We use the same notation as in the proof of Theorem
5.4. With this, we define
$$\begin{aligned} \varvec{I}_k&= \varvec{a}_{k,>}^* \varvec{a}_{k,>}^* \varvec{a}_{k,>} \varvec{a}_{k,>} = \bigotimes _{\ell =1}^k I, \\ \mathsf {a}_k^*&= \left[ \begin{array}{cc} \varvec{a}_{k,i_1}^* \varvec{a}_{k,>}^* \varvec{a}_{k,>} \varvec{a}_{k,>} \end{array}\right] ^{i_1=1,\ldots ,k}, \\ \mathsf {a}_k&= \left[ \begin{array}{cc} \varvec{a}_{k,>}^* \varvec{a}_{k,>}^* \varvec{a}_{k,j_1} \varvec{a}_{k,>} \end{array}\right] ^{j_1=1,\ldots ,k}, \\ \mathsf {b}_k&= \left[ \begin{array}{cc} \varvec{a}_{k,i_1}^* \varvec{a}_{k,>}^* \varvec{a}_{k,j_1} \varvec{a}_{k,>} \end{array}\right] ^{i_1=1,\ldots ,k; j_1=1,\ldots ,k},&\\ \mathsf {c}_k^*&= \left[ \begin{array}{cc} \varvec{a}_{k,i_1}^* \varvec{a}_{k,i_1}^* \varvec{a}_{k,>} \varvec{a}_{k,>} \end{array}\right] ^{i_1=1,\ldots ,k; i_2=i_1,\ldots ,k}, \end{aligned}$$
as well as
$$\begin{aligned} \mathsf {c}_k&= \left[ \begin{array}{cc} \varvec{a}_{k,>}^* \varvec{a}_{k,>}^* \varvec{a}_{k,j_1} \varvec{a}_{k,j_2} \end{array}\right] ^{j_1=1,\ldots ,k; j_2=j_1,\ldots ,k}, \\ \mathsf {e}_k^*&= \left[ \begin{array}{cc} \displaystyle \sum _{\begin{array}{c} i_1,j_1,j_2=1 \\ j_1<j_2 \end{array} }^k \tilde{v}_{i_1 i_2 j_1 j_2} \varvec{a}_{k,i_1}^* \varvec{a}_{k,i_2}^* \varvec{a}_{k,j_1} \varvec{a}_{k,j_2} \end{array}\right] ^{i_2=k+1,\ldots ,K}, \\ \mathsf {e}_k&= \left[ \begin{array}{cc} \displaystyle \sum _{\begin{array}{c} i_1,i_2,j_1=1 \\ i_1<i_2 \end{array} }^k \tilde{v}_{i_1 i_2 j_1 j_2} \varvec{a}_{k,i_1}^* \varvec{a}_{k,i_2}^* \varvec{a}_{k,j_1} \varvec{a}_{k,j_2} \end{array}\right] ^{j_2=k+1,\ldots ,K},\\ \varvec{D}_n&= \sum _{\begin{array}{c} i_1,i_2,j_1,j_2=1 \\ i_1<i_2,\,j_1<j_2 \end{array}}^k \tilde{v}_{i_1 i_2 j_1 j_2} \varvec{a}_{k,i_1}^*\varvec{a}_{k,i_2}^*\varvec{a}_{k,j_1}\varvec{a}_{k,j_2}. \end{aligned}$$
We again proceed by induction and show that
$$\begin{aligned} V_1 {{\,\mathrm{\bowtie }\,}}\dots {{\,\mathrm{\bowtie }\,}}V_k = \left[ \begin{array}{ccccccccc} \varvec{I}_{k}&\mathsf {a}_{k}^*&\mathsf {a}_{k}&\mathsf {b}_{k}&\mathsf {c}_{k}^*&\mathsf {c}_{k}&\mathsf {e}_{k}&\mathsf {e}_{k}^*&\varvec{D}_{k} \end{array}\right] . \end{aligned}$$
This holds for
\(k=1\) and for
\(k=2,\ldots , \frac{K}{2}\), we want to show
$$\begin{aligned} \left[ \begin{array}{ccccccccc} \varvec{I}_{k-1}&\mathsf {a}_{k-1}^*&\mathsf {a}_{k-1}&\mathsf {b}_{k-1}&\mathsf {c}_{k-1}^*&\mathsf {c}_{k-1}&\mathsf {e}_{k-1}&\mathsf {e}_{k-1}^*&\varvec{D}_{k-1} \end{array}\right] {{\,\mathrm{\bowtie }\,}}\left[ \begin{array}{cc} V_k^{1,1} &{} V_k^{1,2} \\ 0 &{}V_k^{2,2} \end{array}\right] \\ = \left[ \begin{array}{ccccccccc} \varvec{I}_{k}&\mathsf {a}_{k}^*&\mathsf {a}_{k}&\mathsf {b}_{k}&\mathsf {c}_{k}^*&\mathsf {c}_{k}&\mathsf {e}_{k}&\mathsf {e}_{k}^*&\varvec{D}_{k} \end{array}\right] . \end{aligned}$$
The product with
\(V_k^{1,1}\) is rather straightforward. For the products with
\(V_k^{1,2}\) and
\(V_k^{2,2}\), we calculate
$$\begin{aligned} \mathsf {a}_{k-1}^* {{\,\mathrm{\bowtie }\,}}\, (W_{V,k}^1\uparrow A^*A)&= \left[ \begin{array}{cc} \displaystyle \sum _{i_1=1}^{k-1} {\tilde{v}}_{i_1kkj_2} \varvec{a}_{k,i_1}^* \varvec{a}_{k,>}^* \varvec{a}_{k,>} \varvec{a}_{k,>} \otimes A^*A \end{array}\right] ^{j_2=k+1,\ldots , K} \\&= \left[ \begin{array}{cc} \displaystyle \sum _{i_1=1}^{k-1} {\tilde{v}}_{i_1kkj_2} \varvec{a}_{k,i_1}^* \varvec{a}_{k,k}^* \varvec{a}_{k,k} \varvec{a}_{k,>} \end{array}\right] ^{j_2=k+1,\ldots , K},\\ \mathsf {b}_{k-1} {{\,\mathrm{\bowtie }\,}}\, (W_{V,k}^3\uparrow A^*)&= \left[ \begin{array}{cc} \displaystyle \sum _{i_1,j_1=1}^{k-1} {\tilde{v}}_{i_1kj_1j_2} \varvec{a}_{k,i_1}^* \varvec{a}_{k,>}^* \varvec{a}_{k,j_1} \varvec{a}_{k,>} \otimes A^* \end{array}\right] ^{j_2=k+1\cdots K} \\&= \left[ \begin{array}{cc} \displaystyle \sum _{i_1,j_1=1}^{k-1} {\tilde{v}}_{i_1kj_1j_2} \varvec{a}_{k,i_1}^* \varvec{a}_{k,k}^* \varvec{a}_{k,j_1} \varvec{a}_{k,>} \end{array}\right] ^{j_2=k+1\cdots K} \end{aligned}$$
and
$$\begin{aligned} \mathsf {c}_{k-1}^* {{\,\mathrm{\bowtie }\,}}\, (W_{V,k}^6\uparrow A)&= \left[ \begin{array}{cc} \displaystyle \sum _{\begin{array}{c} i_1,i_2=1 \\ i_1< i_2 \end{array}}^{k-1} {\tilde{v}}_{i_1i_2kj_2} \varvec{a}_{k,i_1}^* \varvec{a}_{k,i_1}^* \varvec{a}_{k,>} \varvec{a}_{k,>} \otimes A \end{array}\right] ^{j_2=k+1,\ldots , K} \\&= \left[ \begin{array}{cc} \sum _{\begin{array}{c} i_1,i_2=1 \\ i_1 < i_2 \end{array}}^{k-1} {\tilde{v}}_{i_1i_2kj_2} \varvec{a}_{k,i_1}^* \varvec{a}_{k,i_1}^* \varvec{a}_{k,k} \varvec{a}_{k,>} \end{array}\right] ^{j_2=k+1,\ldots , K} \end{aligned}$$
as well as
$$\begin{aligned} \mathsf {e}_{k-1} {{\,\mathrm{\bowtie }\,}}\mathsf {S}_{K-k}&= \left[ \begin{array}{cc} \displaystyle \sum _{\begin{array}{c} i_1,i_2,j_1=1 \\ i_1< i_2 \end{array}}^{k-1} {\tilde{v}}_{i_1i_2j_1j_2} \varvec{a}_{k,i_1}^* \varvec{a}_{k,i_2}^* \varvec{a}_{k,j_1} \varvec{a}_{k,j_2} \otimes S \end{array}\right] ^{j_2=k+1,\ldots , K} \\&= \left[ \begin{array}{cc} \displaystyle \sum _{\begin{array}{c} i_1,i_2,j_1=1 \\ i_1 < i_2 \end{array}}^{k-1} {\tilde{v}}_{i_1i_2j_1j_2} \varvec{a}_{k,i_1}^* \varvec{a}_{k,i_2}^* \varvec{a}_{k,j_1} \varvec{a}_{k,j_2} \end{array}\right] ^{j_2=k+1,\ldots , K} , \\ \end{aligned}$$
and with this
$$\begin{aligned}&\mathsf {a}_{k-1}^* {{\,\mathrm{\bowtie }\,}}\,(W_{V,k}^1\uparrow A^*A) + \mathsf {b}_{k-1} {{\,\mathrm{\bowtie }\,}}\,(W_{V,k}^3\uparrow A^*) + \mathsf {c}_{k-1}^* {{\,\mathrm{\bowtie }\,}}\, (W_{V,k}^6\uparrow A) \\&\quad + \mathsf {e}_{k-1} {{\,\mathrm{\bowtie }\,}}\mathsf {S}_{K-k}= \mathsf {e}_k. \end{aligned}$$
Similarly, we show
$$\begin{aligned}&\mathsf {a}_{k-1} {{\,\mathrm{\bowtie }\,}}\,(W_{V,k}^2\uparrow A^*A) + \mathsf {b}_{k-1} {{\,\mathrm{\bowtie }\,}}\,(W_{V,k}^4\uparrow A) \\&+ \mathsf {c}_{k-1} {{\,\mathrm{\bowtie }\,}}\,(W_{V,k}^7\uparrow A^*) + \mathsf {e}_{k-1}^* {{\,\mathrm{\bowtie }\,}}\mathsf {S}_{K-k} = \mathsf {e}_k^* \end{aligned}$$
and
$$\begin{aligned}&\mathsf {b}_{k-1} {{\,\mathrm{\bowtie }\,}}\,(W_{V,k}^5\uparrow A^*A) \\&\quad + \sum _{\begin{array}{c} i_1,i_2,j_1=1 \\ i_1<i_2 \end{array} }^k \tilde{v}_{i_1 i_2 j_1 j_2} \varvec{a}_{k,i_1}^* \varvec{a}_{k,i_2}^* \varvec{a}_{k,j_1} \varvec{a}_{k,k} \otimes A \\&\quad + \sum _{\begin{array}{c} i_1,j_1,j_2=1 \\ j_1<j_2 \end{array} }^k \tilde{v}_{i_1 i_2 j_1 j_2} \varvec{a}_{k,i_1}^* \varvec{a}_{k,k}^* \varvec{a}_{k,j_1} \varvec{a}_{k,j_2} \otimes A^* \\&\quad + \sum _{\begin{array}{c} i_1,i_2,j_1,j_2=1 \\ i_1<i_2,\,j_1<j_2 \end{array}}^k \tilde{v}_{i_1 i_2 j_1 j_2} \varvec{a}_{k,i_1}^*\varvec{a}_{k,i_2}^*\varvec{a}_{k,j_1}\varvec{a}_{k,j_2} \otimes I = \varvec{D}_k. \end{aligned}$$
We proceed in the same fashion for
\(k = \frac{K}{2} +1 ,\ldots , K\). We define
\(\varvec{{\tilde{I}}}_{k}\),
\({\tilde{\mathsf {a}}}_{k}^{*}\),
\({\tilde{\mathsf {a}}}_{k}\),
\({\tilde{\mathsf {b}}}_{k}\),
\({\tilde{\mathsf {c}}}_{k}^*\),
\({\tilde{\mathsf {c}}}_{k}\),
\({\tilde{\mathsf {e}}}_{k}\),
\({\tilde{\mathsf {e}}}_{k}^*\), and
\({\tilde{\mathsf {D}}}_{k}\) accordingly, such that
$$\begin{aligned} V_k {{\,\mathrm{\bowtie }\,}}\left[ \begin{array}{c} \tilde{\varvec{I}}_{k+1} \\ \tilde{\mathsf {a}}_{k+1}^* \\ \tilde{\mathsf {a}}_{k+1} \\ \tilde{\mathsf {b}}_{k+1} \\ \tilde{\mathsf {c}}_{k+1}^* \\ \tilde{\mathsf {c}}_{k+1} \\ \tilde{\mathsf {e}}_{k+1} \\ \tilde{\mathsf {e}}_{k+1}^* \\ \tilde{\varvec{D}}_{k+1} \end{array}\right] =&\left[ \begin{array}{c} {\tilde{\varvec{I}}}_{k} \\ \tilde{{\mathsf {a}}}_{k}^*\\ \tilde{\mathsf {a}}_{k} \\ \tilde{\mathsf {b}}_{k} \\ {\tilde{\mathsf {c}}}_{k}^* \\ {\tilde{\varvec{c}}}_{k} \\ \tilde{\mathsf {e}}_{k} \\ \tilde{\mathsf {e}}_{k}^* \\ \tilde{\varvec{D}}_{k} \end{array}\right] . \end{aligned}$$
It remains to show that
$$\begin{aligned} \varvec{D}{=}\left[ \begin{array}{ccccccccc} \varvec{I}_{K/2}&\mathsf {a}_{K/2}^*&\mathsf {a}_{K/2}&\mathsf {b}_{K/2}&\mathsf {c}_{K/2}^*&\mathsf {c}_{K/2}&\mathsf {e}_{K/2}&\mathsf {e}_{K/2}^*&\varvec{D}_{K/2} \end{array}\right] M_V{{\,\mathrm{\bowtie }\,}}\left[ \begin{array}{c} \varvec{\tilde{I}}_{K/2+1} \\ \tilde{\mathsf {a}}_{K/2+1}^*\\ \tilde{\mathsf {a}}_{K/2+1} \\ \tilde{\mathsf {b}}_{K/2+1} \\ \mathsf {c}_{K/2+1}^* \\ \mathsf {c}_{K/2+1} \\ \tilde{\mathsf {e}}_{K/2+1} \\ \tilde{\mathsf {e}}_{K/2+1}^* \\ \varvec{\tilde{D}}_{K/2+1} \end{array}\right] .\qquad \end{aligned}$$
(9.1)
This holds due to the anti-diagonal structure of
\(M_V\) and with an explicit calculation of
\(\mathsf {b}_{K/2} W_{V,k}^8 \mathsf {b}_{K/2 + 1}\),
\(\mathsf {c}_{K/2}^* W_{V,k}^9 \mathsf {c}_{K/2 + 1}\), and
\(\mathsf {c}_{K/2} W_{V,k}^{10} \mathsf {c}_{K/2 + 1}^*\).
For the sparse case (
5.3), we only consider at the matrix
\(M_V\), since it determines the highest rank in the representation of
\(\varvec{D}\). In this case, the contributions to the rank of the left and right components in (
9.1) are as follows: Each of the two combinations of components
\(\varvec{I}_{K/2}\) and
\(\varvec{\tilde{D}}_{K/2+1}\) as well as
\(\varvec{D}_{K/2}\) and
\(\varvec{\tilde{I}}_{K/2+1}\) contributes one to the rank. Each of the four combinations of
\(\mathsf {a}_{K/2}^*\) and
\(\tilde{\mathsf {e}}^*_{K/2+1}\);
\(\mathsf {a}_{K/2}\) and
\(\tilde{\mathsf {e}}_{K/2+1}\);
\(\mathsf {e}_{K/2}\) and
\(\tilde{\mathsf {a}}_{K/2+1}\);
\(\mathsf {e}_{K/2}^*\) and
\(\tilde{\mathsf {a}}_{K/2+1}^*\) contributes
\(d-1\). In the case of
\(\mathsf {b}_{K/2}\) and
\(\tilde{\mathsf {b}}_{K/2+1}\), we obtain
$$\begin{aligned} \sum _{\ell =1}^{d-1}\min \{2\ell -1,2(d-\ell )-1\}={\left\{ \begin{array}{ll} \frac{1}{2} (d-1)^2 +d &{} d \text { odd, }\\ \frac{1}{2} d^2 &{} d \text { even; } \end{array}\right. } \end{aligned}$$
in the case of
\(\mathsf {c}_{K/2}^*\) and
\(\tilde{\mathsf {c}}_{K/2+1}\),
$$\begin{aligned} \sum _{\ell =2}^{d-1}\min \{\ell -1,d-\ell \}={\left\{ \begin{array}{ll} 2\left( {\begin{array}{c}\frac{1}{2} (d-1)\\ 2\end{array}}\right) +\frac{1}{2} (d-1) &{} d \text { odd, }\\ 2\left( {\begin{array}{c}\frac{1}{2} d\\ 2\end{array}}\right)&d \text { even; } \end{array}\right. } \end{aligned}$$
and in the case of
\(\mathsf {c}_{K/2}\) and
\(\tilde{\mathsf {c}}_{K/2+1}^*\),
$$\begin{aligned} \sum _{\ell =2}^{d-1}\min \{\ell -1,d-\ell \}={\left\{ \begin{array}{ll} 2\left( {\begin{array}{c}\frac{1}{2} (d-1)\\ 2\end{array}}\right) +\frac{1}{2} (d-1) &{} d \text { odd, }\\ 2\left( {\begin{array}{c}\frac{1}{2} d\\ 2\end{array}}\right)&d \text { even, } \end{array}\right. } \end{aligned}$$
and summing up these contributions completes the proof.
\(\square \)