In this section we consider a special kind of transformation of Sudoku puzzles and study its spectral properties. To this end, we define the *k*-fold blow up of a free-form Sudoku. This is formed by replacing each cell of the original Sudoku by a *k* × *k* arrangement of cells. The block partition *T*^{ʹ} of the new graph is derived from the original partition *T* as follows. For each block *B* ϵ *T* we create a block *B*^{ʹ} ϵ *T*^{ʹ} by collecting the replacement cells of all the cells in *B*. In terms of graphs we see that the *k*-fold blow up transforms FSud(*n*, *T*) into FSud(*kn*, *T*^{ʹ}). Let us denote the latter graph by FSud^{↑k}(*n*, *T*). In Figure 2 the 3-fold blow up of a free-form 2-Sudoku is illustrated. It also sketches the neighborhood of an exemplary vertex of the blown up graph.

Figure 2 3-fold blow up of the free-form 2-Sudoku puzzle from Figure 1 (b)

Let us now investigate how the adjacency matrix of a blown up Sudoku graph can be determined from the adjacency matrix of the original graph. To this end, we a assume that a fixed but otherwise arbitrary numbering of the cells of the original free-form Sudoku is given. The vertices of the associated graph shall be numbered accordingly. Further, we number the cells of a *k*-fold blow up of the given puzzle as follows. Let *S*_{i} denote the *k* × *k* subsquare of the blown up puzzle containing exactly the replacement cells of the original cell number *i*.We number the *k*^{2}*n*^{2} cells by subsequently numbering all vertices in *S*_{1}, then *S*_{2} and so on. Inside each *S*_{i} we number the cells by starting in the top left corner and proceeding from left to right, advancing row by row.

Next, note the following facts about the blow up operation:

#### Observation 3.1

–

*If two cells i and j lie in the same row (resp. column) of the original puzzle, then all cells sharing the same relative row (resp. column) index inside S*_{i} and/or S_{j} lie in the same row (resp. column) of the blown up puzzle

–

*If two cells i and j lie in the same block of the original puzzle, then all cells from S*_{i} and S_{j} lie in the same block of the blown up puzzle

In view of these facts and due to the chosen cell numbering of the blown up puzzle we can construct the adjacency matrix *A↑* of FSud^{↑k}(*n*, *T*) from the adjacency matrix *A* = (*a*_{ij}) of FSud(*n*, *T*) by replacing each entry *a*_{ij} of *A* by a (*k*^{2} × *k*^{2})-matrix that solely depends on whether (in the original puzzle) cell number *j* is in the same block as cell number *i*, or otherwise in the same row or column as *j* or lies somewhere else.

For this purpose we define the symbolic template adjacency matrix *T*(*A*) = (*t*_{ij}) as follows:

–

*t*_{ii} = D,

–

*t*_{ij} = B if *i*/ = *j* and cells *i*, *j* are in the same block of the original puzzle,

–

*t*_{ij} = H if cells *i*, *j* are not in the same block but in the same row of the original puzzle,

–

*t*_{ij} = V if cells *i*, *j* are not in the same block but in the same column of the original puzzle,

–

*t*_{ij} = N else.

The next step is to define the matrices that will be substituted into the template matrix. But first we need some building blocks. Let *I*_{r} denote the identity matrix, *J*_{r} the all-ones matrix and *N*_{r} the zero matrix of size *r* × *r*. Further, we will make use of the Kronecker product of real matrices, cf. [18].

Fixing the given blow up factor *k*, we define the following (*k*^{2} × *k*^{2})-matrices:

$$H={I}_{k}\otimes {J}_{k},\text{\hspace{0.17em}}\text{\hspace{0.17em}}V={J}_{k}\otimes {I}_{k},\text{\hspace{0.17em}}\text{\hspace{0.17em}}B={J}_{{k}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}D={J}_{{k}^{2}}-{I}_{{k}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}N={N}_{{k}^{2}}.$$(2)

The following result should now be self-evident:

#### Proposition 3.2

*Given an n* × *n free-form Sudoku puzzle with block partition T, the adjacency matrix of* FSud^{↑k}(*n*, *T*) *(with respect to the vertex numbering mentioned in the second paragraph of this section) can be obtained from the symbolic template matrix T*(*A*) = (*t*_{ij}) *of the adjacency matrix A* = (*a*_{ij}) *of the graph* FSud(*n*, *T*) *by replacing each entry t*_{ij} of T(*A*) *by the contents of the matching matrix from* (2) *that has the same name as the symbolic value of t*_{ij} suggests

The replacement process can be expressed by means of the Kronecker product. Let us take the template matrix *T*(*A*) and use it to partition the non-zero entries *a*_{ij} of *A* according to their respective symbols *t*_{ij}. Setting the blown up adjacency matrix *A↑* can now be expressed as follows:

#### Proposition 3.3

$${A}^{\uparrow}={L}_{B}\otimes B+{L}_{H}\otimes H+{L}_{V}\otimes V+{L}_{D}\otimes D.$$(3)

Our goal is to study the eigenvalues of blown up free-form Sudokus and express them in terms of the eigenvalues of the original puzzle.

For the rest of this section, we consider an arbitrary but fixed free-form *n* × *n* Sudoku puzzle with tiling *T*. Let *A* be the adjacency matrix of its associated graph FSud(*n*, *T*) and let *A↑* be the adjacency matrix of FSud^{↑k}(*n*, *T*).We assume that *L*_{V}, *L *_{H}, *L *_{B} etc. denote the matrices appearing in equations (1) and (3). Further, let Eig(*M*) represent the set of all eigenvectors of a given matrix (or even a graph) *M* and let ker(*M*) be the set of all eigenvectors of *M* associated with eigenvalue 0. Note that neither set contains the null vector.

In the following, we will make use of the following properties of the Kronecker product:

#### Theorem 3.4

(see [18]).

(*αA*) ⊗ *B* = *A* ⊗ (*αB*) = *α*(*A* ⊗ *B*) *for all α* ϵ ℝ, *A* ϵ ℝ^{p}^{×q} , *B* ϵ ℝ^{r}^{×s}

(*A* ⊗ *B*) ⊗ *C* = *A* ⊗ (*B* ⊗ *C*) *for all A* ϵ ℝ^{m}^{×n} , *B* ϵ ℝ^{p}^{×q} , *C* ϵ ℝ^{r}^{×s}

(*A* + *B*) ⊗ *C* = *A* ⊗ *C* + *B* ⊗ *C for all A*, *B* ϵ ℝ^{p}^{×q} , *C* ϵ ℝ^{r}^{×s}

*A* ⊗ (*B* + *C*) = *A* ⊗ *B* + *A* ⊗ *C for all A* ϵ ℝ^{p}^{×q} , *B*, *C* ϵ ℝ^{r}^{×s}

(*A* ⊗ *B*)(*C* ⊗ *D*) = *AC* ⊗ *BD for all A* ϵ ℝ^{p}^{×q} , *B* ϵ ℝ^{r}^{×s} , *C* ϵ ℝ^{q}^{×k} , *D* ϵ ℝ^{s}^{×l}

Note that the Kronecker product formally includes the case where one or both factors are vectors.

#### Lemma 3.5

*Suppose that x and y are eigenvectors of L*_{V} and J_{k} corresponding to the eigenvalues λ and α, respectively. Then, for any z ϵ ker(*J*_{k})*, x* ⊗ *y* ⊗ *z is an eigenvector of A*^{↑} corresponding to eigenvalue λα − 1.

*Proof*. We use equations (2), (3) and the facts *L*_{V} x = *λx*, *J*_{k}y = *αy*, *J*_{k}z = 0.

$$\begin{array}{l}{A}^{\uparrow}\left(x\otimes y\otimes z\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left({L}_{B}\otimes B+{L}_{H}\otimes H+{L}_{V}\otimes V+{L}_{D}\otimes D\right)\left(x\otimes y\otimes z\right)\\ =({L}_{B}\otimes {J}_{{K}^{2}}+{L}_{H}\otimes {I}_{k}\otimes {J}_{k}+{L}_{V}\otimes {J}_{k}\otimes {I}_{k}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+{L}_{D}\otimes {J}_{{k}^{2}}-{L}_{D}\otimes {I}_{{k}^{2}})\left(x\otimes y\otimes z\right)\\ =\left(\left({L}_{B}x\right)\otimes \left({J}_{k}y\right)\otimes \left({J}_{k}z\right)\right)+\left(\left({L}_{H}x\right)\otimes \left({I}_{k}y\right)\otimes \left({J}_{k}z\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\left(\left({L}_{V}x\right)\otimes \left({J}_{k}y\right)\otimes \left({I}_{k}z\right)\right)+\left(\left({L}_{D}x\right)\otimes \left({J}_{k}y\right)\otimes \left({J}_{k}z\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\left(\left({L}_{D}x\right)\otimes \left({I}_{k}y\right)\otimes \left({I}_{k}z\right)\right)\\ =\left(\lambda x\right)\otimes \left(\alpha y\right)\otimes z-x\otimes y\otimes z\\ =\left(\lambda \alpha -1\right)\left(x\otimes y\otimes z\right).\end{array}$$

#### Lemma 3.6

*Suppose that x and z are eigenvectors of L*_{H} and J_{k} corresponding to the eigenvalues λ and α, respectively. Then, for any y ϵ ker(*J*_{k})*, x* ⊗ *y* ⊗ *z is an eigenvector of A↑ corresponding to eigenvalue λα* − 1.

*Proof*. Similar to the proof of Lemma 3.5.

The previous two lemmas will play a role in the construction of a basis of eigenvectors of a blown up Sudoku graph. To this end, we need to know the intersection of the spans of the two vector sets mentioned there. In the following, let 1_{r} denote the all-ones vector of dimension *r*. Now note the following obvious facts:

#### Proposition 3.7

*1. The spectrum of J*_{k} is {*k*^{(1)}, 0^{(k−1)}}.

*2. The set*

$$\begin{array}{l}{\mathcal{K}}_{J}\text{\hspace{0.17em}}:=\{{\left(1,-1,0,0,\mathrm{...},0,0\right)}^{T},{\left(1,0,-1,0,\mathrm{...},0,0\right)}^{T},\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{...},{\left(1,0,0,0,\mathrm{...},0,-1\right)}^{T}\}\end{array}$$

*is a maximal linearly independent subset of* ker(*J*_{k}).

*3. The set* {1_{k}} ∪ 𝒦_{J} is a maximal linearly independent subset of Eig(*J*_{k}).

*4*. 1_{k} ┴ ker(*J*_{n}).

For the next lemma we define the following sets:

$$\begin{array}{rl}{X}_{H}& =\{x\otimes y\otimes z:\text{\hspace{0.17em}}x\in \text{Eig}({L}_{H}),y\in \text{Ker}({J}_{k}),z\in \text{Eig}({J}_{k})\},\\ {X}_{V}& =\{x\otimes y\otimes z:\text{\hspace{0.17em}}x\in \text{Eig}({L}_{V}),y\in \text{Eig}({J}_{k}),z\in \text{Ker}({J}_{k})\},\\ {\stackrel{~}{X}}_{H}& =\{x\otimes y\otimes z:\text{\hspace{0.17em}}x\in \text{Eig}({L}_{H}),\text{\hspace{0.17em}}y,z\in \text{Ker}({J}_{k})\},\\ {\stackrel{~}{X}}_{V}& =\{x\otimes y\otimes z:\text{\hspace{0.17em}}x\in \text{Eig}({L}_{V}),\text{\hspace{0.17em}}y,z\in \text{Ker}({J}_{k})\}.\end{array}$$

#### Lemma 3.8

$$\begin{array}{r}\text{span}({X}_{H})\cap \text{span}({X}_{V})=\text{span}({\stackrel{~}{X}}_{H})=\text{span}({\stackrel{~}{X}}_{V}).\end{array}$$

*Proof*. Since ker(*J*_{k}) ⊆ Eig(*J*_{k}) it is obvious that

$$\begin{array}{r}\text{span}({\stackrel{~}{X}}_{H})\subseteq \text{span}({X}_{H})\cap \text{span}({X}_{V}).\end{array}$$

Conversely, note that both *L*_{H} and *L*_{V} are symmetric and therefore diagonizable, i.e. span(Eig(*L*_{H})) = $\text{span}(\text{Eig}({L}_{H}))=\text{span}(\text{Eig}({L}_{V}))={\mathbb{R}}^{{n}^{2}}.$ Using this we conclude

$$\begin{array}{rl}\text{span}({X}_{H})& \subseteq {\mathbb{R}}^{{n}^{2}}\otimes \text{span}(\text{Ker}({J}_{k}))\otimes \text{span}(\text{Eig}({J}_{k})),\\ \text{span}({X}_{V})& \subseteq {\mathbb{R}}^{{n}^{2}}\otimes \text{span}(\text{Eig}({J}_{k}))\otimes \text{span}(\text{Ker}({J}_{k})),\\ \text{span}({\stackrel{~}{X}}_{H})& ={\mathbb{R}}^{{n}^{2}}\otimes \text{span}(\text{Ker}({J}_{k}))\otimes \text{span}(\text{Ker}({J}_{k})).\end{array}$$

Once again employing the fact that ker(*J*_{k}) ⊆ Eig(*J*_{k}), we arrive at

$$\begin{array}{r}\text{span}({X}_{H})\cap \text{span}({X}_{V})\subseteq \text{span}({\stackrel{~}{X}}_{H}).\end{array}$$

#### Lemma 3.9

*Let x* ϵ ℝ^{n}^{2} *and y*, *z* ϵ ker(*J*_{k})*. Then x* ⊗ *y* ⊗ *z is an eigenvector of A*^{↑} corresponding to eigenvalue −1.

*Proof*.

$$\begin{array}{l}{A}^{\uparrow}\left(x\otimes y\otimes z\right)\text{\hspace{0.17em}}=({L}_{B}\otimes {J}_{{K}^{2}}+{L}_{H}\otimes {I}_{k}\otimes {J}_{k}+{L}_{V}\otimes {J}_{k}\otimes {I}_{k}\\ \text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}+{L}_{D}\otimes {J}_{{k}^{2}}-{L}_{D}\otimes {I}_{{k}^{2}})\left(x\otimes y\otimes z\right)\\ =\left(\left({L}_{B}x\right)\otimes \left({J}_{k}y\right)\otimes \left({J}_{k}z\right)\right)+\left(\left({L}_{H}x\right)\otimes \left({I}_{k}y\right)\otimes \left({J}_{k}z\right)\right)\\ \text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}+\left(\left({L}_{V}x\right)\otimes \left({J}_{k}y\right)\otimes \left({I}_{k}z\right)\right)+\left(\left({L}_{D}x\right)\otimes \left({J}_{k}y\right)\otimes \left({J}_{k}z\right)\right)\\ \text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}\text{\hspace{0.05em}}-\left(\left({L}_{D}x\right)\otimes \left({I}_{k}y\right)\otimes \left({I}_{k}z\right)\right)\\ =-\left(\left({I}_{{n}^{2}}x\right)\otimes \left({I}_{k}y\right)\otimes \left({I}_{k}z\right)\right)\\ =-\left(x\otimes y\otimes z\right).\end{array}$$

#### Lemma 3.10

*For any eigenvector x of the matrix k*^{2}*L*_{B} + *kL*_{H} + *kL*_{V} corresponding to eigenvalue λ, the vector x ⊗ ${1}_{{k}^{2}}$*is an eigenvector of A*^{↑} corresponding to eigenvalue λ + *k*^{2} − 1.

*Proof*.

$$\begin{array}{rl}& {A}^{\uparrow}\left(x\otimes {1}_{{k}^{2}}\right)\\ & =\left({L}_{B}\otimes B+{L}_{H}\otimes H+{L}_{V}\otimes V+{L}_{D}\otimes D\right)\left(x\otimes {1}_{{k}^{2}}\right)\\ & =\left(\left({L}_{B}x\right)\otimes \left({J}_{{k}^{2}}{1}_{{k}^{2}}\right)\right)+\left(\left({L}_{H}x\right)\otimes \left(H{1}_{{k}^{2}}\right)\right)+\left(\left({L}_{V}x\right)\otimes \left(V{1}_{{k}^{2}}\right)\right)\\ & +\left(\left({L}_{D}x\right)\otimes \left({J}_{{k}^{2}}{1}_{{k}^{2}}\right)\right)-\left(\left({L}_{D}x\right)\otimes \left({J}_{{k}^{2}}{1}_{{k}^{2}}\right)\right)\\ & =\left(\left({L}_{B}x\right)\otimes \left({k}^{2}{1}_{{k}^{2}}\right)\right)+\left(\left({L}_{H}x\right)\otimes \left(k{1}_{{k}^{2}}\right)\right)+\left(\left({L}_{V}x\right)\otimes \left(k{1}_{{k}^{2}}\right)\right)\\ & +\left(\left({L}_{D}x\right)\otimes \left({k}^{2}{1}_{{k}^{2}}\right)\right)-\left(\left({L}_{D}x\right)\otimes {1}_{{k}^{2}}\right)\\ & \left(\left({k}^{2}{L}_{B}x\right)\otimes \left({1}_{{k}^{2}}\right)+\left(\left(k{L}_{H}x\right)\otimes {1}_{{k}^{2}}\right)+\left(\left(k{L}_{V}x\right)\otimes {1}_{{k}^{2}}\right)\right.\\ & \left.+\left(\left({k}^{2}-1\right)\right)\left({L}_{D}x\right)\otimes 1{k}^{2}\right)\\ & =\left(\left({k}^{2}{L}_{B}+k{L}_{v}+k{L}_{v}\right)x\otimes {1}_{{k}^{2}}\right)+\left(\left({k}^{2}-1\right)x\otimes {1}_{{k}^{2}}\right)\\ & =\left(\lambda +{k}^{2}-1\right)\left(x\otimes {1}_{{k}^{2}}\right).\end{array}$$

Before we present the main result of this section we need one more technical lemma.

#### Lemma 3.11

*Let* $\{{Y}_{i}{\}}_{i=1}^{n}$*be a set of linearly independent vectors. Then, for any set* $\{{X}_{i}{\}}_{i=1}^{m}$*of nonzero vectors and any function*

$$\varphi :\{1,2,3,\dots ,n\}\to \{1,2,3,\dots ,m\},$$

*the set* $\{{X}_{\varphi (i)}\otimes {Y}_{i}{\}}_{i=1}^{n}$*is linearly independent*

*Proof*. Suppose otherwise that

$$\begin{array}{r}\sum _{i=1}^{n}{c}_{i}({X}_{\varphi (i)}\otimes {Y}_{i})=0\end{array}$$

for suitable numbers *c*_{i} ϵ ℝ. Let *X *_{ϕ}_{(i)} = (*x*_{1,ϕ(i)}, . . , *x*_{r}_{,ϕ(i)})^{T}. By the definition of the Kronecker product we have

$$\sum _{i=1}^{n}{c}_{i}\left({X}_{\varphi \left(i\right)}\otimes {Y}_{i}\right)}=\left(\begin{array}{l}{\displaystyle \sum _{i=1}^{n}{c}_{i}{x}_{1,\varphi \left(i\right)}{Y}_{i}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\vdots \\ {\displaystyle \sum _{i=1}^{n}{c}_{i}{x}_{r,\varphi \left(i\right)}{Y}_{i}}\end{array}\right)=0.$$

Thus, for each *j* we have

$$\begin{array}{r}\sum _{i=1}^{n}{c}_{i}{x}_{j,\varphi (i)}{Y}_{i}=0.\end{array}$$

Due to the linear independence of the vectors *Y*_{i} we see that *c*_{i}x_{j}_{,ϕ(i)} = 0 for every *j* = 1, . . , *r* and *i* = 1, . . , *n*. Now assume that *c*_{i}_{*}/ = 0 for some index *i*^{*}. Then *x*_{j}_{,}= 0 for all *j*, therefore *X* = 0. But this is impossible $\{{X}_{i}{\}}_{i=1}^{m}$_{ϕ}_{(i*) ϕ(i*)} since is a set of nonzero vectors.

For what follows, let 𝓑_{V}, 𝓑_{H} and 𝓑_{M} denote arbitrary maximal linearly independent subsets of Eig(*L*_{V})^{2} , Eig(*L*_{H}) and Eig(*k*^{2}*L*_{B} + *kL*_{V} + *kL*_{H}), respectively. Further, let ℰ = *{e*_{1}, . . , *e*_{n}_{2}*}* be the standard basis of ℝ^{n}, where *e*_{i} denotes the *i*-th unit vector.

#### Theorem 3.12

*Given a graph* FSud(*n*, *T*)*, define the sets*

$$\begin{array}{l}{\chi}_{V}=\left\{x\otimes {1}_{k}\otimes y:x\in {\mathcal{B}}_{V},y\in {\mathcal{K}}_{J}\right\},\\ {\chi}_{H}=\left\{x\otimes y\otimes {1}_{k}:x\in {\mathcal{B}}_{H},y\in {\mathcal{K}}_{J}\right\},\\ {\chi}_{E}=\left\{x\otimes y\otimes z:x\in \epsilon ,y,z\in {\mathcal{K}}_{J}\right\},\\ {\chi}_{M}=\left\{x\otimes {1}_{{k}^{2}}:x\in {\mathcal{B}}_{M}\right\}.\end{array}$$

*Then, their union* 𝒳_{V} ∪ 𝒳_{H} ∪ 𝒳_{E} ∪ 𝒳_{M} forms a maximal linearly independent subset of Eig(FSud^{↑k}(*n*, *T*)).

*Proof*. By construction and Lemma 3.1., each of the sets 𝒳_{V}, 𝒳_{H}, 𝒳_{E}, 𝒳_{M} in itself is linearly independent. Further, by construction and Proposition 3.7, the spans of these four sets are mutually disjoint (neglecting the null vector). Moreover, Lemmas 3.5, 3.6, 3.9 and 3.11 guarantee that the union contains only eigenvectors of FSud^{↑k}(*n*, *T*). Finally, note that

$$\left|{\mathcal{B}}_{V}\right|=\left|{\mathcal{B}}_{H}\right|=\left|\epsilon \right|={n}^{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left|{\mathcal{K}}_{j}\right|=k-1$$

so that

$$\left|{\chi}_{V}\right|=\left|{\chi}_{H}\right|=\left(k-1\right){n}^{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left|{\chi}_{E}\right|=\left|{\chi}_{H}\right|=\left(k-1\right){n}^{2},\text{\hspace{0.17em}}\left|{\chi}_{M}\right|={n}^{2}$$

and thus

$$\left|{\chi}_{V}\right|+\left|{\chi}_{H}\right|+\left|{\chi}_{E}\right|+\left|{\chi}_{M}\right|=2\left(k-1\right){n}^{2}+{\left(k-1\right)}^{2}{n}^{2}+{n}^{2}={k}^{2}{n}^{2}\square $$

Looking closer at Theorem 3.12, we see that if *L*_{V}, *L *_{H}, *k*^{2}*L*_{B} + *kL *_{H} + *kL*_{V} were all integral, then FSud^{↑k}(*n*, *T*) would be integral as well.

#### Corollary 3.13

*Under the conditions stated in Theorem 2.7 it follows that the blown up graph* FSud^{↑k}(*n*, *T*) *is integral for every k*

*Proof*. From the proof of Theorem 2.7 it follows that *L*_{V}, *L *_{H} and *L *_{B} are all integral and commute with each other, hence they are simultaneously diagonizable. Consequently, *k*^{2}*L*_{B} + *kL *_{H} + *kL*_{V} is integral and therefore so is FSud^{↑k}(*n*, *T*).

Owing to Theorem 3.12, we can use Lemmas 3.5, 3.6, 3.9 and 3.11 to establish the spectrum of a *k*-fold blow up from its original. Interestingly, we can predict explicitly that the largest eigenvalue stems from the set 𝒳_{M}:

#### Theorem 3.14

*Given a graph* FSud(*n*, *T*)*, let λ be the largest eigenvalue of the associated matrix k*^{2}*L*_{B} + *kL*_{V} + *kL *_{H}. Then λ + *k*^{2} − 1 *is the largest eigenvalue of* FSud^{↑k}(*n*, *T*).

*Proof*. For the purposes of this proof we renumber the vertices of FSud(*n*, *T*) such that we sequentially number the vertices with one block, then continue with the next block and so on. The order in which the vertices are numbered with a single block is arbitrary. With respect to this vertex order the matrix *L *_{B} assumes the form *I*_{n} J_{n} − *I*_{n}_{2} . Clearly, this matrix contains *J*_{n} − *I*_{n} as a principal submatrix. Consequently, the matrix *M* := *k*^{2}*L*_{B} + *kL*_{V} + *kL *_{H} contains the matrix *k*^{2}*J*_{n} − *k*^{2}*I*_{n} as a principal submatrix, the latter having maximum eigenvalue *k*^{2}(*n* − 1). By virtue of eigenvalue interlacing (see e.g. [19]) we conclude that *λ* > (*n* − 1)*k*^{2}. So, according to Lemma 3.10, the largest eigenvalue of FSud^{↑k}(*n*, *T*) originating from the set 𝒳_{M} is at least (*n* − 1)*k*^{2} + *k*^{2} − 1 = *nk*^{2} − 1. We will now show that the largest eigenvalues originating from 𝒳_{V}, 𝒳_{H} and 𝒳_{B} are smaller than this bound.

Since the largest eigenvalue of a matrix is bounded from above by the maximum row sum of the matrix, it is clear that the maximum eigenvalue both of *L *_{H} and *L*_{V} can be at most *n* − 1. Now recall from Proposition 3.7 that *k* is the maximum eigenvalue of *J*_{k}. Combining these findings, it now follows from Lemmas 3.5 and 3.6 that none of the eigenvalues associated with the vectors of the sets 𝒳_{V} and 𝒳_{H} exceeds (*n* − 1)*k* − 1, which is less than the lower bound given for 𝒳_{M}. Finally, Lemma 3.9 tells us that no positive eigenvalue of FSud^{↑k}(*n*, *T*) originates from 𝒳_{B}.

## Comments (0)

General note:By using the comment function on degruyter.com you agree to our Privacy Statement. A respectful treatment of one another is important to us. Therefore we would like to draw your attention to our House Rules.