *a*_{31} = 0 implies *K*_{13} and when *a*_{31} ≠ 0, we obtain *K*_{14}.

Now assume that *a*_{11} ≠ 0. Then *a*_{22} = −*a*_{13}*a*_{42}/*a*_{11} and *a*_{24}= −*a*_{13}*a*_{44}/*a*_{11}, so *a*_{31}*a*_{22} + *a*_{33}*a*_{42} = 0 and *a*_{31}*a*_{24} + *a*_{33}*a*_{44} = 0, or equivalently,
$$\begin{array}{}\left\{\begin{array}{l}{a}_{42}\left({a}_{33}-\frac{{a}_{13}{a}_{31}}{{a}_{11}}\right)=0,\\ {a}_{44}\left({a}_{33}-\frac{{a}_{13}{a}_{31}}{{a}_{11}}\right)=0.\end{array}\right.\end{array}$$

In the case that *a*_{33} = *a*_{13}*a*_{31}/*a*_{11}, the numbers *a*_{42} and *a*_{44} are arbitrary and the matrix *K*_{16} appears, and otherwise *a*_{42} = *a*_{44} = 0, so *a*_{22} = *a*_{24} = 0, resulting in *K*_{17}. This completes the proof. □

Lemma 4.1 gives all the solutions *K*_{11}, …, *K*_{17} of the last equation of (8) with Λ = Λ_{3}, which can be used to prove the next result.

#### Theorem 4.2

*Suppose* Λ = Λ_{3} *in the Jordan form (4)* *of* *A* = *PQ*^{T}. *Then all the solutions of (1)* *are* *X* = *WYW*^{−1}, *where* *Y* *is partitioned as* (7) *in which* *M* *is an arbitrary* (*n* − 4) × (*n* − 4) *matrix such that Y* =
$$\begin{array}{}\left[\begin{array}{ccccc}M& 0& {u}_{2}& 0& {u}_{4}\\ {v}_{1}^{T}& 0& {a}_{12}& 0& {a}_{14}\\ {v}_{2}^{T}& 0& {a}_{22}& 0& {a}_{24}\\ {v}_{3}^{T}& 0& {a}_{32}& 0& {a}_{34}\\ {v}_{4}^{T}& 0& {a}_{42}& 0& {a}_{44}\end{array}\right],\left[\begin{array}{ccccc}M& 0& {u}_{2}& {u}_{3}& {u}_{4}\\ {v}_{1}^{T}& 0& {a}_{12}& 0& {a}_{14}\\ {v}_{2}^{T}& 0& {a}_{22}& 0& {a}_{24}\\ {v}_{3}^{T}& 0& {a}_{32}& 0& {a}_{34}\\ {0}^{T}& 0& 0& 0& 0\end{array}\right],\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{u}_{3}\phantom{\rule{thinmathspace}{0ex}}\ne \phantom{\rule{thinmathspace}{0ex}}0,\end{array}$$(25)
$$\begin{array}{}\left[\begin{array}{ccccc}M& {u}_{1}& {u}_{2}& {u}_{3}& {u}_{4}\\ {v}_{1}^{T}& 0& {a}_{12}& 0& {a}_{14}\\ -\frac{{u}_{1}^{H}\phantom{\rule{thinmathspace}{0ex}}{u}_{3}}{\parallel {u}_{1}{\parallel}^{2}}{v}_{4}^{T}& 0& -\frac{{u}_{1}^{H}\phantom{\rule{thinmathspace}{0ex}}{u}_{3}}{\parallel {u}_{1}{\parallel}^{2}}{a}_{42}& 0& -\frac{{u}_{1}^{H}\phantom{\rule{thinmathspace}{0ex}}{u}_{3}}{\parallel {u}_{1}{\parallel}^{2}}{a}_{44}\\ {v}_{3}^{T}& 0& {a}_{32}& 0& {a}_{34}\\ {v}_{4}^{T}& 0& {a}_{42}& 0& {a}_{44}\end{array}\right]\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}{u}_{1}\ne \phantom{\rule{thinmathspace}{0ex}}0,\end{array}$$(26)
$$\begin{array}{}\left[\begin{array}{ccccc}M& 0& {u}_{2}& {u}_{3}& {u}_{4}\\ {v}_{1}^{T}& 0& {a}_{12}& 0& {a}_{14}\\ {v}_{2}^{T}& 0& {a}_{22}& 0& {a}_{24}\\ {v}_{3}^{T}& 0& {a}_{32}& {a}_{33}& {a}_{34}\\ {0}^{T}& 0& 0& 0& 0\end{array}\right],\left[\begin{array}{ccccc}M& {u}_{1}& {u}_{2}& {u}_{3}& {u}_{4}\\ {v}_{1}^{T}& 0& {a}_{12}& 0& {a}_{14}\\ {0}^{T}& 0& 0& 0& 0\\ {v}_{3}^{T}& 0& {a}_{32}& {a}_{33}& {a}_{34}\\ {0}^{T}& 0& 0& 0& 0\end{array}\right],\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{u}_{1}\phantom{\rule{thinmathspace}{0ex}}\ne \phantom{\rule{thinmathspace}{0ex}}0,\phantom{\rule{thinmathspace}{0ex}}{a}_{33}\phantom{\rule{thinmathspace}{0ex}}\ne \phantom{\rule{thinmathspace}{0ex}}0,\end{array}$$(27)
$$\begin{array}{}\left[\begin{array}{ccccc}M& 0& {u}_{2}& {u}_{3}& {u}_{4}\\ {v}_{1}^{T}& 0& {a}_{12}& {a}_{13}& {a}_{14}\\ {v}_{2}^{T}& 0& {a}_{22}& 0& {a}_{24}\\ {v}_{3}^{T}& 0& {a}_{32}& {a}_{33}& {a}_{34}\\ {0}^{T}& 0& 0& 0& 0\end{array}\right],\left[\begin{array}{ccccc}M& {u}_{1}& {u}_{2}& {u}_{3}& {u}_{4}\\ {v}_{1}^{T}& 0& {a}_{12}& {a}_{13}& {a}_{14}\\ {0}^{T}& 0& 0& 0& 0\\ {v}_{3}^{T}& 0& {a}_{32}& {a}_{33}& {a}_{34}\\ {0}^{T}& 0& 0& 0& 0\end{array}\right],\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{u}_{1}\phantom{\rule{thinmathspace}{0ex}}\ne \phantom{\rule{thinmathspace}{0ex}}0,\phantom{\rule{thinmathspace}{0ex}}{a}_{13}\phantom{\rule{thinmathspace}{0ex}}\ne \phantom{\rule{thinmathspace}{0ex}}0,\end{array}$$(28)
$$\begin{array}{}\left[\begin{array}{ccccc}M& {u}_{1}& {u}_{2}& {u}_{3}& {u}_{4}\\ {v}_{1}^{T}& 0& {a}_{12}& {a}_{13}& {a}_{14}\\ {0}^{T}& 0& 0& 0& 0\\ {v}_{3}^{T}& {a}_{31}& {a}_{32}& {a}_{33}& {a}_{34}\\ {0}^{T}& 0& 0& 0& 0\end{array}\right],\left[\begin{array}{ccccc}M& {u}_{1}& {u}_{2}& \frac{{a}_{33}}{{a}_{31}}{u}_{1}& {u}_{4}\\ {v}_{1}^{T}& 0& {a}_{12}& 0& {a}_{14}\\ -\frac{{a}_{33}}{{a}_{31}}{v}_{4}^{T}& 0& -\frac{{a}_{33}{a}_{42}}{{a}_{31}}& 0& -\frac{{a}_{33}{a}_{44}}{{a}_{31}}\\ {v}_{3}^{T}& {a}_{31}& {a}_{32}& {a}_{33}& {a}_{34}\\ {v}_{4}^{T}& 0& {a}_{42}& 0& {a}_{44}\end{array}\right]\end{array}$$(29)
*with* *a*_{13} ≠ 0; *a*_{31} ≠ 0,
$$\begin{array}{}\left[\begin{array}{ccccc}M& {u}_{1}& {u}_{2}& {u}_{3}& {u}_{4}\\ {v}_{1}^{T}& 0& {a}_{12}& 0& {a}_{14}\\ {0}^{T}& 0& 0& 0& 0\\ {v}_{3}^{T}& {a}_{31}& {a}_{32}& {a}_{33}& {a}_{34}\\ {0}^{T}& 0& 0& 0& 0\end{array}\right],\left[\begin{array}{ccccc}M& {u}_{1}& {u}_{2}& {u}_{3}& {u}_{4}\\ {v}_{1}^{T}& {a}_{11}& {a}_{12}& {a}_{13}& {a}_{14}\\ {0}^{T}& 0& 0& 0& 0\\ {v}_{3}^{T}& {a}_{31}& {a}_{32}& \frac{{a}_{13}{a}_{31}}{{a}_{11}}& {a}_{34}\\ {0}^{T}& 0& 0& 0& 0\end{array}\right]\end{array}$$(30)

*with* *a*_{31} ≠ 0, *u*_{3} ≠ *a*_{33}*u*_{1}/*a*_{31} *in the left matrix, and* *a*_{11} ≠ 0, *u*_{3} ≠ *a*_{13}*u*_{1}/*a*_{11} *in the second one*,
$$\begin{array}{}\left[\begin{array}{ccccc}M& {u}_{1}& {u}_{2}& \frac{{a}_{13}}{{a}_{11}}{u}_{1}& {u}_{4}\\ {v}_{1}^{T}& {a}_{11}& {a}_{12}& {a}_{13}& {a}_{14}\\ -\frac{{a}_{13}}{{a}_{11}}{v}_{4}^{T}& 0& -\frac{{a}_{13}{a}_{42}}{{a}_{11}}& 0& -\frac{{a}_{13}{a}_{44}}{{a}_{11}}\\ {v}_{3}^{T}& {a}_{31}& {a}_{32}& \frac{{a}_{13}{a}_{31}}{{a}_{11}}& {a}_{34}\\ {v}_{4}^{T}& 0& {a}_{42}& 0& {a}_{44}\end{array}\right]\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left[\begin{array}{ccccc}M& {u}_{1}& {u}_{2}& {u}_{3}& {u}_{4}\\ {v}_{1}^{T}& {a}_{11}& {a}_{12}& {a}_{13}& {a}_{14}\\ {0}^{T}& 0& 0& 0& 0\\ {v}_{3}^{T}& {a}_{31}& {a}_{32}& {a}_{33}& {a}_{34}\\ {0}^{T}& 0& 0& 0& 0\end{array}\right]\end{array}$$(31)

*with* *a*_{11} ≠ 0,*a*_{33} ≠ *a*_{13}*a*_{31}/*a*_{11}.

#### Proof

Clearly the system (8) does not involve *u*_{2}, *u*_{4}, *v*_{1},*v*_{3}, so they are free vectors in all the solutions. The first equation of (8) is
${u}_{1}{v}_{2}^{T}+{u}_{3}{v}_{4}^{T}=0.$
Now we solve the first three equations of (8) with *K* = *K*_{11}, …, *K*_{17} separately.

When *K* = *K*_{11}, the system (8) is reduced to
$$\begin{array}{}\left\{\begin{array}{l}{u}_{1}{v}_{2}^{T}+{u}_{3}{v}_{4}^{T}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0,\\ {a}_{{22}^{\mathcal{U}}1}+{a}_{42}{u}_{3}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0,\\ {a}_{24}{u}_{1}+{a}_{44}{u}_{3}\phantom{\rule{thinmathspace}{0ex}}=0.\end{array}\right.\end{array}$$

Assume that *u*_{1} = 0. Then *u*_{3}*v*_{4}^{T} = 0, *a*_{42}*u*_{3} = 0, *a*_{44}*u*_{3} = 0. If we let *u*_{3} = 0, then *v*_{4}, *a*_{42}, *a*_{44} can be arbitrary, getting the first matrix of (25). Otherwise *v*_{4} = 0, *a*_{42} = *a*_{44} = 0, which results in the second matrix in (25). Now assume that *u*_{1} ≠ 0. Then
${v}_{2}=-({u}_{1}^{H}{u}_{3}){v}_{4}/\parallel {u}_{1}{\parallel}^{2},{a}_{22}=-({u}_{1}^{H}{u}_{3}){a}_{42}/\parallel {u}_{1}{\parallel}^{2},{a}_{24}=-({u}_{1}^{H}{u}_{3}){a}_{44}/\parallel {u}_{1}{\parallel}^{2}.$ This gives (26).

For *K* = *K*_{12}, (8) becomes
$$\begin{array}{}\left\{\begin{array}{l}{u}_{1}{v}_{2}^{T}+{u}_{3}{v}_{4}^{T}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0,\\ {a}_{{22}^{\mathcal{U}}1}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}=0,\\ {a}_{{24}^{\mathcal{U}}1}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}=0,\\ {a}_{33}{v}_{4}^{T}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0.\end{array}\right.\end{array}$$

Since *a*_{33}≠ 0, we have *v*_{4} = 0, so the above is simplified to
${u}_{1}{v}_{2}^{T}=0,{a}_{22}{u}_{1}=0,{a}_{24}{u}_{1}=0.$ Letting *u*_{1} = 0 gives the first matrix of (27), and otherwise we have *v*_{2} = 0, *a*_{22} = *a*_{24} = 0, leading to the second matrix in (27).

If *K* = *K*_{13}, then (8) is simplified to
$$\begin{array}{}\left\{\begin{array}{l}{v}_{4}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0,\\ {u}_{1}{v}_{2}^{T}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0,\\ {a}_{{22}^{\mathcal{U}}1}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0,\\ {a}_{24}{u}_{1}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0.\end{array}\right.\end{array}$$

*u*_{1} = 0 produces the left matrix of (28) and otherwise, *v*_{2} = 0, *a*_{22} = *a*_{24} = 0, so the second matrix in (28).

With *K* = *K*_{14}, we have
$$\begin{array}{}\left\{\begin{array}{l}{u}_{1}{v}_{2}^{T}+{u}_{3}{v}_{4}^{T}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0,\\ {a}_{13}{v}_{4}^{T}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0,\\ {a}_{31}{v}_{2}^{T}+{a}_{33}{v}_{4}^{T}=0.\end{array}\right.\end{array}$$

Since *a*_{13} ≠ 0 and *a*_{31} ≠ 0, we have *v*_{2} = *v*_{4} = 0, and the first matrix of (29) is obtained.

The choice of *K* = *K*_{15} gives the system
$$\begin{array}{}\left\{\begin{array}{l}{u}_{1}{v}_{2}^{T}+{u}_{3}{v}_{4}^{T}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0,\\ {a}_{42}\left({u}_{3}-\frac{{a}_{33}}{{a}_{31}}{u}_{1}\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0,\\ {a}_{44}\left({u}_{3}-\frac{{a}_{33}}{{a}_{31}}{u}_{1}\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0,\\ {a}_{31}{v}_{2}^{T}+{a}_{33}{v}_{4}^{T}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0.\end{array}\right.\end{array}$$

Since *a*_{31} ≠ 0, from the last equation, *v*_{2} = −*a*_{33}*v*_{4}/*a*_{31}. Substituting into the first equation, we obtain (*u*_{3} − *a*_{33}*u*_{1}/*a*_{31})
${v}_{4}^{T}=0.$ So if *u*_{3} − *a*_{33}*u*_{1}/*a*_{31} = 0, then *v*_{4}, *a*_{42}, *a*_{44} are arbitrary and the second matrix of (29) is true. Otherwise, *v*_{4} = 0,*a*_{42} = *a*_{44} = 0, which gives the first matrix of (30).

In the case *K* = *K*_{16}, (8) can be written as
$$\begin{array}{}\left\{\begin{array}{l}{u}_{1}{v}_{2}^{T}+{u}_{3}{v}_{4}^{T}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0,\\ {a}_{42}\left({u}_{3}-\frac{{a}_{13}}{{a}_{11}}{u}_{1}\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0,\\ {a}_{44}\left({u}_{3}-\frac{{a}_{13}}{{a}_{11}}{u}_{1}\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0,\\ {a}_{11}{v}_{2}^{T}+{a}_{13}{v}_{4}^{T}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0,\\ {a}_{31}\left({v}_{2}^{T}+\frac{{a}_{13}}{{a}_{11}}{v}_{4}^{T}\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0.\end{array}\right.\end{array}$$

By the forth equation, *v*_{2} = −*a*_{13}*v*_{4}/*a*_{11}. Substituting into the first one gives
$({u}_{3}-{a}_{13}{u}_{1}/{a}_{11}){v}_{4}^{T}=0.$ Hence, if *U*_{3} ≠ *a*_{13}*u*_{1}/*a*_{11}, then *v*_{4} = 0, *a*_{42} = *a*_{44} = 0 and we get the right matrix of (30). Otherwise *v*_{4}, *a*_{42} *a*_{44} are arbitrary, so the left matrix of (31).

The last case is *K* = *K*_{17}. Then
$$\begin{array}{}\left\{\begin{array}{l}{u}_{1}{v}_{2}^{T}+{u}_{3}{v}_{4}^{T}\phantom{\rule{1em}{0ex}}=0,\\ {a}_{11}{v}_{2}^{T}+{a}_{13}{v}_{4}^{T}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0,\\ {a}_{31}{v}_{2}^{T}+{a}_{33}{v}_{4}^{T}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=0,\end{array}\right.\end{array}$$

the solutions of which are given by the last matrix of (31). □

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