Generalized persistence algorithm for decomposing multiparameter persistence modules

Abstract

The classical persistence algorithm computes the unique decomposition of a persistence module implicitly given by an input simplicial filtration. Based on matrix reduction, this algorithm is a cornerstone of the emergent area of topological data analysis. Its input is a simplicial filtration defined over the integers ${\mathbb {Z}}$ giving rise to a 1-parameter persistence module. It has been recognized that multiparameter version of persistence modules given by simplicial filtrations over d-dimensional integer grids ${\mathbb {Z}}^d$ is equally or perhaps more important in data science applications. However, in the multiparameter setting, one of the main challenges is that topological summaries based on algebraic structure such as decompositions and bottleneck distances cannot be as efficiently computed as in the 1-parameter case because there is no known extension of the persistence algorithm to multiparameter persistence modules. We present an efficient algorithm to compute the unique decomposition of a finitely presented persistence module M defined over the multiparameter ${\mathbb {Z}}^d$. The algorithm first assumes that the module is presented with a set of N generators and relations that are distinctly graded. Based on a generalized matrix reduction technique it runs in $O(N^{2\omega +1})$ time where $\omega <2.373$ is the exponent of matrix multiplication. This is much better than the well known algorithm called Meataxe which runs in ${\tilde{O}}(N^{6(d+1)})$ time on such an input. In practice, persistence modules are usually induced by simplicial filtrations. With such an input consisting of n simplices, our algorithm runs in $O(n^{(d-1)(2\omega + 1)})$ time for $d\ge 2$. For the special case of zero dimensional homology, it runs in time $O(n^{2\omega +1})$.

Appendices

Appendices

Free resolution and graded Betti numbers

Here we introduce free resolutions and graded Betti numbers of graded modules. Based on these tools, we give a proof of our Theorem 1.

Definition 13

For a graded module M, a free resolution ${\mathcal {F}}\rightarrow M$ is an exact sequence:

where each $F^i$ is a free graded R-module.

We say two free resolutions ${\mathcal {F}}, {\mathcal {G}}$ of M are isomorphic, denoted as ${\mathcal {F}}\simeq {\mathcal {G}}$, if there exists a collection of isomorphisms $\{h^i:F^i\rightarrow G^i\}_{i=0,1,\dots }$ which commutes with $f^i$’s and $g^i$’s. That is,for all $i=0,1,\dots$, $g^i \circ h^i = h^{i-1} \circ f^{i}$ where $h^{-1}$ is the identity map on M. See the following commutative diagram as an illustration.

For two free resolutions ${\mathcal {F}}\rightarrow M$ and ${\mathcal {G}}\rightarrow N$, by taking direct sums of free modules $F^i\oplus G^i$ and morphisms $f^i\oplus g^i$, we get a free resolution of $M\oplus N$, denoted as ${\mathcal {F}}\oplus {\mathcal {G}}$.

Note that a presentation of M can be viewed as the tail part

of a free resolution ${\mathcal {F}}\rightarrow M$. Free resolutions and presentations are not unique. But there exists a unique minimal free resolution in the following sense:

Fact 7

For a graded module M, there exists a unique free resolution such that $\forall i \ge 0, \text {im\,}f_{i+1}\subseteq {\mathfrak {m}}F_{i}$, where ${\mathfrak {m}}=(x_1,\ldots , x_d)$ is the unique maximal ideal of the graded ring $R={\mathbb {k}}[x_1,\ldots , x_d]$.

Definition 14

In a minimal free resolution ${\mathcal {F}}\rightarrow M$, the tail part

is called the minimal presentation of M and $f^1$ is called the minimal presentation map of M.

Here we briefly state the construction of the unique free resolution without formal proof. More details can be found in Bruns and Herzog (1998) and Römer (2001):

Construction A1

Choose a minimal set of homogeneous generators $g_1, \ldots , g_n$ of M. Let $F^0=\bigoplus _{i=1}^{n} R_{\rightarrow \text {gr}(g_i)}$ with standard basis $e_1^{\text {gr}(g_1)}, \ldots , e_n^{\text {gr}(g_n)}$ of $F^0$. The homogeneous R-map $f^0: F^0 \rightarrow M$ is determined by $f^0(e_i)=g_i$. Now the 1st syzygy module of M,

is again a finitely generated graded R-module. We choose a minimal set of homogeneous generators $s_1, \ldots , s_m$ of $S_1$ and let $F^1=\bigoplus _{j=1}^{m} R_{\rightarrow \text {gr}(s_j)}$ with standard basis $e_1'^{\text {gr}(s_1)}, \ldots , e_m'^{\text {gr}(s_m)}$ of $F^1$. The homogeneous R-map $f^1: F^1 \rightarrow F^0$ is determined by $f^1(e_j')=s_j$. By repeating this procedure for $S_2=\ker f^1$ and moving backward further, one gets a graded free resolution of M.

Fact 8

Any free resolution of M can be obtained (up to isomorphism) from the minimal free resolution by summing it with free resolutions of trivial modules, each with the following form

Note that the only nontrivial morphism

is the identity map $\mathbb {1}$.

From the above constructions, it is not hard to see that this unique free resolution is a minimal one in the sense that each free module $F^j$ has smallest possible size of basis.

For this unique free resolution, for each j, we can write $F^j\simeq \bigoplus _{\mathbf{u }\in {\mathbb {Z}}^d} \bigoplus ^{\beta ^{M}_{j,\mathbf{u }}} R_{\rightarrow \mathbf{u }}$ (the notation $\bigoplus ^{\beta ^{M}_{j,\mathbf{u }}} R_{\rightarrow \mathbf{u }}$ means the direct sum of ${\beta ^{M}_{j,\mathbf{u }}}$ copies of $R_{\rightarrow \mathbf{u }}$). The set $\{\beta ^{M}_{j,\mathbf{u }}\mid j\in {\mathbb {N}}, \mathbf{u }\in {\mathbb {Z}}^d\}$ is called the graded Betti numbers of M. When M is clear, we might omit the upper index in Betti number. For example, the graded Betti number of the persistence module for our working Example 1 is listed as Table 2.

Table 2 All the nonzero graded Betti numbers $\beta _{i,\mathbf{u }}$ are listed in the table

Note that the graded Betti number of a module is uniquely determined by the unique minimal free resolution. On the other hand, if a free resolution ${\mathcal {G}}\rightarrow M$ with $G^j\simeq \bigoplus _{\mathbf{u }\in {\mathbb {Z}}^d} \bigoplus ^{\gamma ^{M}_{j,\mathbf{u }}} R_{\rightarrow \mathbf{u }}$ satisfies ${\gamma ^{M}_{j,\mathbf{u }}} = {\beta ^{M}_{j,\mathbf{u }}}$ everywhere, then ${\mathcal {G}}\simeq {\mathcal {F}}$ is also a minimal free resolution of M.

Fact 9

$\beta ^{M\oplus N}_{*,*}=\beta ^{M}_{*,*}+ \beta ^{N}_{*,*}$

Proposition 6

Given a graded module M with a decomposition $M\simeq M^1\oplus M^2$, let ${\mathcal {F}}\rightarrow M$ be the minimal resolution of M, and ${\mathcal {G}}\rightarrow M^1$ and ${\mathcal {H}}\rightarrow M^2$ be the minimal resolution of $M^1$ and $M^2$ respectively, then ${\mathcal {F}}\simeq {\mathcal {G}}\oplus {\mathcal {H}}$.

Proof

${\mathcal {G}}\oplus {\mathcal {H}}\rightarrow M$ is a free resolution. We need to show it is a minimal free resolution. By previous argument, we just need to show that the graded Betti numbers of ${\mathcal {G}}\oplus {\mathcal {H}}\rightarrow M^1\oplus M^2$ coincide with graded Betti numbers of ${\mathcal {F}}\rightarrow M$. This is true by the fact 9. $\square$

Note that the free resolution is an extension of free presentation. So the above proposition applies to free presentation, which immediately results in the following Corollary.

Corollary 1

Given a graded module M with a decomposition $M\simeq M^1\oplus M^2$, let f be its minimal presentation map, and g, h be the minimal presentation maps of $M^1, M^2$ respectively, then $f\simeq g\oplus h$.

We also have the following fact relating morphisms:

Fact 10

$\ker (f^1\oplus f^2)=\ker f^1\oplus \ker f^2$; $\text {coker}(f^1\oplus f^2)=\text {coker}f^1\oplus \text {coker}f^2$.

Based on the above statements, now we can prove Theorem 1

Proof (proof of Theorem 1)

With the obvious correspondence $[f_i]\leftrightarrow [f]_i$, ($2\leftrightarrow 3$) easily follows from our arguments about matrix diagonalization in the main context.

($1\rightarrow 2$) Given $H\simeq \bigoplus H^i$ with the minimal presentation maps f of H: For each $H^i$, there exists a minimal presentation map $f_i$. By Corollary 1, we have $f\simeq \bigoplus f_i$.

($2\rightarrow 1$) Given $f\simeq \bigoplus f_i$: Since $H=\text {coker}f= \text {coker}(\bigoplus f_i)=\bigoplus \text {coker}f_i$, let $H^i=\text {coker}f_i$, we have the decomposition $H\simeq \bigoplus H^i$.

It follows that the above two constructions together give the desired 1-1 correspondence. $\square$

Proof (proof of Proposition 1)

We start with (2). Consider the total decomposition $f\simeq \bigoplus f^i$. By Remark 2, any presentation is isomorphic to a direct sum of the minimal presentation and some trivial presentations. Let $f\simeq g\oplus h$ with g being the minimal presentation. So $\text {coker}h=0$. Let $g\simeq \bigoplus g^j$ and $h\simeq \bigoplus h^k$ be the total decomposition of g and h. Note that $\forall k, \text {coker}h^k=0$. Now we have $\text {coker}f\simeq \bigoplus \text {coker}f^i$ with $\text {coker}f^i$ being either $\text {coker}g^j$ or 0, by the essentially uniqueness of total decomposition. With $H\simeq \bigoplus \text {coker}g^j$ being a total decomposition of H by Remark 3, and $\bigoplus \text {coker}f^i=\bigoplus \text {coker}g^j \bigoplus 0$, we can say that $H\simeq \bigoplus \text {coker}f^i$ is also a total decomposition.

Now for (1). For any decomposition $H\simeq \bigoplus H^i$, it is not hard to see that each $H^i$ can be written as a direct sum of a subset of $H_*^j$’s with $H\simeq \bigoplus H_*^j$ being the total decomposition of H. One just need to combine the $f^i$’s correspondingly in the total decomposition of $f\simeq \bigoplus f^i$ to get the desired decomposition of f. $\square$

Missing proofs in Sect. 4

Proposition

(4) The target block $\mathbf{A }|_{{T}}$ can be reduced to 0 while preserving the prior if and only if $\mathbf{A }|_{{T}}$ can be written as a linear combination of independent operations. That is,

$$\begin{aligned} \mathbf{A }|_{{T}}= \sum _{{\begin{array}{c} l\notin \textsf {Row}(T)\\ k\in \textsf {Row}({T}) \end{array}}}\alpha _{k,l} \mathbf{X }^{k,l}|_{{T}}+\sum _{{\begin{array}{c} i\notin \textsf {Col}(T)\\ j\in \textsf {Col}({T}) \end{array}}} \beta _{i,j}\mathbf{Y }^{i,j}|_{{T}} \end{aligned}$$

(6)

where $\alpha _{k,l}$’s and $\beta _{i,j}$’s are coefficient in $\mathbb {k}={\mathbb {F}}_2$.

Proof

Everything in the statement of the proposition is restricted to T. For simplicity of notations, we omit the lower script ${\le t}$ by assuming $\mathbf{A }_{\le t}=\mathbf{A }$, i.e., $t=m$ is the last column index. It can be verified that this omission does not affect the proof. The simple reason is that because of the admissible rules of column operations, entries beyond column t carried by any admissible operations will never affect entires in $\mathbf{A }_{\le t}$.

Recall that $\mathbf{Y }^{i,j}=\mathbf{A }\varvec{\cdot }[\delta _{i,j}]$ for some $(i,j)\in \textsf {Colop}$ and $\mathbf{X }^{k,l}=[\delta _{k,l}]\varvec{\cdot }\mathbf{A }$ for some $(l,k)\in \textsf {Rowop}$ where

$$\begin{aligned} \textsf {Colop}= & {} \{(i,j)\mid c_i\rightarrow c_j \text{ is } \text{ an } \text{ admissible } \text{ column } \text{ operation } \}\subseteq \textsf {Col}(\mathbf{A })\\&\times \textsf {Col}(\mathbf{A }) \text{ and } \\ \textsf {Rowop}= & {} \{(l,k)\mid r_l\rightarrow r_k \text{ is } \text{ an } \text{ admissible } \text{ row } \text{ operation } \}\subseteq \textsf {Row}(\mathbf{A })\times \textsf {Row}(\mathbf{A }) \end{aligned}$$

Let $\mathbf{I }$ be the identity matrix. We say a matrix $\mathbf{P }$ is an admissible left multiplication matrix if $\mathbf{P }=\mathbf{I }+\sum _{\textsf {Rowop}} \alpha _{k,l}[\delta _{k,l}]$ for some $(l,k)\in \textsf {Rowop}, \alpha _{k,l}\in \{0,1\}$. Similarly, we say a matrix $\mathbf{Q }$ is an admissible right multiplication matrix if $\mathbf{Q }=\mathbf{I }+\sum _{\mathcal {\textsf {Colop}}} \beta _{i,j}[\delta _{i,j}]$ for some $(i,j)\in \textsf {Colop}, \beta _{i,j}\in \{0,1\}$. In short, we just say $\mathbf{P }$ and $\mathbf{Q }$ are admissible. $\square$

It is not difficult to observe the following properties of admissible matrices:

Fact 11

Matrix $\mathbf{A }'\sim \mathbf{A }$ is an equivalent matrix transformed from $\mathbf{A }$ by a sequence of admissible operations if and only if $\mathbf{A }'=\mathbf{P }\mathbf{A }\mathbf{Q }$ for some admissible $\mathbf{P }$ and $\mathbf{Q }$.

Fact 12

Admissible matrices are closed under multiplication and taking inverse.

Fact 13

For any admissible $\mathbf{P }$, let $S\subseteq \textsf {Row}(\mathbf{P })$ be any subset of row indices. Then $\mathbf{P }|_{S\times S}$ is invertible.

For the last fact, observe that the matrix $\mathbf{P }|_{S\times S}$ can be embedded as a block of an admissible matrix $\mathbf{P }'$ constructed by making all off-diagonal entries of $\mathbf{P }$ whose indices are not in $S\times S$ to be zero. The matrix $\mathbf{P }'$ is obviously admissible. So by the second fact, it is invertible. Also, $\mathbf{P }'$ can be written in block diagonal form with two blocks $\mathbf{P }'|_{S\times S} \text{ and } \mathbf{P }'|_{{\bar{S}}\times {\bar{S}}}=\mathbf{I }$ where ${\bar{S}}=\textsf {Row}(\mathbf{P }')-S$. Therefore, if $\mathbf{P }'$ is invertible, so is $\mathbf{P }|_{S\times S}= \mathbf{P }'|_{S\times S}$.

Fig. 13

(Left) $\mathbf{A }$ at iteration t during reduction of the sub-column $c_t|_{\textsf {Row}(B)}$ for the block $B=B_2$. (Right) Target block T shown in magenta includes the sub-column of $c_t$. It does not include $B:=B_2$. All rows external to T have zeros in the columns external to T. All columns external to T have zeros in the rows external to T. Red regions combined form R

We write the matrix $\mathbf{A }$ in the following block forms with respect to B and T with necessary reordering of rows and columns (see Fig. 13 for a simple illustration without reordering rows and columns):

$$\begin{aligned} \mathbf{A }= \left[ \begin{array}{c|c} R &{} 0 \\ \hline T &{} B \end{array} \right] \end{aligned}$$

Here we abuse the notations of block and index block to make the expression more legible. In the above block forms of $\mathbf{A }$, for example, T represents the entries of $\mathbf{A }$ on the index block T, that is the block $\mathbf{A }|_{T}$, which is the target block we want to reduce. Note that

$$\begin{aligned} \begin{aligned} R&=\big [\textsf {Row}(\mathbf{A })\setminus \textsf {Row}(B),\, \textsf {Col}(\mathbf{A }_{\le t})\setminus \textsf {Col}(B))\big ]\\&=[\bigoplus _{B_i\ne B} B_i]\cup \big [\textsf {Row}(\mathbf{A })\setminus \textsf {Row}(B),\, \{t\}\big ] \end{aligned} \end{aligned}$$

which is the block obtained by merging all other previous index blocks together with the sub-column of t excluding entries on $\textsf {Row}(B)$. The right top block is zero since it belongs to the intersections of rows and columns from different blocks.

Observe that, the target block T can be reduced to 0 in $\mathbf{A }$ with prior preserved if and only if

$$\begin{aligned} \mathbf{P }\mathbf{A }\mathbf{Q }:= \mathbf{P }\varvec{\cdot }\left[ \begin{array}{c|c} R &{} 0 \\ \hline T &{} B \end{array} \right] \varvec{\cdot }\mathbf{Q }= \left[ \begin{array}{c|c} R &{} 0 \\ \hline 0 &{} B \end{array} \right] \end{aligned}$$

(7)

for some admissible $\mathbf{P }$ and $\mathbf{Q }$.

For $\Leftarrow$ direction, consider $\mathbf{P }= \mathbf{I }+\sum \alpha _{k,l}[\delta _{k,l}]$ and $\mathbf{Q }= \mathbf{I }+\sum \beta _{i,j}[\delta _{i,j}]$ with binary coefficients $\alpha _{k,l}$’s and $\beta _{i,j}$’s given in Equation 6. Then, we have

$$\begin{aligned} \mathbf{P }\mathbf{A }\mathbf{Q }&= (\mathbf{I }+\sum \alpha _{k,l}[\delta _{k,l}]) \mathbf{A }(\mathbf{I }+\sum \beta _{i,j}[\delta _{i,j}]) \end{aligned}$$

(8)

$$\begin{aligned}&= \mathbf{A }+ \sum \alpha _{k,l}[\delta _{k,l}]\mathbf{A }+ \sum \beta _{i,j}\mathbf{A }[\delta _{i,j}] + \sum \sum \alpha _{k,l}\beta _{i,j}[\delta _{k,l}]\mathbf{A }[\delta _{i,j}] \end{aligned}$$

(9)

$$\begin{aligned}&= \mathbf{A }+ \sum \alpha _{k,l}[\delta _{k,l}]\mathbf{A }+ \sum \beta _{i,j}\mathbf{A }[\delta _{i,j}] \end{aligned}$$

(10)

$$\begin{aligned}&=\mathbf{A }+ \sum \alpha _{k,l}\mathbf{X }^{k,l} + \sum \beta _{i,j}\mathbf{Y }^{i,j} \end{aligned}$$

(11)

The third Eq. (10) follows from Observations 3. After restriction to T, by the assumption that $\sum \alpha _{k,l}\mathbf{X }^{k,l} + \sum \beta _{i,j}\mathbf{Y }^{i,j}=\mathbf{A }|_T$, we get $\mathbf{P }\mathbf{A }\mathbf{Q }|_T=0$. By the definition of independent operations and Observation 2, one can see that our $\mathbf{P }, \mathbf{Q }$ solves Equation 7.

For $\Rightarrow$, we will show that if the above equation is solvable, then there always exist solutions $\mathbf{P }'$ and $\mathbf{Q }'$ in a simpler forms as stated in the following proposition.

Proposition 7

Equation (7) is solvable for some admissible $\mathbf{P }$ and $\mathbf{Q }$ if and only if it is solvable for some admissible $\mathbf{P }'$ and $\mathbf{Q }'$ in the following form:

$$\begin{aligned} \mathbf{P }'= \left[ \begin{array}{c|c} I &{} 0 \\ \hline U &{} I \end{array} \right] \text{ and } \mathbf{Q }'= \left[ \begin{array}{c|c} I &{} 0 \\ \hline V &{} I \end{array} \right] \end{aligned}$$

(12)

Before we prove Proposition 7, we show how one can prove the $\Rightarrow$ direction in Proposition 4 from it. Based on the equivalent condition Eq. 7 and Proposition 7, we can write $\mathbf{P }'$ and $\mathbf{Q }'$ in formula 12 as

$$\begin{aligned} \mathbf{P }'=\mathbf{I }+\sum _{{\begin{array}{c} (l,k)\in \\ \textsf {Rowop}_{R\rightarrow T} \end{array}}} \alpha _{k,l}[\delta _{k,l}] \;\quad \mathbf{Q }'=\mathbf{I }+\sum _{{\begin{array}{c} (i,j)\in \\ \textsf {Colop}_{B\rightarrow T} \end{array}}} \beta _{i,j}[\delta _{i,j}] \end{aligned}$$

where $\textsf {Rowop}_{R\rightarrow T}=\{(l,k)\in \textsf {Rowop}\mid (l,k)\in \textsf {Row}(R)\times \textsf {Row}(T)\}$ and $\textsf {Colop}_{B\rightarrow T}=\{(i,j)\in \textsf {Colop}\mid (i,j)\in \textsf {Col}(B)\times \textsf {Col}(T)\}$, and $\alpha _{k,l}, \beta _{i,j}\in \{0,1\}$. Then, similar to Equation 11, we get

$$\begin{aligned} \begin{aligned} \mathbf{P }'\mathbf{A }\mathbf{Q }'&=(\mathbf{I }+\sum _{{\begin{array}{c} (l,k)\in \\ \textsf {Rowop}_{R\rightarrow T} \end{array}}} \alpha _{k,l}[\delta _{k,l}])\varvec{\cdot }\mathbf{A }\varvec{\cdot }(\mathbf{I }+\sum _{{\begin{array}{c} (i,j)\in \\ \textsf {Colop}_{B\rightarrow T} \end{array}}} \beta _{i,j}[\delta _{i,j}])\\&=\mathbf{A }+\sum _{{\begin{array}{c} (l,k)\in \\ \textsf {Rowop}_{R\rightarrow T} \end{array}}}\alpha _{k,l} [\delta _{k,l}]\varvec{\cdot }\mathbf{A }+\sum _{{\begin{array}{c} (i,j)\in \\ \textsf {Colop}_{B\rightarrow T} \end{array}}} \beta _{i,j}\mathbf{A }\varvec{\cdot }[\delta _{i,j}] \\&\quad + \sum _{{\begin{array}{c} (l,k)\in \\ \textsf {Rowop}_{R\rightarrow T} \end{array}}}\sum _{{\begin{array}{c} (i,j)\in \\ \textsf {Colop}_{B\rightarrow T} \end{array}}} \alpha _{k,l}\beta _{i,j} [\delta _{k,l}]\varvec{\cdot }\mathbf{A }\varvec{\cdot }[\delta _{i,j}]\\&=\mathbf{A }+\sum _{{\begin{array}{c} (l,k)\in \\ \textsf {Rowop}_{R\rightarrow T} \end{array}}}\alpha _{k,l} [\delta _{k,l}]\varvec{\cdot }\mathbf{A }+\sum _{{\begin{array}{c} (i,j)\in \\ \textsf {Colop}_{B\rightarrow T} \end{array}}} \beta _{i,j}\mathbf{A }\varvec{\cdot }[\delta _{i,j}]\\&=\mathbf{A }+\sum _{{\begin{array}{c} (l,k)\in \\ \textsf {Rowop}_{R\rightarrow T} \end{array}}}\alpha _{ k,l }\mathbf{X }^{k,l}+\sum _{{\begin{array}{c} (i,j)\in \\ \textsf {Colop}_{B\rightarrow T} \end{array}}} \beta _{i,j}\mathbf{Y }^{i,j} \end{aligned} \end{aligned}$$

By restriction on T we have

$$\begin{aligned} \mathbf{P }'\mathbf{A }\mathbf{Q }'|_{T}= =\mathbf{A }|_{T}+\sum _{{\begin{array}{c} (l,k)\in \\ \textsf {Rowop}_{R\rightarrow T} \end{array}}}\alpha _{ k,l }\mathbf{X }^{k,l}|_{T}+\sum _{{\begin{array}{c} (i,j)\in \\ \textsf {Colop}_{B\rightarrow T} \end{array}}} \beta _{i,j}\mathbf{Y }^{i,j}|_{T} \end{aligned}$$

(13)

With $\mathbf{P }'\mathbf{A }\mathbf{Q }'|_{T}=0$ by our assumption, we get

$$\begin{aligned} \mathbf{A }|_{T}=\sum _{{\begin{array}{c} (l,k)\in \\ \textsf {Rowop}_{R\rightarrow T} \end{array}}}\alpha _{ k,l } \mathbf{X }^{ k,l }|_T+\sum _{{\begin{array}{c} (i,j)\in \\ \textsf {Colop}_{B\rightarrow T} \end{array}}} \beta _{i,j}\mathbf{Y }^{i,j}|_T \end{aligned}$$

This is exactly what we want

$$\begin{aligned} \mathbf{A }|_{{T}}= \sum _{{\begin{array}{c} l\notin \textsf {Row}(T)\\ k\in \textsf {Row}({T}) \end{array}}}\alpha _{k,l} \mathbf{X }^{k,l}|_{{T}}+\sum _{{\begin{array}{c} i\notin \textsf {Col}(T)\\ j\in \textsf {Col}({T}) \end{array}}} \beta _{i,j}\mathbf{Y }^{i,j}|_{{T}} \end{aligned}$$

(14)

$\square$

Now we give the proof of Proposition 7.

Proof of Proposition 7

The $\Leftarrow$ direction is trivial. For the $\Rightarrow$ direction, we want to show that, if Eq. (7) is solvable for some admissible $\mathbf{P }$ and $\mathbf{Q }$, then there exists admissible $\mathbf{P }'$ and $\mathbf{Q }'$ so that

$$\begin{aligned} \mathbf{P }'= & {} \left[ \begin{array}{c|c} I &{} 0 \\ \hline U &{} I \end{array} \right] , \mathbf{Q }'= \left[ \begin{array}{c|c} I &{} 0 \\ \hline V &{} I \end{array} \right] , \text{ and } \mathbf{P }^{'}\varvec{\cdot }\left[ \begin{array}{c|c} R &{} 0 \\ \hline T &{} B \end{array} \right] \varvec{\cdot }\mathbf{Q }^{'}\\= & {} \left[ \begin{array}{c|c} R &{} 0 \\ \hline UR+BV+T &{} B \end{array} \right] = \left[ \begin{array}{c|c} R &{} 0 \\ \hline 0 &{} B \end{array} \right] \end{aligned}$$

We write $\mathbf{P }$ and $\mathbf{Q }$ in corresponding block forms as follows:

$$\begin{aligned} \mathbf{P }= \left[ \begin{array}{c|c} P_1 &{} P_2 \\ \hline P_3 &{} P_4 \end{array} \right] \text{ and } \mathbf{Q }= \left[ \begin{array}{c|c} Q_1 &{} Q_2 \\ \hline Q_3 &{} Q_4 \end{array} \right] \end{aligned}$$

(15)

From Eq. (7) one can get a set of equations

$$\begin{aligned} P_1 R Q_2 + P_2 B Q_4= & {} 0 \end{aligned}$$

(16)

$$\begin{aligned} P_1 R Q_1 + P_2 B Q_3= & {} R \end{aligned}$$

(17)

$$\begin{aligned} P_3 R Q_2 + P_4 B Q_4= & {} B \end{aligned}$$

(18)

$$\begin{aligned} P_3 R Q_1 + P_4 B Q_3= & {} T \end{aligned}$$

(19)

From Fact 13, we know that $P_1, P_4, Q_1, Q_4$ are invertible. By left multiplication with $P_1^{-1}$ and right multiplication with $Q_4^{-1}$ on both sides of Eq. (16), one can get :

$$\begin{aligned}&P_1^{-1}P_1 R Q_2 Q_4^{-1} + P_1^{-1} P_2 B Q_4 Q_4^{-1} \nonumber \\&\quad = R Q_2 Q_4^{-1} + P_1^{-1} P_2 B = 0 \implies - R Q_2 Q_4^{-1} = P_1^{-1} P_2 B \end{aligned}$$

(20)

Similarly, by left multiplication with $P_1^{-1}$ on both sides of Eq. (17) and by right multiplication with $Q_4^{-1}$ on both sides of Eq. (18), one can get the following equations:

$$\begin{aligned} P_1 R Q_1 + P_2 B Q_3= & {} R \implies R Q_1 = P_1^{-1}R - P_1^{-1}P_2 B Q_3 \end{aligned}$$

(21)

$$\begin{aligned} P_3 R Q_2 + P_4 B Q_4= & {} B \implies P_4 B = B Q_4^{-1} - P_3 R Q_2 Q_4^{-1} \end{aligned}$$

(22)

Now from Eq. 19, we have:

Letting $U=P_3 P_1^{-1}$ and $V=Q_4^{-1}Q_3$, we get the desired equation. Now we just need to show that $\mathbf{P }', \mathbf{Q }'$ are both admissible. We prove it for $\mathbf{Q }'$. Similar proof holds for $\mathbf{P }'$. We want to show that for any $(i,j)\in \textsf {Row}(V)\times \textsf {Col}(V)$, if $\mathbf{Q }'_{i,j}=1$, then $(i,j)\in \textsf {Colop}$. From equality, $V=Q_4^{-1}Q_3$, which implies $\mathbf{Q }'_{i,j}=\sum _k (Q_4^{-1})_{i,k} \varvec{\cdot }(Q_3)_{k,j}=1$, we know that $(Q_4^{-1})_{i,k}= (Q_3)_{k,j}=1$ for some k. Since $Q_4^{-1} \text{ and } Q_3$ are both blocks in the admissible matrix $\mathbf{Q }$, by the definition of admissible left multiplication matrix, we have $(i,k), (k,j)\in \textsf {Colop}$. Note that $\textsf {Colop}$ is closed under transitive relation by Proposition 2. So we have $(i,j)\in \textsf {Colop}$. $\square$