In this section, first, we present two lemmas which are important tools for the convergence of Algorithm 2.1.
Lemma 3.1 Assume that the sequences {Ri}, {Pi} and{Qi} are obtained by Algorithm 2.1, if there exists an integer number s > 1, such that \( {R}_i\ne \mathbf{0}, \) for all i = 1, 2, …, s, then we have
$$ tr\left({R}_j^T{R}_i\right)=0\kern0.50em and\ tr\left({P}_j^T{P}_i+{Q}_j^T{Q}_i\right)=0,i,j=1,2,\dots, s,i\ne j $$
(3.1)
Proof In view of the fact that tr(Y) = tr(YT) for arbitrary matrix Y. Therefore, we only need to prove that
$$ tr\left({R}_j^T{R}_i\right)=0, tr\left({P}_j^T{P}_i+{Q}_j^T{Q}_i\right)=0,\mathrm{for}\ 1\le i<j\le s $$
(3.2)
We prove the conclusion (3.2) by induction through the following two steps.
Step 1: First, we show that
$$ tr\left({R}_{i+1}^T{R}_i\right)=0\ \mathrm{and}\ tr\left({P}_{i+1}^T{P}_i+{Q}_{i+1}^T{Q}_i\right)=0,i=1,2,\dots, s $$
(3.3)
To prove (3.3), we also use induction.
For i = 1, noting that P1 = PP1P, and Q1 = SQ1S, from the iterative Algorithm 2.1, we can write \( tr\left({R}_2^T{R}_1\right)= tr\left({\left[{R}_1-\frac{{\left\Vert {R}_1\right\Vert}^2}{{\left\Vert {P}_1\right\Vert}^2+{\left\Vert {Q}_1\right\Vert}^2}\left({AP}_1-{EP}_1F+{BQ}_1\right)\right]}^T{R}_1\right) \)
$$ {\displaystyle \begin{array}{l}={\left\Vert {R}_1\right\Vert}^2-\frac{{\left\Vert {R}_1\right\Vert}^2}{{\left\Vert {P}_1\right\Vert}^2+{\left\Vert {Q}_1\right\Vert}^2} tr\left({P}_1^T{A}^T{R}_1-{F}^T{P}_1^T{E}^T{R}_1+{Q}_1^T{B}^T{R}_1\right)\\ {}={\left\Vert {R}_1\right\Vert}^2-\frac{{\left\Vert {R}_1\right\Vert}^2}{{\left\Vert {P}_1\right\Vert}^2+{\left\Vert {Q}_1\right\Vert}^2} tr\left({P}_1^T{A}^T{R}_1-{P}_1^T{E}^T{R}_1{F}^T+{Q}_1^T{B}^T{R}_1\right)\\ {}={\left\Vert {R}_1\right\Vert}^2-\frac{{\left\Vert {R}_1\right\Vert}^2}{{\left\Vert {P}_1\right\Vert}^2+{\left\Vert {Q}_1\right\Vert}^2}\left[ tr\left({P}_1^T\left[\frac{A^T{R}_1+{PA}^T{R}_1P-{E}^T{R}_1{F}^T-{PE}^T{R}_1{F}^TP}{2}\right]\right.\right.\\ {}+{Q}_1^T\left[\frac{B^T{R}_1+{SB}^T{R}_1S}{2}\right]+{Q}_1^T\left[\frac{B^T{R}_1-{SB}^T{R}_1S}{2}\right]\\ {}\left.\left.+{P}_1^T\left[\frac{A^T{R}_1-{PA}^T{R}_1P-{E}^T{R}_1{F}^T+{PE}^T{R}_1{F}^TP}{2}\right]\right)\right]\\ {}={\left\Vert {R}_1\right\Vert}^2-\frac{{\left\Vert {R}_1\right\Vert}^2}{{\left\Vert {P}_1\right\Vert}^2+{\left\Vert {Q}_1\right\Vert}^2}\left[ tr\left({P}_1^T\left[\frac{A^T{R}_1+{PA}^T{R}_1P-{E}^T{R}_1{F}^T-{PE}^T{R}_1{F}^TP}{2}\right]\right.\right.\\ {}\left.\left.+{Q}_1^T\left[\frac{B^T{R}_1+{SB}^T{R}_1S}{2}\right]\right)\right]\\ {}={\left\Vert {R}_1\right\Vert}^2-\frac{{\left\Vert {R}_1\right\Vert}^2}{{\left\Vert {P}_1\right\Vert}^2+{\left\Vert {Q}_1\right\Vert}^2}\left[ tr\left({P}_1^T{P}_1+{Q}_1^T{Q}_1\Big)\right)\right]=0.\end{array}} $$
(3.4)
Similarly, we can write
$$ {\displaystyle \begin{array}{c}\begin{array}{l} tr\left({P}_2^T{P}_1+{Q}_2^T{Q}_1\right)= tr\left({\left[\frac{A^T{R}_2+{PA}^T{R}_2P-{E}^T{R}_2{F}^T-{PE}^T{R}_2{F}^TP}{2}+\frac{{\left\Vert {R}_2\right\Vert}^2}{{\left\Vert {R}_1\right\Vert}^2}{P}_1\right]}^T{P}_1\right.\\ {}\left.+{\left[\frac{B^T{R}_2+{SB}^T{R}_2S}{2}+\frac{{\left\Vert {R}_2\right\Vert}^2}{{\left\Vert {R}_1\right\Vert}^2}{Q}_1\right]}^T{Q}_1\right)\end{array}\\ {}=\frac{{\left\Vert {R}_2\right\Vert}^2}{{\left\Vert {R}_1\right\Vert}^2} tr\left({P}_1^T{P}_1+{Q}_1^T{Q}_1\right)+ tr\left(\begin{array}{l}{P}_1^T\left[\frac{A^T{R}_2+{PA}^T{R}_2P-{E}^T{R}_2{F}^T-{PE}^T{R}_2{F}^TP}{2}\right]\\ {}+{Q}_1^T\left[\frac{B^T{R}_2+{SB}^T{R}_2S}{2}\right]\end{array}\right)\\ {}\begin{array}{l}=\frac{{\left\Vert {R}_2\right\Vert}^2}{{\left\Vert {R}_1\right\Vert}^2}\left({\left\Vert {P}_1\right\Vert}^2+{\left\Vert {Q}_1\right\Vert}^2\right)+ tr\left({P}_1^T\left[\frac{A^T{R}_2+{A}^T{R}_2-{E}^T{R}_2{F}^T-{E}^T{R}_2{F}^T}{2}\right]\right.\\ {}\left.+{Q}_1^T\left[\frac{B^T{R}_2+{B}^T{R}_2}{2}\right]\right)\end{array}\\ {}=\frac{{\left\Vert {R}_2\right\Vert}^2}{{\left\Vert {R}_1\right\Vert}^2}\left({\left\Vert {P}_1\right\Vert}^2+{\left\Vert {Q}_1\right\Vert}^2\right)+ tr\left({R}_2^T\left[{AP}_1-{EP}_1F+{BQ}_1\right]\right)\\ {}=\frac{{\left\Vert {R}_2\right\Vert}^2}{{\left\Vert {R}_1\right\Vert}^2}\left({\left\Vert {P}_1\right\Vert}^2+{\left\Vert {Q}_1\right\Vert}^2\right)+ tr\left({R}_2^T\frac{{\left\Vert {P}_1\right\Vert}^2+{\left\Vert {Q}_1\right\Vert}^2}{{\left\Vert {R}_1\right\Vert}^2}\left({R}_1-{R}_2\right)\right)\\ {}=\frac{{\left\Vert {R}_2\right\Vert}^2}{{\left\Vert {R}_1\right\Vert}^2}\left({\left\Vert {P}_1\right\Vert}^2+{\left\Vert {Q}_1\right\Vert}^2\right)+\frac{{\left\Vert {P}_1\right\Vert}^2+{\left\Vert {Q}_1\right\Vert}^2}{{\left\Vert {R}_1\right\Vert}^2} tr\left({R}_2^T{R}_1\right)-{\left\Vert {R}_2\right\Vert}^2=0.\end{array}} $$
(3.5)
Now, assume that (3.3) holds for 1 < i ≤ t − 1 < s, noting that Pt = PPtP, and Qt = SQtS, then we have for i = t
$$ {\displaystyle \begin{array}{c} tr\left({R}_{t+1}^T{R}_t\right)= tr\left({\left[{R}_t-\frac{{\left\Vert {R}_t\right\Vert}^2}{{\left\Vert {P}_t\right\Vert}^2+{\left\Vert {Q}_t\right\Vert}^2}\left\{{AP}_t-{EP}_tF+{BQ}_t\right\}\right]}^T{R}_t\right)\\ {}={\left\Vert {R}_t\right\Vert}^2-\frac{{\left\Vert {R}_t\right\Vert}^2}{{\left\Vert {P}_t\right\Vert}^2+{\left\Vert {Q}_t\right\Vert}^2} tr\left({P}_t{A}^T{R}_t-{F}^T{P}_t{E}^T{R}_t+{Q}_t{B}^T{R}_t\right)\\ {}={\left\Vert {R}_t\right\Vert}^2-\frac{{\left\Vert {R}_t\right\Vert}^2}{{\left\Vert {P}_t\right\Vert}^2+{\left\Vert {Q}_t\right\Vert}^2} tr\left({P}_t^T{A}^T{R}_t-{P}_t^T{E}^T{R}_t{F}^T+{Q}_t^T{B}^T{R}_t\right)\\ {}\begin{array}{l}={\left\Vert {R}_t\right\Vert}^2-\frac{{\left\Vert {R}_t\right\Vert}^2}{{\left\Vert {P}_t\right\Vert}^2+{\left\Vert {Q}_t\right\Vert}^2}\left[ tr\left({P}_t^T\left[\frac{A^T{R}_t+{PA}^T{R}_tP-{E}^T{R}_t{F}^T-{PE}^T{R}_t{F}^TP}{2}\right]\right.\right.\\ {}\left.\left.+{Q}_t^T\left[\frac{B^T{R}_t+{SB}^T{R}_tS}{2}\right]\right)\right]\end{array}\\ {}={\left\Vert {R}_t\right\Vert}^2-\frac{{\left\Vert {R}_t\right\Vert}^2}{{\left\Vert {P}_t\right\Vert}^2+{\left\Vert {Q}_t\right\Vert}^2}\left[ tr\left({P}_t^T\left({P}_t-\frac{{\left\Vert {R}_t\right\Vert}^2}{{\left\Vert {R}_{t-1}\right\Vert}^2}{P}_{t-1}\right)+{Q}_t^T\left({Q}_t-\frac{{\left\Vert {R}_t\right\Vert}^2}{{\left\Vert {R}_{t-1}\right\Vert}^2}{Q}_{t-1}\right)\right)\right]\\ {}={\left\Vert {R}_t\right\Vert}^2-\frac{{\left\Vert {R}_t\right\Vert}^2}{{\left\Vert {P}_t\right\Vert}^2+{\left\Vert {Q}_t\right\Vert}^2}\left[{\left\Vert {P}_t\right\Vert}^2+{\left\Vert {Q}_t\right\Vert}^2-\frac{{\left\Vert {R}_t\right\Vert}^2}{{\left\Vert {R}_{t-1}\right\Vert}^2} tr\left({P}_t^T{P}_{t-1}+{Q}_t^T{Q}_{t-1}\right)\right]=0.\end{array}} $$
(3.6)
Also, we have
$$ {\displaystyle \begin{array}{c} tr\left({P}_{t+1}^T{P}_t+{Q}_{t+1}^T{Q}_t\right)= tr\left({\left[\frac{A^T{R}_{t+1}+{PA}^T{R}_{t+1}P-{E}^T{R}_{t+1}{F}^T-{PE}^T{R}_{t+1}{F}^TP}{2}+\frac{{\left\Vert {R}_{t+1}\right\Vert}^2}{{\left\Vert {R}_t\right\Vert}^2}{P}_t\right]}^T{P}_t\right.\\ {}\left.+{\left[\frac{B^T{R}_{t+1}+{SB}^T{R}_{t+1}S}{2}+\frac{{\left\Vert {R}_{t+1}\right\Vert}^2}{{\left\Vert {R}_t\right\Vert}^2}{Q}_t\right]}^T{Q}_t\right)\\ {}=\frac{{\left\Vert {R}_{t+1}\right\Vert}^2}{{\left\Vert {R}_t\right\Vert}^2} tr\left({P}_t^T{P}_t+{Q}_t^T{Q}_t\right)+ tr\left(\begin{array}{l}{P}_t^T\left[\frac{A^T{R}_{t+1}+{PA}^T{R}_{t+1}P-{E}^T{R}_{t+1}{F}^T-{PE}^T{R}_{t+1}{F}^TP}{2}\right]\\ {}+{Q}_t^T\left[\frac{B^T{R}_{t+1}+{SB}^T{R}_{t+1}S}{2}\right]\end{array}\right)\\ {}=\frac{{\left\Vert {R}_{t+1}\right\Vert}^2}{{\left\Vert {R}_t\right\Vert}^2} tr\left({P}_t^T{P}_t+{Q}_t^T{Q}_t\right)+ tr\left({R}_{t+1}^T\left[{AP}_t-{EP}_tF+{BQ}_t\right]\right)\\ {}\begin{array}{l}=\frac{{\left\Vert {R}_{t+1}\right\Vert}^2}{{\left\Vert {R}_t\right\Vert}^2} tr\left({P}_t^T{P}_t+{Q}_t^T{Q}_t\right)+ tr\left({R}_{t+1}^T\frac{{\left\Vert {P}_t\right\Vert}^2+{\left\Vert {Q}_t\right\Vert}^2}{{\left\Vert {R}_t\right\Vert}^2}\left({R}_t-{R}_{t+1}\right)\right)\\ {}=\frac{{\left\Vert {R}_{t+1}\right\Vert}^2}{{\left\Vert {R}_t\right\Vert}^2} tr\left({P}_t^T{P}_t+{Q}_t^T{Q}_t\right)+\frac{{\left\Vert {P}_t\right\Vert}^2+{\left\Vert {Q}_t\right\Vert}^2}{{\left\Vert {R}_t\right\Vert}^2} tr\left({R}_{t+1}^T{R}_t\right)-{\left\Vert {R}_{t+1}\right\Vert}^2=0.\end{array}\end{array}} $$
(3.7)
Therefore, the conclusion (3.3) holds for i = t. Hence, (3.3) holds by the principle of induction.
Step 2: In this step, we show for 1 ≤ i ≤ s − 1
$$ tr\left({R}_{i+l}^T{R}_i\right)=0\ \mathrm{and}\ tr\left({P}_{i+l}^T{P}_i+{Q}_{i+l}^T{Q}_i\right)=0 $$
(3.8)
for l = 1, 2, …, s. The case of l = 1 has been proven in step 1. Assume that (3.8) holds for l ≤ ν.
Now, we prove that \( tr\left({R}_{i+\nu +1}^T{R}_i\right)=0 \) and \( tr\left({P}_{i+\nu +1}^T{P}_i+{Q}_{i+\nu +1}^T{Q}_i\right)=0 \) through the following two substeps.
Substep 2.1: In this substep, we show that
$$ tr\left({R}_{\nu +2}^T{R}_1\right)=0 $$
(3.9)
$$ tr\left({P}_{\nu +2}^T{P}_1+{Q}_{\nu +2}^T{Q}_1\right)=0 $$
(3.10)
By Algorithm 2.1 and the induction assumptions, we have
$$ {\displaystyle \begin{array}{c} tr\left({R}_{\nu +2}^T{R}_1\right)= tr\left({\left[{R}_{\nu +1}-\frac{{\left\Vert {R}_{\nu +1}\right\Vert}^2}{{\left\Vert {P}_{\nu +1}\right\Vert}^2+{\left\Vert {Q}_{\nu +1}\right\Vert}^2}\left({AP}_{\nu +1}-{EP}_{\nu +1}F+{BQ}_{\nu +1}\right)\right]}^T{R}_1\right)\\ {}= tr\left({R}_{\nu +1}^T{R}_1\right)-\frac{{\left\Vert {R}_{\nu +1}\right\Vert}^2}{{\left\Vert {P}_{\nu +1}\right\Vert}^2+{\left\Vert {Q}_{\nu +1}\right\Vert}^2} tr\left({P}_{\nu +1}^T\left[{A}^T{R}_1-{E}^T{R}_1{F}^T\right]+{Q}_{\nu +1}^T{B}^T{R}_1\right)\\ {}\begin{array}{l}=-\frac{{\left\Vert {R}_{\nu +1}\right\Vert}^2}{{\left\Vert {P}_{\nu +1}\right\Vert}^2+{\left\Vert {Q}_{\nu +1}\right\Vert}^2} tr\left({P}_{\nu +1}^T\left[\frac{A^T{R}_1-{E}^T{R}_1{F}^T+{PA}^T{R}_1P-{PE}^T{R}_1{F}^TP}{2}\right]\right.\\ {}\left.+{Q}_{\nu +1}^T\left[\frac{B^T{R}_1+{SB}^T{R}_1S}{2}\right]\right)\end{array}\\ {}=-\frac{{\left\Vert {R}_{\nu +1}\right\Vert}^2}{{\left\Vert {P}_{\nu +1}\right\Vert}^2+{\left\Vert {Q}_{\nu +1}\right\Vert}^2} tr\left({P}_{\nu +1}^T{P}_1+{Q}_{\nu +1}^T{Q}_1\right)=0,\end{array}} $$
and \( tr\left({P}_{\nu +2}^T{P}_1+{Q}_{\nu +2}^T{Q}_1\right) \)
$$ {\displaystyle \begin{array}{l}= tr\left({\left[\frac{A^T{R}_{\nu +2}+{PA}^T{R}_{\nu +2}P-{E}^T{R}_{\nu +2}{F}^T-{PE}^T{R}_{\nu +2}{F}^TP}{2}+\frac{{\left\Vert {R}_{\nu +2}\right\Vert}^2}{{\left\Vert {R}_{\nu +1}\right\Vert}^2}{P}_{\nu +1}\right]}^T{P}_1\right.\\ {}\left.+{\left[\frac{B^T{R}_{\nu +2}+{SB}^T{R}_{\nu +2}S}{2}+\frac{{\left\Vert {R}_{\nu +2}\right\Vert}^2}{{\left\Vert {R}_{\nu +1}\right\Vert}^2}{Q}_{\nu +1}\right]}^T{Q}_1\right)\\ {}=\frac{{\left\Vert {R}_{\nu +2}\right\Vert}^2}{{\left\Vert {R}_{\nu +1}\right\Vert}^2} tr\left({P}_{\nu +1}^T{P}_1+{Q}_{\nu +1}^T{Q}_1\right)+ tr\left({R}_{\nu +2}^T\left[{AP}_1-{EP}_1F+{BQ}_1\right]\right)\\ {}=\frac{{\left\Vert {P}_1\right\Vert}^2+{\left\Vert {Q}_1\right\Vert}^2}{{\left\Vert {R}_1\right\Vert}^2} tr\left({R}_{\nu +2}^T\left({R}_1-{R}_2\right)\right)=0\end{array}} $$
Thus, (3.9) and (3.10) hold.
Substep 2.2: By Algorithm 2.1 and the induction assumptions, we can write
$$ {\displaystyle \begin{array}{c} tr\left({R}_{i+\nu +1}^T{R}_i\right)= tr\left({\left[{R}_{i+\nu }-\frac{{\left\Vert {R}_{i+\nu}\right\Vert}^2}{{\left\Vert {P}_{i+\nu}\right\Vert}^2+{\left\Vert {Q}_{i+\nu}\right\Vert}^2}\left({AP}_{i+\nu }-{EP}_{i+\nu }F+{BQ}_{i+\nu}\right)\right]}^T{R}_i\right)\\ {}= tr\left({R}_{\nu +1}^T{R}_i\right)-\frac{{\left\Vert {R}_{i+\nu}\right\Vert}^2}{{\left\Vert {P}_{i+\nu}\right\Vert}^2+{\left\Vert {Q}_{i+\nu}\right\Vert}^2} tr\left({P}_{i+\nu}^T\left[{A}^T{R}_i-{E}^T{R}_i{F}^T\right]+{Q}_{i+\nu}^T{B}^T{R}_i\right)\\ {}\begin{array}{l}=-\frac{{\left\Vert {R}_{i+\nu}\right\Vert}^2}{{\left\Vert {P}_{i+\nu}\right\Vert}^2+{\left\Vert {Q}_{i+\nu}\right\Vert}^2} tr\left({P}_{i+\nu}^T\left[\frac{A^T{R}_i-{E}^T{R}_i{F}^T+{PA}^T{R}_iP-{PE}^T{R}_i{F}^TP}{2}\right]\right.\\ {}\left.+{Q}_{i+\nu}^T\left[\frac{B^T{R}_i+{SB}^T{R}_iS}{2}\right]\right)\end{array}\\ {}=-\frac{{\left\Vert {R}_{i+\nu}\right\Vert}^2}{{\left\Vert {P}_{i+\nu}\right\Vert}^2+{\left\Vert {Q}_{i+\nu}\right\Vert}^2} tr\left({P}_{i+\nu}^T\left[{P}_i-\frac{{\left\Vert {R}_i\right\Vert}^2}{{\left\Vert {R}_{i-1}\right\Vert}^2}{P}_{i-1}\right]+{Q}_{i+\nu}^T\left[{Q}_i-\frac{{\left\Vert {R}_i\right\Vert}^2}{{\left\Vert {R}_{i-1}\right\Vert}^2}{Q}_{i-1}\right]\right)\\ {}=-\frac{{\left\Vert {R}_{i+\nu}\right\Vert}^2}{{\left\Vert {P}_{i+\nu}\right\Vert}^2+{\left\Vert {Q}_{i+\nu}\right\Vert}^2}\left[ tr\left({P}_{i+\nu}^T{P}_i+{Q}_{i+v}^T{Q}_i\right)-\frac{{\left\Vert {R}_i\right\Vert}^2}{{\left\Vert {R}_{i-1}\right\Vert}^2} tr\left({P}_{i+\nu}^T{P}_{i-1}+{Q}_{i+\nu}^T{Q}_{i-1}\right)\right]\\ {}=\frac{{\left\Vert {R}_i\right\Vert}^2{\left\Vert {R}_{i+\nu}\right\Vert}^2}{{\left\Vert {R}_{i-1}\right\Vert}^2\left({\left\Vert {P}_{i+\nu}\right\Vert}^2+{\left\Vert {Q}_{i+\nu}\right\Vert}^2\right)} tr\left({P}_{i+\nu}^T{P}_{i-1}+{Q}_{i+\nu}^T{Q}_{i-1}\right)\end{array}} $$
(3.11)
Also, we have
$$ {\displaystyle \begin{array}{c} tr\left({P}_{i+\nu +1}^T{P}_i+{Q}_{i+\nu +1}^T{Q}_i\right)\\ {}= tr\left({\left[\frac{A^T{R}_{i+\nu +1}+{PA}^T{R}_{i+\nu +1}P-{E}^T{R}_{i+\nu +1}{F}^T-{PE}^T{R}_{i+\nu +1}{F}^TP}{2}+\frac{{\left\Vert {R}_{i+\nu +1}\right\Vert}^2}{{\left\Vert {R}_{i+\nu}\right\Vert}^2}{P}_{i+\nu}\right]}^T{P}_i\right.\\ {}\left.+{\left[\frac{B^T{R}_{i+\nu +1}+{SB}^T{R}_{i+\nu +1}S}{2}+\frac{{\left\Vert {R}_{i+\nu +1}\right\Vert}^2}{{\left\Vert {R}_{i+\nu}\right\Vert}^2}{Q}_{i+\nu}\right]}^T{Q}_i\right)\\ {}=\frac{{\left\Vert {R}_{i+\nu +1}\right\Vert}^2}{{\left\Vert {R}_{i+\nu}\right\Vert}^2} tr\left({P}_{i+\nu}^T{P}_i+{Q}_{i+\nu}^T{Q}_i\right)+ tr\left({R}_{i+\nu +1}^T\left[{AP}_i-{EP}_iF+{BQ}_i\right]\right)\\ {}=\frac{{\left\Vert {P}_i\right\Vert}^2+{\left\Vert {Q}_i\right\Vert}^2}{{\left\Vert {R}_i\right\Vert}^2} tr\left({R}_{i+\nu +1}^T\left[{R}_i-{R}_{i+1}\right]\right)=\frac{{\left\Vert {P}_i\right\Vert}^2+{\left\Vert {Q}_i\right\Vert}^2}{{\left\Vert {R}_i\right\Vert}^2} tr\left({R}_{i+\nu +1}^T{R}_i\right)\\ {}=\frac{{\left\Vert {P}_i\right\Vert}^2+{\left\Vert {Q}_i\right\Vert}^2}{{\left\Vert {R}_i\right\Vert}^2}\frac{{\left\Vert {R}_i\right\Vert}^2{\left\Vert {R}_{i+\nu}\right\Vert}^2}{{\left\Vert {R}_{i-1}\right\Vert}^2\left({\left\Vert {P}_{i+\nu}\right\Vert}^2+{\left\Vert {Q}_{i+\nu}\right\Vert}^2\right)} tr\left({P}_{i+\nu}^T{P}_{i-1}+{Q}_{i+\nu}^T{Q}_{i-1}\right)\end{array}} $$
(3.12)
Repeating the above process (3.11) and (3.12), we can obtain, for certain α and β
\( tr\left({R}_{i+\nu +1}^T{R}_i\right)=\alpha tr\left({P}_{\nu +2}^T{P}_1+{Q}_{\nu +2}^T{Q}_1\right) \), and \( tr\left({P}_{i+\nu +1}^T{P}_i+{Q}_{i+\nu +1}^T{Q}_i\right)=\beta tr\left({P}_{\nu +2}^T{P}_1+{Q}_{\nu +2}^T{Q}_1\right) \).
Combining these two relations with (3.10) implies that (3.8) holds for l = ν + 1.
From steps 1 and 2, the conclusion (3.1) holds by the principle of induction.
Lemma 3.2
Let Problem 2.1 be consistent over reflexive matrices, and V
∗
and W
∗
be arbitrary reflexive solutions of Problem 2.1. Then for any initial reflexive matrices V
1
andW
1
, we have
$$ tr\left({\left({V}^{\ast }-{V}_i\right)}^T{P}_i+{\left({W}^{\ast }-{W}_i\right)}^T{Q}_i\right)={\left\Vert {R}_i\right\Vert}^2\ for\ i=1,2,\dots $$
(3.13)
where the Sequences {Ri}, {Pi}, {Qi}, {Vi} and{Wi} are generated by Algorithm 2.1.
Proof We can prove the conclusion (3.13) by using the induction as follows
For i = 1, noting that V∗ − V1 = P(V∗ − V1)P, and W∗ − W1 = Z(W∗ − W1)Z, we have
$$ {\displaystyle \begin{array}{c} tr\left({\left({V}^{\ast }-{V}_1\right)}^T{P}_1+{\left({W}^{\ast }-{W}_1\right)}^T{Q}_1\right)\\ {}= tr\left({\left({V}^{\ast }-{V}_1\right)}^T\left[\frac{A^T{R}_1+{PA}^T{R}_1P-{E}^T{R}_1{F}^T-{PE}^T{R}_1{F}^TP}{2}\right]+{\left({W}^{\ast }-{W}_1\right)}^T\left[\frac{B^T{R}_1+{SB}^T{R}_1S}{2}\right]\right)\\ {}\begin{array}{l}= tr\left({\left({V}^{\ast }-{V}_1\right)}^T\left[\frac{A^T{R}_1+{PA}^T{R}_1P-{E}^T{R}_1{F}^T-{PE}^T{R}_1{F}^TP+{A}^T{R}_1-{PA}^T{R}_1P-{E}^T{R}_1{F}^T}{2}\right.\right.\\ {}\left.+\frac{PE^T{R}_1{F}^TP}{2}\right]\left.+{\left({W}^{\ast }-{W}_1\right)}^T\left[\frac{B^T{R}_1+{SB}^T{R}_1S+{B}^T{R}_1-{SB}^T{R}_1S}{2}\right]\right)\end{array}\\ {}= tr\left({\left({V}^{\ast }-{V}_1\right)}^T\left[{A}^T{R}_1-{E}^T{R}_1{F}^T\right]+{\left({W}^{\ast }-{W}_1\right)}^T\left[{B}^T{R}_1\right]\right)\\ {}\begin{array}{l}= tr\left({R}_1^T\left[A\left({V}^{\ast }-{V}_1\right)-E\left({V}^{\ast }-{V}_1\right)F+B\left({W}^{\ast }-{W}_1\right)\right]\right)\\ {}= tr\left({R}_1^T\left[C-{AV}_1+{EV}_1F-{BW}_1\right]\right)={\left\Vert {R}_1\right\Vert}^2.\end{array}\end{array}} $$
(3.14)
Assume that the conclusion (3.13) holds for i = t. Now, for i = t + 1, we have
$$ {\displaystyle \begin{array}{l} tr\left({\left({V}^{\ast }-{V}_{t+1}\right)}^T{P}_{t+1}+{\left({W}^{\ast }-{W}_{t+1}\right)}^T{Q}_{t+1}\right)\\ {}= tr\left({\left({V}^{\ast }-{V}_{t+1}\right)}^T\left[\frac{A^T{R}_{t+1}+{PA}^T{R}_{t+1}P-{E}^T{R}_{t+1}{F}^T-{PE}^T{R}_{t+1}{F}^TP}{2}+\frac{{\left\Vert {R}_{t+1}\right\Vert}^2}{{\left\Vert {R}_t\right\Vert}^2}{P}_t\right]\right.\\ {}\left.+{\left({W}^{\ast }-{W}_{t+1}\right)}^T\left[\frac{B^T{R}_{t+1}+{SB}^T{R}_{t+1}S}{2}+\frac{{\left\Vert {R}_{t+1}\right\Vert}^2}{{\left\Vert {R}_t\right\Vert}^2}{Q}_t\right]\right)\\ {}= tr\left({\left({V}^{\ast }-{V}_{t+1}\right)}^T\left[{A}^T{R}_{t+1}-{E}^T{R}_{t+1}{F}^T\right]+{\left({W}^{\ast }-{W}_{t+1}\right)}^T\left[{B}^T{R}_{t+1}\right]\right.\\ {}+\frac{{\left\Vert {R}_{t+1}\right\Vert}^2}{{\left\Vert {R}_t\right\Vert}^2}\left[{\left({V}^{\ast }-{V}_{t+1}\right)}^T{P}_t+{\left({W}^{\ast }-{W}_{t+1}\right)}^T{Q}_t\right]\\ {}= tr\left({R}_{t+1}^T\left[C-{AV}_{t+1}+{EV}_{t+1}F-{BW}_{t+1}\right]\right)={\left\Vert {R}_{t+1}\right\Vert}^2\end{array}} $$
(3.15)
Hence, Lemma 3.2 holds for all i = 1, 2, … by the principle of induction.
Theorem 3.1
Assume that Problem 2.1 is consistent over reflexive matrices, then by using Algorithm 2.1 for any arbitrary initial reflexive matrices
\( {V}_1\in {R}_r^{n\times n}(P) \)
and
\( {W}_1\in {R}_r^{n\times n}(S) \)
, reflexive solutions of Problem 2.1 can be obtained within a finite iterative steps by Algorithm 2.1 in absence of roundoff errors.
Proof Assume that \( {R}_i\ne \mathbf{0} \) for i = 1, 2, …, mn. From Lemma 3.2, we get \( {P}_i\ne \mathbf{0} \) or \( {Q}_i\ne \mathbf{0} \) for i = 1, 2, …, mn. Therefore, we can compute Rmn + 1, Vmn + 1 and Wmn + 1 by Algorithm 2.1. Also from Lemma 3.1, we have
$$ tr\left({R}_{mn+1}^T{R}_i\right)=0\ \mathrm{for}\ i=1,2,\dots, mn, $$
(3.16)
and
$$ tr\left({R}_i^T{R}_j\right)=0\ \mathrm{for}\ i,j=1,2,\dots, mn,\left(i\ne j\right). $$
(3.17)
Therefore, the set {R1, R2, …, Rmn} is an orthogonal basis of the matrix space Rm × n, which implies that \( {R}_{mn+1}=\mathbf{0} \), i.e., Vmn + 1, and Wmn + 1 are reflexive matrices solutions of Problem 2.1. Hence, the proof is completed.
To obtain least Frobenius norm solution of the generalized solution pair of Problem 2.1, we first present the following lemma.
Lemma 3.3 [4] Assume that the consistent system of linear equations Ax = b has a solution x∗ ∈ R(AT), then x∗ is a unique least Frobenius norm solution of the system of linear equations.
Theorem 3.2
Suppose that Problem 2.1 is consistent over reflexive matrices. Let the initial iteration matrices
\( {V}_1={A}^TG+{PA}^T\tilde{G}P-{E}^T{GF}^T-{PE}^T\tilde{G}{F}^TP \)
and
\( {W}_1={B}^TG+{SB}^T\tilde{G}S \)
where G and
\( \tilde{G} \)
are arbitrary, or especially
\( {V}_1=\mathbf{0} \)
and
\( {W}_1=\mathbf{0} \)
, then the reflexive solutions V
∗
andW
∗
obtained by Algorithm 2.1, are the least Frobenius norm reflexive solutions of Eq. (
2.1
).
Proof The solvability of the matrix Eq. (
2.1
) over reflexive matrices is equivalent to the solvability of the system of equations
$$ \left\{\begin{array}{l} AV- EVF+ BW=C,\\ {} APVP- EPVPF+ BSWS=C.\end{array}\right. $$
(3.18)
And the system of equations (
3.18
) is equivalent to
$$ \left(\begin{array}{cc}\left(I\otimes A\right)-\left({F}^T\otimes E\right)& \left(I\otimes B\right)\\ {}\left(P\otimes AP\right)-\left({F}^TP\otimes EP\right)& \left(S\otimes BS\right)\end{array}\right)\left(\begin{array}{c} vec(V)\\ {} vec(W)\end{array}\right)=\left(\begin{array}{c} vec(C)\\ {} vec(C)\end{array}\right) $$
(3.19)
Now, assume that G and
\( \tilde{G} \)
are arbitrary matrices, we can write
$$ {\displaystyle \begin{array}{c}\left(\begin{array}{c} vec\left({A}^TG+{PA}^T\tilde{G}P-{E}^T{GF}^T-{PE}^T\tilde{G}{F}^TP\right)\\ {} vec\left({B}^TG+{SB}^T\tilde{G}S\right)\end{array}\right)\\ {}=\left(\begin{array}{cc}\left(I\otimes {A}^T\right)-\left(F\otimes {E}^T\right)& \left(P\otimes {PA}^T\right)-\left( PF\otimes {PE}^T\right)\\ {}\left(I\otimes {B}^T\right)& \left(S\otimes {SB}^T\right)\end{array}\right)\ \left(\begin{array}{c} vec(G)\\ {} vec\left(\tilde{G}\right)\end{array}\right)\\ {}={\left(\begin{array}{cc}\left(I\otimes A\right)-\left({F}^T\otimes E\right)& \left(I\otimes B\right)\\ {}\left(P\otimes AP\right)-\left({F}^TP\otimes EP\right)& \left(S\otimes BS\right)\end{array}\right)}^T\left(\begin{array}{c} vec(G)\\ {} vec\left(\tilde{G}\right)\end{array}\right)\\ {}\in R\left({\left(\begin{array}{cc}\left(I\otimes A\right)-\left({F}^T\otimes E\right)& \left(I\otimes B\right)\\ {}\left(P\otimes AP\right)-\left({F}^TP\otimes EP\right)& \left(S\otimes BS\right)\end{array}\right)}^T\right)\end{array}} $$
If we consider \( {V}_1={A}^TG+{PA}^T\tilde{G}P-{E}^T{GF}^T-{PE}^T\tilde{G}{F}^TP \) and \( {W}_1={B}^TG+{SB}^T\tilde{G}S \) then all Vk And Wk generated by Algorithm 2.1 satisfy
$$ \left(\begin{array}{c} vec\left({V}_k\right)\\ {} vec\left({W}_k\right)\end{array}\right)\in R\left({\left(\begin{array}{cc}\left(I\otimes A\right)-\left({F}^T\otimes E\right)& \left(I\otimes B\right)\\ {}\left(P\otimes AP\right)-\left({F}^TP\otimes EP\right)& \left(S\otimes BS\right)\end{array}\right)}^T\right) $$
By applying Lemma 3.3 with the initial iteration matrices \( {V}_1={A}^TG+{PA}^T\tilde{G}P-{E}^T{GF}^T-{PE}^T\tilde{G}{F}^TP \) and \( {W}_1={B}^TG+{SB}^T\tilde{G}S \) where G and \( \tilde{G} \) are arbitrary, or especially \( {V}_1=\mathbf{0} \) and \( {W}_1=\mathbf{0} \), the reflexive solutions V∗ and W∗ obtained by Algorithm 2.1 are the least Frobenius norm reflexive solutions of Eq. (2.1).