A. Proof of Main Theorem
In this section, we prove Main Theorem using above lemmas.
Proof
(Main Theorem) Summarizing the terms, we have
$$\begin{aligned} \varPhi (A,B)&= \Vert AB-A_0B_0 \Vert ^2 \nonumber \\&= \sum _{j=1}^N \sum _{i=1}^{M-1} \left\{ \!\sum _{k=1}^{H_0-1} (\!a_{ik}b_{kj}{-}a^0_{ik}b^0_{kj})+a_{iH_0}b_{H_0j}\right. \nonumber \\&\quad \left. -a^0_{iH_0}b^0_{H_0j} +\sum _{k=H_0+1}^{H-1}a_{ik}b_{kj} + a_{iH} b_{Hj} \!\right\} ^2 \nonumber \\&\quad + \sum _{j=1}^N \left\{ \!\sum _{k=1}^{H_0-1} (\!a_{Mk}b_{kj} {-}a^0_{Mk}b^0_{kj}){+}a_{MH_0}b_{H_0j}\right. \nonumber \\&\quad \left. {-}a^0_{MH_0}b^0_{H_0j} {+}\sum _{k=H_0+1}^{H-1}a_{Mk}b_{kj} {+} a_{MH} b_{Hj}\! \right\} ^2\!\!. \end{aligned}$$
(15)
Put
$$\begin{aligned}&K_{ij}:=\sum _{k=1}^{H_0-1} (a_{ik}b_{kj} -a^0_{ik}b^0_{kj})+a_{iH_0}b_{H_0j} \\&\qquad -a^0_{iH_0}b^0_{H_0j} +\sum _{k=H_0+1}^{H-1}a_{ik}b_{kj} + a_{iH} b_{Hj} , \\&L_j:= \sum _{k=1}^{H_0-1} (a_{Mk}b_{kj} -a^0_{Mk}b^0_{kj})+a_{MH_0}b_{H_0j}\\&\qquad -a^0_{MH_0}b^0_{H_0j} +\sum _{k=H_0+1}^{H-1}a_{Mk}b_{kj} + a_{MH} b_{Hj}, \end{aligned}$$
then we get
$$\begin{aligned} \Vert AB-A_0 B_0\Vert ^2 = \sum _{j=1}^N \sum _{i=1}^{M-1}K_{ij}^2 + \sum _{j=1}^N L_j^2. \end{aligned}$$
Using \(a_{Mk}=1-\sum _{i=1}^{M-1}a_{ik}\), \(b_{Hj}=1-\sum _{k=1}^{H-1} b_{kj}\), \(a^0_{Mk}=1-\sum _{i=1}^{M-1}a^0_{ik}\), and \(b^0_{H_0j}=1-\sum _{k=1}^{H_0-1} b^0_{kj}\), we have
$$\begin{aligned} \sum _{i=1}^{M-1}K_{ij}=\, & \sum _{i=1}^{M-1} \sum _{k=1}^{H-1} (a_{ik}-a_{iH})b_{kj} \\&-\sum _{i=1}^{M-1} \sum _{k=1}^{H_0-1} \left( a^0_{ik}-a^0_{iH_0}\right) b^0_{kj}+ \sum _{i=1}^{M-1} \left( a_{iH}-a^0_{iH_0}\right) ,\\ L_j=\, & -\sum _{i=1}^{M-1} \sum _{k=1}^{H-1} (a_{ik}-a_{iH})b_{kj}\\&+\sum _{i=1}^{M-1} \sum _{k=1}^{H_0-1}\left( a^0_{ik}-a^0_{iH_0}\right) b^0_{kj} - \sum _{i=1}^{M-1} \left( a_{iH}-a^0_{iH_0}\right) , \end{aligned}$$
thus
$$\begin{aligned} L_j^2=\left( \sum _{i=1}^{M-1} K_{ij} \right) ^2. \end{aligned}$$
Therefore
$$\begin{aligned} \Vert AB-A_0B_0\Vert ^2&= \sum _{j=1}^N \sum _{i=1}^{M-1} K_{ij}^2 +\sum _{j=1}^N L_j^2 \\&= \sum _{j=1}^N \sum _{i=1}^{M-1} K_{ij}^2 +\sum _{j=1}^N \left( \sum _{i=1}^{M-1} K_{ij} \right) ^2. \end{aligned}$$
On account of Corollary 12, we have
$$\begin{aligned} \Vert AB-A_0B_0\Vert ^2 \sim \sum _{j=1}^N \sum _{i=1}^{M-1} K_{ij}^2, \end{aligned}$$
i.e.,
$$\begin{aligned}&\quad \Vert AB-A_0B_0\Vert ^2 \\&\sim \sum _{j=1}^N \sum _{i=1}^{M-1} \left\{ \sum _{k=1}^{H-1} (a_{ik}-a_{iH})b_{kj} \right. \nonumber \\&\qquad \left. - \sum _{k=1}^{H_0-1} (a^0_{ik}-a^0_{iH_0})b^0_{kj}+ (a_{iH}-a^0_{iH_0}) \right\} ^2 \\&= \sum _{j=1}^N \sum _{i=1}^{M-1} \left[ \sum _{k=1}^{H_0-1} \{(a_{ik}-a_{iH})b_{kj}- (a^0_{ik}-a^0_{iH_0})b^0_{kj}\} \right. \nonumber \\&\qquad \left. + \sum _{k=H_0}^{H-1} (a_{ik}-a_{iH})b_{kj}+ (a_{iH}-a^0_{iH_0}) \right] ^2. \end{aligned}$$
$$\begin{aligned} \text{ Let } {\left\{ \begin{array}{ll} a_{ik}=a_{ik}-a_{iH}, &{} k<H \\ c_i = a_{iH}-a^0_{iH_0}, \\ b_{kj}=b_{kj} \end{array}\right. } \end{aligned}$$
and put \(a^0_{ik}=a^0_{ik}-a^0_{iH_0}\). Then we have
$$\begin{aligned}&\quad \Vert AB-A_0B_0\Vert ^2 \\&\sim \sum _{j=1}^N \sum _{i=1}^{M-1} \left[ \sum _{k=1}^{H_0-1} \{(a_{ik}-a_{iH})b_{kj}- (a^0_{ik}-a^0_{iH_0})b^0_{kj}\} \right. \nonumber \\&\quad \left. + \sum _{k=H_0}^{H-1} (a_{ik}-a_{iH})b_{kj}+ (a_{iH}-a^0_{iH_0}) \right] ^2 \\&= \sum _{j=1}^N \sum _{i=1}^{M-1} \left\{ \sum _{k=1}^{H_0-1} (a_{ik}b_{kj}- a^0_{ik}b^0_{kj}) \right. \nonumber \\&\quad \left. + \sum _{k=H_0}^{H-1} a_{ik}b_{kj}+ c_i \right\} ^2. \end{aligned}$$
There is a positive constant \(C>0\), we have
$$\begin{aligned}&\quad C \Vert AB-A_0 B_0 \Vert ^2 \\&\leqq \sum _{j=1}^N \sum _{i=1}^{M-1} \left\{ \sum _{k=1}^{H_0-1} (a_{ik}b_{kj}-a^0_{ik}b^0_{kj}) + c_i \right\} ^2 \\&\quad + \sum _{j=1}^N \sum _{i=1}^{M-1} \left( \sum _{k=H_0}^{H-1} a_{ik}b_{kj} \right) ^2. \end{aligned}$$
Put
$$\begin{aligned} K_1=\, & \sum _{j=1}^N \sum _{i=1}^{M-1} \left[ \sum _{k=1}^{H_0-1} (a_{ik}b_{kj}- a^0_{ik}b^0_{kj}) + c_i\right] ^2, \\ K_2=\, & \sum _{j=1}^N \sum _{i=1}^{M-1} \left\{ \sum _{k=H_0}^{H-1} a_{ik}b_{kj} \right\} ^2. \end{aligned}$$
Let \({\bar{\lambda }}_1\) be the RLCT of \(K_1\) ,\({\bar{\lambda }}_2\) be the RLCT of \(K_2\), and \(\lambda\) be the RLCT of \(\Vert AB-A_0B_0\Vert ^2\). The following inequality holds since an RLCT is order isomorphic and \(K_1\) and \(K_2\) are independent:
$$\begin{aligned} \lambda \leqq {\bar{\lambda }}_1 + {\bar{\lambda }}_2. \end{aligned}$$
According to Lemma 17 in the case of \(H \leftarrow H_0\),
$$\begin{aligned} {\bar{\lambda }}_1 \leqq \frac{M-1}{2} + (H_0-1) \frac{M+N-3}{2}. \end{aligned}$$
In contrast, there exists a positive constant \(D>0\), we have
$$\begin{aligned} K_2&= \sum _{j=1}^N \sum _{i=1}^{M-1} \left( \sum _{k=H_0}^{H-1} a_{ik}b_{kj} \right) ^2 \\&\leqq D\sum _{j=1}^N \sum _{i=1}^{M-1} \sum _{k=H_0}^{H-1} a_{ik}^2 b_{kj}^2 \\&\sim \sum _{j=1}^N \sum _{i=1}^{M-1} \sum _{k=H_0}^{H-1} a_{ik}^2 b_{kj}^2 \\&= \sum _{k=H_0}^{H-1} \sum _{j=1}^N \sum _{i=1}^{M-1} a_{ik}^2 b_{kj}^2 \\&= \sum _{k=H_0}^{H-1} \left( \sum _{j=1}^N b_{kj}^2 \right) \left( \sum _{i=1}^{M-1} a_{ik}^2 \right) . \end{aligned}$$
The RLCT of the last term becomes a sum of each ones about k. Considering blowing-ups of variables \(\{ a_{ik} \}\) and \(\{ b_{kj} \}\) for each k, we obtain
$$\begin{aligned} {\bar{\lambda }}_2 \leqq \frac{(H-H_0)\min \{ M-1,N \}}{2}. \end{aligned}$$
Using the above inequalities about the RLCTs, we have
$$\begin{aligned} \lambda&\leqq {\bar{\lambda }}_1 + {\bar{\lambda }}_2 \leqq \frac{M-1}{2} \\&\quad + (H_0-1) \frac{M+N-3}{2} + \frac{(H-H_0)\min \{ M-1,N \}}{2}.\\&\quad \therefore \quad \lambda \leqq \frac{1}{2}\left[ M-1\right. \\&\quad \left. +(H_0-1)(M+N-3)+(H-H_0)\min \{M-1,N\} \right] . \end{aligned}$$
\(\square\)
B. Proof of Lemmas
In this section, we prove the four lemmas introduced in Section 3: Lemmas 14, 15, 16, and 17.
First, Lemma 14 is proved.
Proof
(Lemma 14) We set \(A=(a_{i})_{i=1}^{M}\), \(B^T=(1)_{j=1}^{N}\), \(A_0=(a^0_{i})^M\),\(B_0^T=(1)_{j=1}^N\), then
$$\begin{aligned} \Vert AB-A_0B_0 \Vert ^2&=\sum _{i=1}^M N(a_i-a^0_i)^2 \\&= \sum _{i=1}^{M-1} N(a_i-a^0_i)^2 \\ &\quad + N \left( 1-\sum _{i=1}^{M-1} a_i -1 + \sum _{i=1}^{M-1} a^0_i \right) ^2 \\&= \sum _{i=1}^{M-1} N(a_i-a^0_i)^2 \\&\quad + N \left\{ \sum _{i=1}^{M-1} (a_i - a^0_i) \right\} ^2. \end{aligned}$$
Using Corollary 12, \(\sum _{i=1}^{M-1} (a_i - a^0_i) \in \langle a_1-a^0_1, \ldots , a_{M-1}-a^0_{M-1} \rangle\) causes that
$$\begin{aligned} \Vert AB-A_0B_0 \Vert ^2 \sim \sum _{i=1}^{M-1} N(a_i-a^0_i)^2. \end{aligned}$$
As an RLCT is not changed by any constant factor, all we have to do is calculating an RLCT of
$$\begin{aligned} \sum _{i=1}^{M-1} (a_i-a^0_i)^2 \end{aligned}$$
and this has no singularity. Thus, the RLCT equals to a half of the parameter dimension:
$$\begin{aligned} \lambda = \frac{M-1}{2}. \end{aligned}$$
\(\square\)
Second, Lemma 15 is derived.
Proof
(Lemma 15) We set \(A_0=(a^0_{i})^M\), \(B_0^T=(1)_{j=1}^N\).
$$\begin{aligned}&AB-A_0B_0 \\&\quad =\left( \begin{matrix} a_{11} &{} a_{12} \\ \vdots &{} \vdots \\ a_{(M-1)1} &{} a_{(M-1)2} \\ a_{M1} &{} a_{M2} \end{matrix} \right) \left( \begin{matrix} b_1 &{} \ldots &{} b_N \\ 1-b_1 &{} \ldots &{} 1-b_N \end{matrix} \right) \\&\quad -\left( \begin{matrix} a^0_1 \\ \vdots \\ a^0_{M-1} \\ a^0_{M} \end{matrix} \right) \left( \begin{matrix} 1&\ldots&1 \end{matrix} \right) \\&\quad =\left( \begin{matrix} (a_{11}-a_{12})b_j +a_{12}-a^0_1 \\ \vdots \\ (a_{(M-1)1}-a_{(M-1)2})b_j+a_{(M-1)2}-a^0_{M-1} \\ (a_{M1}-a_{M2})b_j+a_{M2}-a^0_{M} \\ \end{matrix} \right) _{j=1}^N \\&\quad =\left( \begin{matrix} (a_{11}-a_{12})b_j+a_{12}-a^0_1 \\ \vdots \\ (a_{(M-1)1}-a_{(M-1)2})b_j+a_{(M-1)2}-a^0_{M-1} \\ -\sum _{i=1}^{M-1}(a_{i1}-a_{i2})b_j-\sum _{i=1}^{M-1}(a_{i2}-a^0_{i}) \\ \end{matrix} \right) _{j=1}^N . \end{aligned}$$
Thus,
$$\begin{aligned} \Vert AB-A_0B_0 \Vert ^2&= \sum _{j=1}^N \,\,\left(\vphantom{ \left[ \sum _{i=1}^{M-1} \{(a_{i1}-a_{i2})b_j +a_{i2}-a^0_i\} \right] ^2} \sum _{i=1}^{M-1} \{(a_{i1}-a_{i2})b_j +a_{i2}-a^0_i\}^2 \right. \\&\quad \left. + \left[ \sum _{i=1}^{M-1} \{(a_{i1}-a_{i2})b_j +a_{i2}-a^0_i\} \right] ^2 \right) . \end{aligned}$$
Put \(I=\langle \{(a_{i1}-a_{i2})b_j +a_{i2}-a^0_i\}_{i=1}^{M-1} \rangle\). Because of Corollary 11 and
$$\begin{aligned} \sum _{i=1}^{M-1} \{(a_{i1}-a_{i2})b_j +a_{i2}-a^0_i\} \in I, \end{aligned}$$
we get
$$\begin{aligned}&\Vert AB-A_0B_0 \Vert ^2 \sim \sum _{j=1}^N \sum _{i=1}^{M-1} \{(a_{i1}-a_{i2})b_j +a_{i2}-a^0_i\}^2.\\&\text{ Let } {\left\{ \begin{array}{ll} a_{i}=a_{i1}-a_{i2}, \\ a_{i2} = a_{i2}, \\ b_j=b_j \end{array}\right. }. \end{aligned}$$
Then we get
$$\begin{aligned}&\sum _{j=1}^N \sum _{i=1}^{M-1} \{(a_{i1}-a_{i2})b_j +a_{i2}-a^0_i\}^2\\&\quad =\sum _{j=1}^N \sum _{i=1}^{M-1} \{a_{i}b_j +a_{i2}-a^0_i\}^2. \end{aligned}$$
Moreover,
$$\begin{aligned} \text{ Let } {\left\{ \begin{array}{ll} a_{i}=a_{i}, \\ b_j=b_j ,\\ c_{i} = a_{i2}-a^0_i \end{array}\right. } \end{aligned}$$
and
$$\begin{aligned} \sum _{j=1}^N \sum _{i=1}^{M-1} \{a_{i}b_j +a_{i2}-a^0_i\}^2=\sum _{j=1}^N \sum _{i=1}^{M-1} \{a_{i}b_j +c_{i}\}^2 \end{aligned}$$
holds.
$$\begin{aligned} \text{ Let } {\left\{ \begin{array}{ll} a_{i}=a_{i}, \\ b_j=b_j ,\\ x_{i} = a_{i}b_1+c_i, \\ \end{array}\right. }. \end{aligned}$$
If \(j>1\), then we have \(a_i b_j +c_i = x_i -a_i b_1 +a_i b_j\) and obtain
$$\begin{aligned}&\sum _{j=1}^N \sum _{i=1}^{M-1} \{a_{i}b_j +c_{i}\}^2\\&\quad =\sum _{i=1}^{M-1} \left[ x_i^2 + \sum _{j=2}^{N} \{x_i-(a_i b_1-a_i b_j)\}^2 \right] . \end{aligned}$$
Consider the following generated ideal:
$$\begin{aligned} J:=\left\langle (x_i)_{i=1}^{M-1}, (a_i b_1 -a_i b_j)_{(i,j)=(1,2)}^{(M-1,N)} \right\rangle . \end{aligned}$$
We expand the square terms
$$\begin{aligned} \{x_i-(a_i b_1-a_i b_j)\}^2=x_i^2 + (a_i b_1-a_i b_j)^2 -2 x_i (a_i b_1 -a_i b_j) \end{aligned}$$
and \(x_i (a_i b_1 - a_i b_j) \in J\) holds. Hence, owing to Corollary 12, we have
$$\begin{aligned}&\Vert AB-A_0 B_0 \Vert ^2 \sim \sum _{i=1}^{M-1} \left[ x_i^2 + \sum _{j=2}^{N} \{x_i-(a_i b_1-a_i b_j)\}^2 \right] \\&\sim \sum _{i=1}^{M-1} \left[ x_i^2 + \sum _{j=2}^{N} (a_i b_1-a_i b_j)^2 \right] \\&= \sum _{i=1}^{M-1} \left[ x_i^2 + \sum _{j=2}^{N} a_i^2(b_j-b_1)^2 \right] .\\&\text{ Let } {\left\{ \begin{array}{ll} a_{i}=a_i, \\ b_1=b_1, \\ b_j=b_j-b_1, &{} (j>1) \\ x_i =x_i \end{array}\right. }, \end{aligned}$$
then we have
$$\begin{aligned} \Vert AB-A_0 B_0 \Vert ^2&\sim \sum _{i=1}^{M-1} \left\{ x_i^2 + \sum _{j=2}^{N} a_i^2(b_j-b_1)^2 \right\} \\&=\sum _{i=1}^{M-1} \left( x_i^2 + \sum _{j=2}^{N} a_i^2b_j^2 \right) \\&=\sum _{i=1}^{M-1} x_i^2 +\sum _{i=1}^{M-1} \sum _{j=2}^N a_i^2 b_j^2 \\&=\sum _{i=1}^{M-1} x_i^2 +\left( \sum _{i=1}^{M-1}a_i^2 \right) \left( \sum _{j=2}^N b_j^2 \right) . \end{aligned}$$
Since \(a_i\),\(b_j\),\(x_i\) are independent variables for each, we consider blowing-ups of them and get
$$\begin{aligned} \lambda&= \frac{M-1}{2} + \min \left\{ \frac{M-1}{2}, \frac{N-1}{2} \right\} \\& = \min \left\{ M-1,\frac{M+N-2}{2} \right\} . \end{aligned}$$
Therefore,
$$\begin{aligned} \lambda ={\left\{ \begin{array}{ll} M-1 &{} (M \geqq N) \\ \frac{M+N-2}{2} &{} (M<N) \end{array}\right. }. \end{aligned}$$
\(\square\)
Third, we prove Lemma 16.
Proof
(Lemma 16)
$$\begin{aligned}&AB-A_0B_0 \\&\quad =\left( \begin{matrix} a_{11} &{} a_{12} \\ \vdots &{} \vdots \\ a_{(M-1)1} &{} a_{(M-1)2} \\ a_{M1} &{} a_{M2} \end{matrix} \right) \; \left( \begin{matrix} b_1 &{} \!\ldots \! &{} b_N \\ 1{-}b_1 &{} \!\ldots \! &{} 1{-}b_N \end{matrix} \right) \\&\qquad -\left( \begin{matrix} a^0_{11} &{} a^0_{12} \\ \vdots &{} \vdots \\ a^0_{(M-1)1} &{} a^0_{(M-1)2} \\ a^0_{M1} &{} a^0_{M2} \end{matrix} \right) \; \left( \begin{matrix} b^0_1 &{} \ldots &{} b^0_N \\ 1{-}b^0_1 &{} \ldots &{} 1{-}b^0_N \end{matrix} \right) \\&\quad =\left( \begin{matrix} (a_{11}-a_{12})b_j-(a^0_{11}-a^0_{12})b^0_j+a_{12}-a^0_1 \\ \vdots \\ (a_{(M-1)1}-a_{(M-1)2})b_j -(a^0_{(M-1)1}-a^0_{(M-1)2})b^0_j+a_{(M-1)2}-a^0_{M-1} \\ (a_{M1}-a_{M2})b_j-(a^0_{M1}-a^0_{M2})b^0_j+a_{M2}-a^0_{M} \\ \end{matrix} \right) _{j=1}^N\\&\quad =\left( \begin{matrix} (a_{11}-a_{12})b_j-(a^0_{11}-a^0_{12})b^0_j+a_{12}-a^0_1 \\ \vdots \\ (a_{(M-1)1}-a_{(M-1)2})b_j -(a^0_{(M-1)1}-a^0_{(M-1)2})b^0_j+a_{(M-1)2}-a^0_{M-1} \\ -\sum _{i=1}^{M-1}\{(a_{i1}-a_{i2})b_1-(a^0_{i1}-a^0_{i2})b^0_j\}-\sum _{i=1}^{M-1}(a_{i2}-a^0_{i}) \\ \end{matrix} \right) _{j=1}^N. \end{aligned}$$
Then we have
$$\begin{aligned} \Vert AB-A_0B_0 \Vert ^2&= \sum _{j=1}^N \,\,\left(\vphantom{\sum _{i=1^{\int^{\sum}}}^{M-1}} \sum _{i=1}^{M-1} \{(a_{i1}-a_{i2})b_j \right. \\&\quad \left. -(a^0_{i1}-a^0_{i2})b^0_j+a_{i2}-a^0_i\}^ 2 \right. \\&\quad + \left[ \sum _{i=1}^{M-1} \{(a_{i1}-a_{i2})b_j\right. \\&\quad \left. \left. -(a^0_{i1}-a^0_{i2})b^0_j +a_{i2}-a^0_i\} \vphantom{\sum _{i=1}^{M-1}}\right] ^2 \right) . \end{aligned}$$
Put \(I=\langle \{(a_{i1}-a_{i2})b_j -(a^0_{i1}-a^0_{i2})b^0_j+a_{i2}-a^0_i\}_{i=1}^{M-1} \rangle\). Because of Corollary 11 and
$$\begin{aligned} \sum _{i=1}^{M-1} \{(a_{i1}-a_{i2})b_j -(a^0_{i1}-a^0_{i2})b^0_j +a_{i2}-a^0_i\} \in I, \end{aligned}$$
we get
$$\begin{aligned}&\Vert AB-A_0B_0 \Vert ^2 \sim \sum _{j=1}^N \sum _{i=1}^{M-1} \{(a_{i1}-a_{i2})b_j \\&\quad -(a^0_{i1}-a^0_{i2})b^0_j+a_{i2}-a^0_i\}^2.\\&\text{ Let } {\left\{ \begin{array}{ll} a_{i}=a_{i1}-a_{i2}, \\ a_{i2} = a_{i2}, \\ b_j=b_j \end{array}\right. } \end{aligned}$$
and put \(a^0_{i}=a^0_{i1}-a^0_{i2}\). Then we get
$$\begin{aligned}&\sum _{j=1}^N \sum _{i=1}^{M-1} \left\{ (a_{i1}-a_{i2})b_j -(a^0_{i1}-a^0_{i2})b^0_j+a_{i2}-a^0_i\right\} ^2 \\&\quad =\sum _{j=1}^N \sum _{i=1}^{M-1} \left\{ a_{i}b_j -a^0_{i}b^0_j+a_{i2}-a^0_i\right\} ^2. \end{aligned}$$
Moreover,
$$\begin{aligned} \text{ Let } {\left\{ \begin{array}{ll} a_{i}=a_{i}, \\ b_j=b_j ,\\ c_{i} = a_{i2}-a^0_i \end{array}\right. } \end{aligned}$$
and
$$\begin{aligned}&\sum _{j=1}^N \sum _{i=1}^{M-1} \left\{ a_{i}b_j-a^0_{i}b^0_j +a_{i2}-a^0_i\right\} ^2 \\&\quad =\sum _{j=1}^N \sum _{i=1}^{M-1} \left\{ a_{i}b_j -a^0_{i}b^0_j+c_{i}\right\} ^2 \end{aligned}$$
holds.
$$\begin{aligned} \text{ Let } {\left\{ \begin{array}{ll} a_{i}=a_{i}, \\ b_j=b_j ,\\ x_{i} = a_{i}b_1-a^0_{i}b^0_1+c_i, \\ \end{array}\right. }. \end{aligned}$$
If \(j>1\), then we have \(a_i b_j -a^0_{i}b^0_j+c_i = x_i -a_i b_1 +a^0_i b^0_1+a_i b_j-a^0_{i}b^0_j\) and obtain
$$\begin{aligned}&\sum _{j=1}^N \sum _{i=1}^{M-1} \{a_{i}b_j -a^0_i b^0_j+c_{i}\}^2\\&\quad =\sum _{i=1}^{M-1} x_i^2 + \sum _{j=2}^{N} \sum _{i=1}^{M-1}\left\{ x_i-(a_i b_1 -a^0_i b^0_1-a_i b_j +a^0_i b^0_j)\right\} ^2. \end{aligned}$$
Consider the following generated ideal:
$$\begin{aligned}&J:=\left\langle (x_i)_{i=1}^{M-1}, (a_i b_1 -a^0_i b^0_1\right. \\&\qquad \left. -a_i b_j +a^0_i b^0_j)_{(i,j)=(1,2)}^{(M-1,N)} \right\rangle . \end{aligned}$$
We expand the square terms
$$\begin{aligned}&\{x_i-(a_i b_1 -a^0_i b^0_1-a_i b_j +a^0_i b^0_j)\}^2\\&=x_i^2 + (a_i b_1 -a^0_i b^0_1-a_i b_j +a^0_i b^0_j)^2 \\&\quad -2 x_i (a_i b_1 -a^0_i b^0_1-a_i b_j +a^0_i b^0_j) \end{aligned}$$
and \(x_i (a_i b_1 -a^0_i b^0_1-a_i b_j +a^0_i b^0_j) \in J\) holds. Hence, owing to Corollary 12, we have
$$\begin{aligned}&\Vert AB-A_0 B_0 \Vert ^2\\&\quad \sim \sum _{i=1}^{M-1} x_i^2 + \sum _{j=2}^{N} \sum _{i=1}^{M-1}\{x_i-(a_i b_1 -a^0_i b^0_1-a_i b_j +a^0_i b^0_j)\}^2 \\&\quad \sim \sum _{i=1}^{M-1} \left\{ x_i^2 + \sum _{j=2}^{N} (a_i b_1 -a^0_i b^0_1-a_i b_j +a^0_i b^0_j)^2 \right\} \\&\quad = \sum _{i=1}^{M-1} \left[ x_i^2 + \sum _{j=2}^{N} \{a_i(b_j-b_1)-a^0_i(b^0_j-b^0_1)\}^2 \right] .\\&\text{ Let } {\left\{ \begin{array}{ll} a_{i}=a_i, \\ b_1=b_1, \\ b_j=b_j-b_1, &{} (j>1) \\ x_i =x_i \end{array}\right. } \end{aligned}$$
and put \(b^0_j=b^0_j-b^0_1\), then we have
$$\begin{aligned}&\Vert AB-A_0 B_0 \Vert ^2 \\&\quad \sim \sum _{i=1}^{M-1} \left[ x_i^2 + \sum _{j=2}^{N} \{a_i(b_j-b_1)-a^0_i(b^0_j-b^0_1)\}^2 \right] \\&\quad =\sum _{i=1}^{M-1} \left\{ x_i^2 + \sum _{j=2}^{N} (a_i b_j - a^0_i b^0_j)^2 \right\} \\&\quad =\sum _{i=1}^{M-1} x_i^2 +\sum _{i=1}^{M-1} \sum _{j=2}^N (a_i b_j - a^0_i b^0_j)^2 .\end{aligned}$$
Let \(f_{ij}\) be \(a_i b_j - a^0_i b^0_j\). If \(\Vert AB-A_0B_0 \Vert ^2 =0\), \(f_{ij}=0\). Hence, \(a_i \ne 0\) and \(b_j \ne 0\). Owing to Proposition 13
$$\begin{aligned} \sum _{i=1}^{M-1} \sum _{j=2}^N f_{ij}^2 \sim \sum _{i=2}^{M-1} f_{i1}^2 + \sum _{j=3}^N f_{1j}^2 +f_{12}^2, \end{aligned}$$
we have
$$\begin{aligned} \Vert AB-A_0 B_0 \Vert ^2&\sim \sum _{i=1}^{M-1} x_i^2 +\sum _{i=1}^{M-1} \sum _{j=2}^N (a_i b_j - a^0_i b^0_j)^2 \nonumber \\&= \sum _{i=1}^{M-1} x_i^2 +\sum _{i=1}^{M-1} \sum _{j=2}^N f_{ij}^2 \nonumber \\&\sim \sum _{i=1}^{M-1} x_i^2 +\left( f_{12}^2+\sum _{i=2}^{M-1} f_{i2}^2 + \sum _{j=3}^N f_{1j}^2 \right) . \end{aligned}$$
(16)
Thus, all we have to do is calculate an RLCT of the right side. Considering blowing-ups, the RLCT \(\lambda _1\) of the first term is equal to \(\lambda _1=(M-1)/2\). For deriving the RLCT of the second term, we arbitrarily take \(i,j(1 \leqq i \leqq M-1, 2 \leqq j \leqq N, i,j \in {\mathbb {N}})\) and fix them.
$$\begin{aligned} \text{ Let } {\left\{ \begin{array}{ll} a_i = a_i, \\ f_{i2} = a_i b_2 - a^0_i b^0_2, \\ f_{1j} = a_1 b_j -a^0_1 b^0_j, \\ x_i = x_i \end{array}\right. } \end{aligned}$$
and we have that the Jacobi matrix of the above transformation is equal to
$$\begin{aligned} \frac{\partial (a_i ,f_{ij}, x_i)}{\partial (a_i,b_j,x_i)}= \left( \begin{array}{ccc} \frac{\partial a_i}{\partial a_i} &{} \frac{\partial f_{ij}}{\partial a_i} &{} \frac{\partial x_i}{\partial a_i}\\ \frac{\partial a_i}{\partial b_j} &{} \frac{\partial f_{ij}}{\partial b_j} &{} \frac{\partial x_i}{\partial b_j}\\ \frac{\partial a_i}{\partial x_i} &{} \frac{\partial f_{ij}}{\partial x_i} &{} \frac{\partial x_i}{\partial x_i}\\ \end{array} \right) = \left( \begin{array}{ccc} 1 &{} b_j &{} 0 \\ 0 &{} a_i &{} 0 \\ 0 &{} 0 &{} 1 \\ \end{array} \right) . \end{aligned}$$
Because of
$$\begin{aligned} \Biggl | \frac{\partial (a_i ,f_{ij}, x_i)}{\partial (a_i,b_j,x_i)} \Biggr |=a_i \ne 0, \end{aligned}$$
g is an analytic isomorphism. Thus, the RLCT
\(\lambda _2\) of the second term in Eq. (16) is equal to
$$\begin{aligned} \lambda _2 = \frac{M+N-3}{2}. \end{aligned}$$
Let \(\lambda\) be the RLCT of \(\Vert AB-A_0B_0 \Vert ^2\). From the above,
$$\begin{aligned} \lambda =\lambda _1+\lambda _2 = \frac{2M+N-4}{2}. \end{aligned}$$
\(\square\)
Lastly, we derive the inequality of Lemma 17.
Proof
(Lemma 17) We develop the objective function and obtain
$$\begin{aligned}&\Vert AB-A_0B_0 \Vert ^2 \nonumber \\&\quad = \sum _{i=1}^M \sum _{j=1}^N (a_{i1}b_{1j} + \cdots + a_{iH}b_{Hj} - a^0_{i1}b^0_{1j} - a^0_{iH}b^0_{Hj})^2 \nonumber \\&\quad = \sum _{j=1}^N \sum _{i=1}^M \Biggl \{ \sum _{k=1}^{H} (a_{ik}b_{kj} - a^0_{ik}b^0_{kj}) \Biggr \}^2 \nonumber \\&\quad = \sum _{j=1}^N \sum _{i=1}^{M-1} \Biggl \{ \sum _{k=1}^{H} (a_{ik}b_{kj} - a^0_{ik}b^0_{kj}) \Biggr \}^2 \nonumber \\&\qquad + \sum _{j=1}^N \Biggl \{ \sum _{k=1}^{H} (a_{Mk}b_{kj} - a^0_{Mk}b^0_{kj}) \Biggr \}^2. \end{aligned}$$
(17)
Expand the second term in Eq. (17) using \(a_{Mk}=1-\sum _{i=1}^{M-1}a_{ik}\), \(b_{Hj}=1-\sum _{k=1}^{H-1} b_{kj}\), \(a^0_{Mk}=1-\sum _{i=1}^{M-1}a^0_{ik}\), and \(b^0_{Hj}=1-\sum _{k=1}^{H-1} b^0_{kj}\), then we have
$$\begin{aligned}&\sum _{j=1}^N \Biggl \{ \sum _{k=1}^{H} (a_{Mk}b_{kj} - a^0_{Mk}b^0_{kj}) \Biggr \}^2 \\&\quad = \sum _{j=1}^N \Biggl \{ \sum _{k=1}^{H-1} (a_{Mk}b_{kj} - a^0_{Mk}b^0_{kj}) + (a_{MH}b_{Hj} - a^0_{MH}b^0_{Hj}) \Biggr \}^2 \\&\quad =\sum _{j=1}^N \Biggl (- \sum _{i=1}^{M-1} \sum _{k=1}^{H-1} a_{ik}b_{kj} + \sum _{i=1}^{M-1} \sum _{k=1}^{H-1} a^0_{ik}b^0_{kj}\\&\qquad -\sum _{i=1}^{M-1}a_{iH} +\sum _{i=1}^{M-1}\sum _{k=1}^{H-1} a_{iH}b_{kj}+ \sum _{i=1}^{M-1}a^0_{iH} \\&\qquad -\sum _{i=1}^{M-1} \sum _{k=1}^{H-1} a^0_{iH}b^0_{kj} \Biggr )^2 {=}\!:\! \varPhi _2. \end{aligned}$$
Developing the equation, we have
$$\begin{aligned} \varPhi _2&=\sum _{j=1}^N \Bigg \{- \sum _{i=1}^{M-1} \sum _{k=1}^{H-1} (a_{ik}-a_{iH})b_{kj} \\&\quad + \sum _{i=1}^{M-1} \sum _{k=1}^{H-1} (a^0_{ik}-a^0_{iH})b^0_{kj} -\sum _{i=1}^{M-1}(a_{iH}-a^0_{iH}) \Bigg \}^2 \\&=\sum _{j=1}^N \Biggl [\sum _{i=1}^{M-1} \sum _{k=1}^{H-1} \{(a_{ik}{-}a_{iH})b_{kj} \\&\qquad {-} (a^0_{ik}{-}a^0_{iH})b^0_{kj}\}{+}\sum _{i=1}^{M-1}(a_{iH}{-}a^0_{iH}) \Biggr ]^2. \end{aligned}$$
On the other hand, the first term of Eq. (17) is equal to
$$\begin{aligned}&\sum _{j=1}^N \sum _{i=1}^{M-1} \Biggl \{ \sum _{k=1}^{H} (a_{ik}b_{kj} - a^0_{ik}b^0_{kj}) \Biggr \}^2 \\&\quad = \sum _{j=1}^N \sum _{i=1}^{M-1} \Biggl \{ \sum _{k=1}^{H-1} (a_{ik}b_{kj} - a^0_{ik}b^0_{kj}) + (a_{iH}b_{Hj} - a^0_{iH}b^0_{Hj}) \Biggr \}^2 \\&\quad = \sum _{j=1}^N \sum _{i=1}^{M-1} \Biggl \{ \sum _{k=1}^{H-1} (a_{ik}b_{kj} - a^0_{ik}b^0_{kj}) + a_{iH}\\&\qquad -\sum _{k=1}^{H-1} a_{iH}b_{kj} - a^0_{iH} + \sum _{k=1}^{H-1} a^0_{iH} b^0_{kj} \Biggr \}^2 \\&\quad = \sum _{j=1}^N \sum _{i=1}^{M-1} \Biggl \{ \sum _{k=1}^{H-1} (a_{ik}b_{kj} - a^0_{ik}b^0_{kj}) + (a_{iH} - a^0_{iH})\\&\qquad -\sum _{k=1}^{H-1} (a_{iH}b_{kj} - a^0_{iH} b^0_{kj}) \Biggr \}^2 \\&\quad = \sum _{j=1}^N \sum _{i=1}^{M-1} \Biggl [(a_{iH} - a^0_{iH})+\sum _{k=1}^{H-1} \{(a_{ik}-a_{iH})b_{kj} \\&\qquad - (a^0_{ik}-a^0_{iH}) b^0_{kj}\} \Biggr ]^2 . \end{aligned}$$
Consider the following ideal:
$$\begin{aligned} I=\, & \left\langle (a_{iH}-a^0_{iH})_{i=1}^{M-1} , \{(a_{ik}-a_{iH})b_{kj} \right. \\&\left. - (a^0_{ik}-a^0_{iH}) b^0_{kj}\}_{(i,j,k)=(1,1,1)}^{(M-1,N,H-1)} \right\rangle . \end{aligned}$$
Since we have
$$\begin{aligned} \Vert AB-A_0B_0 \Vert ^2&= \sum _{j=1}^N \sum _{i=1}^{M-1} \Biggl [(a_{iH} - a^0_{iH})\\&\quad +\sum _{k=1}^{H-1} \{(a_{ik}-a_{iH})b_{kj} - (a^0_{ik}-a^0_{iH}) b^0_{kj}\} \Biggr ]^2 \\&\quad + \sum _{j=1}^N \Biggl [\sum _{i=1}^{M-1} \sum _{k=1}^{H-1} \{(a_{ik}{-}a_{iH})b_{kj} \\&\quad {-} (a^0_{ik}{-}a^0_{iH})b^0_{kj}\}{+}\sum _{i=1}^{M-1}(a_{iH}{-}a^0_{iH}) \Biggr ]^2 \end{aligned}$$
and
$$\begin{aligned}&\forall j, \sum _{i=1}^{M-1} \sum _{k=1}^{H-1} \{(a_{ik}{-}a_{iH})b_{kj} \\&\quad {-} (a^0_{ik}{-}a^0_{iH})b^0_{kj}\}{+}\sum _{i=1}^{M-1}(a_{iH}{-}a^0_{iH}) \in I, \end{aligned}$$
thus Corollary 12 causes
$$\begin{aligned}&\Vert AB-A_0B_0 \Vert ^2 \sim \sum _{j=1}^N \sum _{i=1}^{M-1} \Biggl [(a_{iH} - a^0_{iH})\\&\quad +\sum _{k=1}^{H-1} \{(a_{ik}-a_{iH})b_{kj} - (a^0_{ik}-a^0_{iH}) b^0_{kj}\} \Biggr ]^2. \end{aligned}$$
We transform the coordinate like the proof of Lemma 16 for resolution singularity of the above polynomial.
$$\begin{aligned} \text{ Let } {\left\{ \begin{array}{ll} a_{ik}=a_{ik}-a_{iH}, &{} (k<H) \\ a_{iH}=a_{iH}, \\ b_{kj}=b_{kj}, \\ \end{array}\right. } \end{aligned}$$
and put \(a^0_{ik}=a^0_{ik}-a^0_{iH}\),
$$\begin{aligned}&\sum _{j=1}^N \sum _{i=1}^{M-1} \Biggl [(a_{iH} - a^0_{iH}) +\sum _{k=1}^{H-1} \{(a_{ik}-a_{iH})b_{kj} - (a^0_{ik}-a^0_{iH}) b^0_{kj}\} \Biggr ]^2 \\&=\sum _{j=1}^N \sum _{i=1}^{M-1} \Biggl [(a_{iH} - a^0_{iH}) +\sum _{k=1}^{H-1} (a_{ik}b_{kj} {-} a^0_{ik}b^0_{kj}) \Biggr ]^2.\\&\text{ Let } {\left\{ \begin{array}{ll} a_{ik}=a_{ik},\\ b_{kj}=b_{kj}, \\ c_i=a_{iH}-a^0_{iH} \end{array}\right. }. \end{aligned}$$
Then we obtain
$$\begin{aligned}&\sum _{j=1}^N \sum _{i=1}^{M-1} \Biggl [(a_{iH} - a^0_{iH}) +\sum _{k=1}^{H-1} (a_{ik}b_{kj} {-} a^0_{ik}b^0_{kj}) \Biggr ]^2 \\&=\sum _{j=1}^N \sum _{i=1}^{M-1} \Biggl [c_i+\sum _{k=1}^{H-1}(a_{ik}b_{kj} {-} a^0_{ik}b^0_{kj}) \Biggr ]^2 \\&=\sum _{i=1}^{M-1} \sum _{j=1}^{N} \Biggl [c_i+\sum _{k=1}^{H-1} (a_{ik}b_{kj} {-} a^0_{ik}b^0_{kj}) \Biggr ]^2 \\&=\sum _{i=1}^{M-1} \left\{ \sum _{k=1}^{H-1} (a_{ik}b_{k1} {-} a^0_{ik}b^0_{k1}){+}c_i \right\} ^2 \\&\quad + \sum _{j=2}^N \sum _{i=1}^{M-1} \left\{ \sum _{k=1}^{H-1} (a_{ik}b_{kj} {-} a^0_{ik}b^0_{kj}){+}c_i \right\} ^2 .\end{aligned}$$
In addition,
$$\begin{aligned} \text{ Let } {\left\{ \begin{array}{ll} a_{ik}=a_{ik},\\ b_{kj}=b_{kj}, \\ x_i=\sum _{k=1}^{H-1} (a_{ik}b_{k1} - a^0_{ik}b^0_{k1}) + c_i \end{array}\right. }. \end{aligned}$$
If \(j>1\), then we have
$$\begin{aligned}&\sum _{k=1}^{H-1} (a_{ik}b_{kj} {-} a^0_{ik}b^0_{kj}){+}c_i \\&= x_i -\sum _{k=1}^{H-1} (a_{ik}b_{k1} - a^0_{ik}b^0_{k1})\\&\quad +\sum _{k=1}^{H-1} (a_{ik}b_{kj} {-} a^0_{ik}b^0_{kj})\\&=x_i +\sum _{k=1}^{H-1}\{ (a_{ik}b_{kj} \\&\quad {-} a^0_{ik}b^0_{kj})- (a_{ik}b_{k1} - a^0_{ik}b^0_{k1})\} \end{aligned}$$
and obtain
$$\begin{aligned}&\sum _{j=1}^N \sum _{i=1}^{M-1} \left\{ \sum _{k=1}^{H-1} (a_{ik}b_{kj} {-} a^0_{ik}b^0_{kj}){+}c_i \right\} ^2 \\&=\sum _{i=1}^{M-1} x_i^2 + \sum _{j=2}^{N}\sum _{i=1}^{M-1} \left[ x_i +\sum _{k=1}^{H-1}\{ (a_{ik}b_{kj} \right. \\&\quad \left. {-} a^0_{ik}b^0_{kj})- (a_{ik}b_{k1} - a^0_{ik}b^0_{k1})\} \vphantom{\sum _{k=1}^{H-1}}\right] ^2. \end{aligned}$$
Put
$$\begin{aligned} g_{ij}:=\sum _{k=1}^{H-1}\left\{ \left( a_{ik}b_{kj} {-} a^0_{ik}b^0_{kj}\right) - \left( a_{ik}b_{k1} - a^0_{ik}b^0_{k1}\right) \right\} . \end{aligned}$$
Consider the following ideal:
$$\begin{aligned} J:=\left\langle (x_i)_{i=1}^{M-1}, (g_{ij})_{(i,j)=(1,2)}^{(M-1,N)} \right\rangle . \end{aligned}$$
We expand the square terms
$$\begin{aligned} (x_i+g_{ij})^2=x_i^2 + (g_{ij})^2 +2x_i g_{ij} \end{aligned}$$
and \(x_i g_{ij} \in J\). Hence, owing to Corollary 12, we get
$$\begin{aligned} &\Vert AB-A_0 B_0 \Vert ^2\\&\quad\sim \sum _{i=1}^{M-1} x_i^2 + \sum _{j=2}^{N} \sum _{i=1}^{M-1} \{x_i+g_{ij}\}^2 \sim \sum _{i=1}^{M-1} \left\{ x_i^2 + \sum _{j=2}^{N} (g_{ij})^2 \right\} \\&\quad= \sum _{i=1}^{M-1} \left( x_i^2 + \sum _{j=2}^{N} \left[ \sum _{k=1}^{H-1}\left\{ \left(a_{ik}b_{kj} {-} a^0_{ik}b^0_{kj}\right) -\left(a_{ik}b_{k1} - a^0_{ik}b^0_{k1}\right)\right\}\right] ^2 \right) \\&\quad= \sum _{i=1}^{M-1} \left(x_i^2 + \sum _{j=2}^{N} \left[ \sum _{k=1}^{H-1}\{ a_{ik}(b_{kj} {-} b_{k1}) -a^0_{ik}(b^0_{kj}- b^0_{k1})\}\right] ^2 \right) . \end{aligned}$$
$$\begin{aligned} \text{ Let } {\left\{ \begin{array}{ll} a_{ik}=a_{ik}, \\ b_{k1}=b_{k1}, \\ b_{kj}=b_{kj}-b_{k1}, &{} (j>1) \\ x_i =x_i \end{array}\right. } \end{aligned}$$
and put \(b^0_{kj}=b^0_{kj}- b^0_{k1}\), then we have
$$\begin{aligned} \Vert AB-A_0 B_0 \Vert ^2&\sim \sum _{i=1}^{M-1} \left( x_i^2 + \sum _{j=2}^{N} \left[ \sum _{k=1}^{H-1}\{ a_{ik}(b_{kj} \right. \right. \\&\quad \left. \left. {-} b_{k1}) -a^0_{ik}(b^0_{kj}- b^0_{k1})\}\right] ^2 \right) \\&=\sum _{i=1}^{M-1} \left[ x_i^2 + \sum _{j=2}^{N} \left\{ \sum _{k=1}^{H-1}(a_{ik}b_{kj} -a^0_{ik}b^0_{kj})\right\} ^2 \right] \\&=\sum _{i=1}^{M-1} x_i^2 +\sum _{i=1}^{M-1} \sum _{j=2}^{N} \left\{ \sum _{k=1}^{H-1}(a_{ik}b_{kj} -a^0_{ik}b^0_{kj})\right\} ^2. \end{aligned}$$
There exists a positive constant \(C>0\), we have
$$\begin{aligned} \Vert AB-A_0B_0 \Vert ^2&\sim \sum _{i=1}^{M-1} x_i^2 +\sum _{i=1}^{M-1} \sum _{j=2}^{N} \left\{ \sum _{k=1}^{H-1}(a_{ik}b_{kj} -a^0_{ik}b^0_{kj})\right\} ^2 \\&\leqq \sum _{i=1}^{M-1} x_i^2 +C\sum _{i=1}^{M-1} \sum _{j=2}^{N} \sum _{k=1}^{H-1}(a_{ik}b_{kj} -a^0_{ik}b^0_{kj})^2 \\&\sim \sum _{i=1}^{M-1} x_i^2 +\sum _{i=1}^{M-1} \sum _{j=2}^{N} \sum _{k=1}^{H-1}(a_{ik}b_{kj} -a^0_{ik}b^0_{kj})^2 \\&= \sum _{i=1}^{M-1} x_i^2 + \sum _{k=1}^{H-1} \sum _{i=1}^{M-1} \sum _{j=2}^N (a_{ik} b_{kj} -a^0_{ik} b^0_{kj})^2. \end{aligned}$$
We blow-up the coordinate like the proof of Lemma 16 for resolution singularity in
$$\begin{aligned} \sum _{i=1}^{M-1} x_i^2 + \sum _{k=1}^{H-1} \sum _{i=1}^{M-1} \sum _{j=2}^N (a_{ik} b_{kj} -a^0_{ik} b^0_{kj})^2. \end{aligned}$$
Let \({\bar{\lambda }}_1\) be the RLCT of the first term and \({\bar{\lambda }}_2\) be the RLCT of the second term. It is immediately proved that \({\bar{\lambda }}_1\) is equal to \((M-1)/2\). For deriving the RLCT of the second term \({\bar{\lambda }}_2\), we use the result of Lemma 16: the RLCT of \(\sum _{i=1}^{M-1} \sum _{j=2}^N (a_{ik} b_{kj} -a^0_{ik} b^0_{kj})^2\) is equal to \((M+N-3)/2\). Thus we have
$$\begin{aligned} {\bar{\lambda }}_2 = (H-1) \frac{M+N-3}{2}. \end{aligned}$$
Let \(\lambda\) be the RLCT of \(\Vert AB-A_0B_0\Vert ^2\). In general, an RLCT is order isomorphic, therefore we get
$$\begin{aligned} \lambda&\leqq \bar{\lambda _1} + {\bar{\lambda }}_2 \\&= \frac{M-1}{2} + (H-1) \frac{M+N-3}{2} \\&= \frac{M-1+(H-1)(M+N-3)}{2}. \end{aligned}$$
\(\square\)
