On the Expressive Power of Linear Algebra on Graphs

Abstract

There is a long tradition in understanding graphs by investigating their adjacency matrices by means of linear algebra. Similarly, logic-based graph query languages are commonly used to explore graph properties. In this paper, we bridge these two approaches by regarding linear algebra as a graph query language. More specifically, we consider MATLANG, a matrix query language recently introduced, in which some basic linear algebra functionality is supported. We investigate the problem of characterising the equivalence of graphs, represented by their adjacency matrices, for various fragments of MATLANG. That is, we are interested in understanding when two graphs cannot be distinguished by posing queries in MATLANG on their adjacency matrices. Surprisingly, a complete picture can be painted of the impact of each of the linear algebra operations supported in MATLANG on their ability to distinguish graphs. Interestingly, these characterisations can often be phrased in terms of spectral and combinatorial properties of graphs. Furthermore, we also establish links to logical equivalence of graphs. In particular, we show that MATLANG-equivalence of graphs corresponds to equivalence by means of sentences in the three-variable fragment of first-order logic with counting. Equivalence with regards to a smaller MATLANG fragment is shown to correspond to equivalence by means of sentences in the two-variable fragment of this logic.

Appendices

Appendix

Proof of Lemma 5.1

Lemma 5.1

Let $\mathsf {ML}({\mathscr{L}})$ be any matrix query language fragment and let G and H be graphs of the same order. Consider expressions e₁(X) and e₂(X) in $\mathsf {ML}({\mathscr{L}})$. If e_i(A_G) and e_i(A_H) are T-similar, for i = 1, 2, for an arbitrary matrix T, then e₁(A_G) ⋅ e₂(A_G) is also T-similar to e₁(A_H) ⋅ e₂(A_H).

Proof

To show this lemma, we distinguish between the following cases, depending on the dimensions of e₁(A_G) and e₂(A_G) (or equivalently, the dimensions of e₁(A_H) and e₂(A_H)). Let e(X) := e₁(X) ⋅ e₂(X). Let n be the order of G (and H).

• (n ×n,n ×n): e₁(A_G) and e₂(A_G) are of dimension n × n. By assumption, e₁(A_G) ⋅ T = T ⋅ e₁(A_H) and e₂(A_G) ⋅ T = T ⋅ e₂(A_H). Hence,

  $$ \begin{array}{@{}rcl@{}} e(A_{G})\cdot T&=&e_{1}(A_{G})\cdot e_{2}(A_{G})\cdot T=e_{1}(A_{G})\cdot T\cdot e_{2}(A_{H})\\ &=&T\cdot e_{1}(A_{H})\cdot e_{2}(A_{H}) =T\cdot e(A_{H}). \end{array} $$

• (n ×n,n ×1): e₁(A_G) is of dimension n × n and e₂(A_G) is of dimension n × 1. By assumption, e₁(A_G) ⋅ T = T ⋅ e₁(A_H) and e₂(A_G) = T ⋅ e₂(A_H). Hence,

  $$ e(A_{G}) = e_{1}(A_{G})\cdot e_{2}(A_{G}) = e_{1}(A_{G})\cdot T\cdot e_{2}(A_{H}) = T\cdot e_{1}(A_{H})\cdot e_{2}(A_{H})=T\cdot e(A_{H}).$$

• (n ×1,1 ×n): e₁(A_G) is of dimension n × 1 and e₂(A_G) is of dimension 1 × n. By assumption, e₁(A_G) = T ⋅ e₁(A_H) and e₂(A_G) ⋅ T = e₂(A_H). Hence,

  $$ \begin{array}{@{}rcl@{}} e(A_{G})\cdot T&=&e_{1}(A_{G})\cdot e_{2}(A_{G})\cdot T =e_{1}(A_{G})\cdot e_{2}(A_{H})\\ &=& T\cdot e_{1}(A_{H})\cdot e_{2}(A_{H})=T\cdot e(A_{H}). \end{array} $$

• (n ×1,1 ×1): e₁(A_G) is of dimension n × 1 and e₂(A_G) is of dimension 1 × 1. By assumption, e₁(A_G) = T ⋅ e₁(A_H) and e₂(A_G) = e₂(A_H). Hence,

  $$ e(A_{G})=e_{1}(A_{G})\cdot e_{2}(A_{G})=e_{1}(A_{G})\cdot e_{2}(A_{H})=T\cdot e_{1}(A_{H})\cdot e_{2}(A_{H}) =T\cdot e(A_{H}).$$

• (1 ×n,n ×n): e₁(A_G) is of dimension 1 × n and e₂(A_G) is of dimension n × n. By assumption, e₁(A_G) ⋅ T = e₁(A_H) and e₂(A_G) ⋅ T = T ⋅ e₂(A_H). Hence,

  $$ e(A_{G})\cdot T = e_{1}(A_{G})\cdot e_{2}(A_{G})\cdot T = e_{1}(A_{H})\cdot T \cdot e_{2}(A_{H}) = e_{1}(A_{H})\cdot e_{2}(A_{H})=e(A_{H}).$$

• (1 ×n,n ×1): e₁(A_G) is of dimension 1 × n and e₂(A_G) is of dimension n × 1. By assumption, e₁(A_G) ⋅ T = e₁(A_H) and e₂(A_G) = T ⋅ e₂(A_H). Hence,

  $$ e(A_{G})=e_{1}(A_{G})\cdot e_{2}(A_{G})=e_{1}(A_{G})\cdot T\cdot e_{2}(A_{H})= e_{1}(A_{H})\cdot e_{2}(A_{H})=e(A_{H}).$$

• (1 ×1,1 ×n): e₁(A_G) is of dimension 1 × 1 and e₂(A_G) is of dimension 1 × n. By assumption, e₁(A_G) = e₁(A_H) and e₂(A_G) ⋅ T = e₂(A_H). Hence,

  $$ e(A_{G})\cdot T=e_{1}(A_{G})\cdot e_{2}(A_{G})\cdot T=e_{1}(A_{G})\cdot e_{2}(A_{H})=e_{1}(A_{G})\cdot e_{2}(A_{H})=e(A_{H}).$$

• (1 ×1,1 ×1): e₁(A) and e₂(A) are of dimension 1 × 1. By assumption, e₁(A_G) = e₁(A_H) and e₂(A_G) = e₂(A_H). Hence, e(A_G) = e₁(A_G) ⋅ e₂(A_G) = e₁(A_H) ⋅ e₂(A_H) = e(A_H).

This concludes the proof. □

Continuation of the proofs of Proposition 7.2 and Theorems 8.1 and 9.1

In all three proofs we relied on the presence of addition and scalar multiplication to compute either equitable partitions or stable edge partitions. Since addition and scalar multiplication are used in the proofs before the proper conjugacy notion was identified, we cannot simply rely on Lemma 5.3. We therefore show that when ${\mathscr{L}}$ is either (for Proposition 7.2), (Theorem 8.1) or (Theorem 9.1), that $G\equiv _{\mathsf {ML}({\mathscr{L}})} H$ implies $G\equiv _{\mathsf {ML}({\mathscr{L}}^{+})} H$, where ${\mathscr{L}}^{+}={\mathscr{L}}\cup \{+,\times \}$.

We show this by verifying that any expression e(X) in $\mathsf {ML}({\mathscr{L}}^{+})$ can be equivalently written as a linear combination of expressions in $\mathsf {ML}({\mathscr{L}})$. We denote equivalence by ≡ and $e(X)\equiv e^{\prime }(X)$ means that $e(A)=e^{\prime }(A)$ for all matrices A. We verify our claim by induction on the structure of expressions.

(base case)

Let e(X) := X. This already has the desired form.

(matrix multiplication)

Let e(X) := e₁(X) ⋅ e₂(X). By induction we have that $e_{1}(X)\equiv {\sum }_{i=1}^{p} a_{i}\times {e_{i}^{1}}(X)$ and $e_{2}(X)\equiv {\sum }_{i=1}^{q} b_{i}\times {e_{i}^{2}}(X)$. Hence, $e(X)\equiv {\sum }_{i=1}^{p}{\sum }_{j=1}^{q} (a_{i}\times b_{i}) \times ({e_{i}^{1}}(X)\cdot {e_{j}^{2}}(X)$).

(complex conjugate transposition)

Let e(X) := (e₁(X))^∗. By induction, $e_{1}(X)\equiv {\sum }_{i=1}^{p} a_{i}\times {e_{i}^{1}}(X)$. Then, $e(X)\equiv {\sum }_{i=1}^{p} \bar {a}_{i}\times (e_{i}(X))^{*}$.

(trace)

Let e(X) := tr(e₁(X)). By induction, $e_{1}(X)\equiv {\sum }_{i=1}^{p} a_{i}\times {e_{i}^{1}}(X)$. Then, $e(X)\equiv {\sum }_{i=1}^{p} a_{i}\times \mathsf {tr}(e_{i}(X))$.

(ones-vector)

Let . By induction, $e_{1}(X)\equiv {\sum }_{i=1}^{p} a_{i}\times {e_{i}^{1}}(X)$. Then, .

(diag)

Let e(X) := diag(e₁(X)). By induction, $e_{1}(X)\equiv {\sum }_{i=1}^{p} a_{i}\times {e_{i}^{1}}(X)$. Then, $e(X)\equiv {\sum }_{i=1}^{p} a_{i}\times \textup {diag}({e_{i}^{1}}(X))$.

(pointwise vector product)

Let e(X) := e₁(X) ⊙_ve₂(X). By induction we have that $e_{1}(X)\equiv {\sum }_{i=1}^{p} a_{i}\times {e_{i}^{1}}(X)$ and $e_{2}(X)\equiv {\sum }_{i=1}^{q} b_{i}\times {e_{i}^{2}}(X)$. Hence, $e(X)\equiv {\sum }_{i=1}^{p}{\sum }_{j=1}^{q} (a_{i}\times b_{i}) \times ({e_{i}^{1}}(X)\odot _{v} {e_{j}^{2}}(X)$).

(Schur-Hadamard) (pointwise vector product)

Let e(X) := e₁(X) ⊙ e₂(X). By induction we have that $e_{1}(X)\equiv {\sum }_{i=1}^{p} a_{i}\times {e_{i}^{1}}(X)$ and $e_{2}(X)\equiv {\sum }_{i=1}^{q} b_{i}\times {e_{i}^{2}}(X)$. Hence, $e(X)\equiv {\sum }_{i=1}^{p}{\sum }_{j=1}^{q} (a_{i}\times b_{i}) \times ({e_{i}^{1}}(X)\odot {e_{j}^{2}}(X)$).

This concludes the proof. $\square$

Proof of Proposition 7.4

Proposition 7.4

, ×,Apply_s[f], f ∈Ω)-vectors are constant on equitable partitions.

Proof

Let ${\mathscr{L}}^{\#}$ denote , ×,Apply_s[f], f ∈Ω}. Consider a graph G of order n with equitable partition $\mathcal {V}=\{V_{1},\ldots ,V_{\ell }\}$. As before, let be the corresponding indicator vectors. We will show that for any expression $e(X)\in \textsf {ML}({\mathscr{L}}^{\#})$ such that e(A_G) is an n × 1-vector, e(A_G) can be uniquely written in the form for scalars $a_{i}\in \mathbb {C}$.

We show, by induction on the structure of expressions in $\mathsf {ML}({\mathscr{L}}^{\#})$, that the following properties hold:

• (a) if e(A_G) returns an n × n-matrix, then for any pair i, j = 1,…, ℓ there exists a scalars $a_{ij}, b_{ij}\in \mathbb {C}$ such that

  

• (b) if e(A_G) returns an n × 1-vector, then for any i = 1,…, ℓ, there exists a scalar a_i ∈ C such that

  

Clearly, if (b) holds for every i = 1,…, ℓ, then, because . We remark these properties can be seen as a generalisation of the known fact that the vector space spanned by indicator vectors of an equitable partition of G is invariant under multiplication by A_G (See e.g., Lemma 5.2 in [16]). That is, for any linear combination we have that . In our setting, (a) and (b) imply that e(A_G) ⋅ v is again a linear combination of indicator vectors, when e(A_G) returns an n × n-matrix. We next verify properties (a) and (b). We often use that and . □

(base case)

Let e(X) := X. The required property is simply a restatement of the being equitable. That is,

for an arbitrary vertex v ∈ V_i. So, we can take a_ij = deg(v, V_j). Similarly, because A_G is a symmetric matrix,

for an arbitrary vertex v ∈ V_i. So, we can take a_ij = deg(v, V_j).

Below, for condition (a) we only verify that holds. The verification of is entirely similar.

(multiplication)

Let e(X) := e₁(X) ⋅ e₂(X). We distinguish between a number of cases, depending on the dimensions of e₁(A_G) and e₂(A_G). We first check the cases when e(A_G) returns an n × n-matrix and need to show that property (a) holds.

• (n ×n,n ×n): e₁(A_G) and e₂(A_G) are of dimension n × n. By induction, and . Then, is equal to

  

  as desired.

• (n ×1,1 ×n): e₁(A_G) is of dimension n × 1 and e₂(A_G) is of dimension 1 × n. By induction we have that and . Hence, is equal to

  

  as desired. Here we used that is either 0, in case that k≠j, or |V_j| in case that j = k.

We next check that condition (b) holds when e(A_G) returns an n × 1-vector.

• (n ×n,n ×1): e₁(A_G) is of dimension n × n and e₂(A_G) is of dimension n × 1. By induction, we have that and . Hence, is equal to

  

  as desired.

• (n ×1,1 ×1): e₁(A_G) is of dimension n × 1 and e₂(A_G) is of dimension 1 × 1. By induction we have that and $e_{2}(A_{G})=b\in \mathbb {C}$. Hence,

  

  as desired.

(ones vector)

. We only need to consider the case when e₁(A_G) is an n × n-matrix or n × 1-vector. In both cases, it suffices to observe that . Indeed,

(conjugate transpose)

e(X) := (e₁(X))^∗. If e₁(A_G) returns a 1 × n-vector, then . Hence, returns an n × n-matrix, then by induction, . Hence,

as desired.

(diag operation)

e(X) := diag(e₁(X)) where e₁(A_G) is an n × 1-vector. By induction, . Hence, in view of the linearity of the diagonal operation,

since is when k = j and the zero vector otherwise.

(addition)

e(X) := e₁(X) + e₂(X). Clearly, when condition (a) or (b) hold for e₁(A_G) and e₂(A_G), they remain to hold for e(A_G).

(scalar multiplication)

e(X) := a × e₁(X). Clearly, when condition (a) or (b) hold for e₁(A_G), they remain to hold for e(A_G).

(trace)

e(X) := tr(e₁(X)). Such sub-expressions do not return matrices or vectors.

(pointwise function applications)

$e(X):=\textsf {Apply}_{\mathsf {s}}[f](e_{1}(X),\dots ,e_{p}(X))$ where each e_i(X) is a sentence. Again, such sub-expressions do not return matrices or vectors.

Proof of Proposition 8.2

Proposition 8.2

,×,Apply_s[f], f ∈Ω)-vectors are constant on equitable partitions

Proof

Given that we verified this property for all operations except for ⊙_v in the proof of Proposition 7.4, we only need to verify that ⊙_v can be added to the list of supported operations. We use the same induction hypotheses as in the proof of Proposition 7.4 and verify that these hypotheses remain to hold for ⊙_v: □

(pointwise vector multiplication)

e(X) := e₁(X) ⊙_ve₂(X) where e₁(X) and e₂(X) return vectors. By induction we have that and . As a consequence,

because is either when i = j, or the zero vector when i≠j.

Continuation of the proof of Theorem 8.1

In the proof in the main body of the paper we left open the verification that , for i = 1,…, ℓ, implies that O preserves equitable partitions of G and H. In particular, we need to verify that , for i = 1,…, ℓ. This can be easily shown, just as in the proof of Theorem 7.2 (based on Lemma 4 in Thüne [67]), in which we verified that J ⋅ O = O ⋅ J implies that .

First, we observe that with and . In other words, where . Furthermore, because is a scalar, . We next show that α = ±n_i. Indeed, since O is an orthogonal matrix

and thus ${\alpha _{i}^{2}}={n_{i}^{2}}$ or α_i = ±n_i. Hence, . We note that . We now argue that either for all i = 1,…, ℓ, or for all i = 1,…, ℓ. Indeed, suppose that we have for i ∈ K ⊂{1,…, ℓ} and for $i\in \bar K=\{1,\ldots ,\ell \}\setminus K$, for some non-empty subset K of {1,…, ℓ}. Then and hence since and ,

This contradicts that . Hence, when for all i = 1,…, ℓ, O satisfies the desired property already. Otherwise, when for all i = 1,…, ℓ, we simply replace O by (− 1) × O to obtain that . This rescaling does not impact that A_G ⋅ O = O ⋅ A_H and we can thus indeed conclude that O preserves equitable partitions of G and H.