Four proofs of a theorem by Vámos

Let $\chi(M)$ be the characteristic set of the matroid $M$; that is, \[\chi(M)=\{\operatorname{char}(\mathbb{F})\colon M\ \text{is representable over the field}\ \mathbb{F}\},\] where $\operatorname{char}(\mathbb{F})$ is the characteristic of $\mathbb{F}$. It is well known that $0\in\chi(M)$ if and only if $\chi(M)$ contains infinitely many primes. One direction is due to Rado [4]. The following theorem contains the other direction.

Theorem 1. If $\chi(M)$ contains infinitely many primes, then it contains zero.

One unusual aspect of Theorem 1 is that there are four proofs of it in the literature. All of them are attractive, and they call upon different (but related) sets of tools: Zorn’s lemma, compactness in first-order logic, ultraproducts, and Gröbner bases. In this post I will review all four proofs.

To start with, let $r$ be $r(M)$, and assume that $E(M)=\{1,\ldots, n\}$. Let $X$ be the set of variables $\{x_{i,j}\}_{1\leq i \leq r,\ 1\leq j \leq n}$, and let $R$ be the polynomial ring $\mathbb{Z}[X]$. We will let $A$ be the $r\times n$ matrix where the entry in row $i$ and column $j$ is $x_{i,j}$. Each $r$-element subset of $E(M)$ corresponds to an $r\times r$ submatrix of $A$, and thereby to a homogeneous polynomial in $R$, namely the determinant of that submatrix. Let $\mathcal{B}$ be the set of polynomials that correspond to bases of $M$, and let $\mathcal{N}$ contain the polynomials that correspond to dependent $r$-element subsets in $M$. Thus a representation of $M$ is a ring homomorphism from $R$ to a field that sends every polynomial in $\mathcal{N}$, but no polynomial in $\mathcal{B}$, to zero.

The first proof of Theorem 1 is due to Peter Vámos. He published two articles in the proceedings entitled Möbius Algebras, produced by the University of Waterloo in 1971. The proof of Theorem 1 has sometimes been cited as coming from [6], but in fact the correct attribution is to [5]. His proof hinges on the following lemma. Recall that an ideal, $P$, is prime, if $ab\in P$ implies that $a\in P$ or $b\in P$.

Lemma 1. Let $I$ be an ideal in a commutative ring, $S$, and let $T$ be a multiplicatively closed set of $S$ that is disjoint from $I$. There exists a prime ideal, $P$, such that $I\subseteq P$, and $P\cap T=\emptyset$.

Proof. We consider the partial order of ideals in $S$ that contain $I$ and that are disjoint from $T$. Assume that $J_{1}\subset J_{2}\subset J_{3}\subset \cdots$ is a chain in the order. It is easy to confirm that the $J_{1}\cup J_{2}\cup J_{3}\cup \cdots$ is itself an ideal in the order. This means that every such chain of ideals has a maximal element, so we can apply Zorn’s Lemma. Therefore the partial order has a maximal element, $P$. It remains to show that $P$ is a prime ideal. Assume otherwise, so that the product $ab$ belongs to $P$, even though $a\notin P$ and $b\notin P$. The maximality of $P$ means that the ideal generated by $P$ and $a$ is not disjoint from $T$, so there is some element, $ca+p$ in $T\cap \langle P, a\rangle$. (Here $p\in P$, and $c$ is an element of $S$.) Similarly, there are elements $q\in P$ and $d\in S$ such that $db+q$ belongs to $T$. Since $T$ is multiplicatively closed, we see that $(ca+p)(db+q)=(cd)(ab)+(ca)q+(db)p+pq$ is in $T$. But this element is also in $P$, since $ab\in P$. Now we have a contradiction to the fact that $P$ and $T$ are disjoint. $\square$

Proof 1. Let $p_{1},p_{2},p_{3},\ldots$ be distinct primes such that $M$ is representable over a field of characteristic $p_{i}$ for each $i$. Let $\phi_{i}$ be a homomorphism from $R$ to a field of characteristic $p_{i}$ that takes all the polynomials in $\mathcal{N}$, and no polynomial in $\mathcal{B}$, to zero. Then the kernel, $\ker(\phi_{i})$, of $\phi_{i}$ is a prime ideal of $R$ containing $\mathcal{N}$, but disjoint from $\mathcal{B}$. Moreover, the integer $p_{i}$ is contained in $\ker(\phi_{i})$. In fact, the only integers in $\ker(\phi_{i})$ are the multiples of $p_{i}$, since if $\ker(\phi_{i})$ contains a non-multiple of $p_{i}$, then it contains the greatest common divisor of that non-multiple and $p_{i}$ (namely $1$). This would imply that $\ker(\phi_{i})$ is equal to $R$, which is impossible.

Assume that there is an element in both the ideal $\langle \mathcal{N}\rangle$ and the multiplicatively closed set generated by $\mathcal{B}\cup\mathbb{Z}$. Such an element is of the form $kf$, where $k$ is an integer, and $f$ is a product of polynomials from $\mathcal{B}$. Since $kf$ is in $\langle \mathcal{N}\rangle$, it is in each $\ker(\phi_{i})$, even though $f$ is not. Thus $k$ is in each ideal $\ker(\phi_{i})$. Thus $k$ is a multiple of each of the distinct primes $p_{1},p_{2},p_{3},\ldots$, and we have a contradiction. Now, by Lemma $1$, there is a prime ideal, $P$, of $R$ which contains $\langle \mathcal{N}\rangle$, and which avoids $\mathcal{B}\cup\mathbb{Z}$. Because $P$ is prime, the quotient ring, $R/P$, is an integral domain. Let $F$ be the field of fractions of $R/P$. The natural homomorphism from $R$ to $F$ gives us a representation of $M$. As $P$ does not contain any integer, it follows that $R/P$ has characteristic zero, and we are done. $\square$

The next proof of Theorem 1 makes use of the Compactness Theorem in first-order logic. It is due to Samuel Wagstaff and appeared two years after Vámos’s proof, in 1973 [7]. Wagstaff does not supply many details — perhaps the account in his PhD thesis (upon which the article based) is more explicit. But I believe that my version below captures his key ideas.

Proof 2. We construct a first-order language of vector spaces. We have a countable supply of variables, $x_{1},x_{2},x_{3},\ldots$, and unary predicates $\operatorname{Scal}$ and $\operatorname{Vec}$. In our interpretation, these predicates will indicate that a variable stands for a scalar or a vector. We also have constants, $0_{S}$, $1_{S}$, and $0_{V}$, which are interpreted as the identities of the vector space. The functions $+_{S}$, $\times_{S}$, $+_{V}$, and $\times_{V}$, are interpreted as the binary operations of the field of scalars, as addition of vectors, and as scalar multiplication. The only binary relation we need is equality. Now, in this language, we can express the axioms for fields and for vector spaces. To illustrate, the compatibility of scalar and vector multiplication would be expressed with the sentence \begin{multline*}\forall x_{1}\forall x_{2}\forall x_{3} (\operatorname{Scal}(x_{1})\land \operatorname{Scal}(x_{2})\land \operatorname{Vec}(x_{3}) \to\\ ((x_{1}\times_{S}x_{2})\times_{V}x_{3}=x_{1}\times_{V}(x_{2}\times_{V}x_{3}))).\end{multline*} We can assert the representability of $M$ by requiring that there exist $x_{1},x_{2},\ldots, x_{n}$ such that $\operatorname{Vec}(x_{i})$ holds for each $i$. For each subset $X\subseteq E(M)=\{1,\ldots, n\}$, we include a sentence asserting that the corresponding subset of $\{x_{1},x_{2},\ldots, x_{n}\}$ is linearly independent if $X$ is independent in $M$, and otherwise asserting that it is linearly dependent. Let us illustrate how this can be accomplished by writing the formula which asserts that $\{x_{1},x_{2},x_{3}\}$ is linearly independent. Let $y_{1}, y_{2}, y_{3}$ be variables that are not used elsewhere in our sentence. The formula we need is as follows: \begin{multline*}\forall y_{1}\forall y_{2}\forall y_{3} (\operatorname{Scal}(y_{1})\land \operatorname{Scal}(y_{2})\land \operatorname{Scal}(y_{3}) \land (y_{1}\ne 0_{S} \lor y_{2}\ne 0_{S} \lor y_{3}\ne 0_{S})\to\\ (y_{1}\times_{V}x_{1}) +_{V} (y_{2}\times_{V} x_{2}) +_{V} (y_{3}\times_{V}x_{3})\ne 0_{V}).\end{multline*} We simply perform this trick for each subset of $E(M)$.

Now we have a finite collection, $\mathcal{T}$, of sentences which can be satisfied if and only if $M$ is representable over a field. For each prime, $p$, let $S_{p}$ be the sentence asserting that the sum of $p$ copies of $1_{S}$ is equal to $0_{S}$. Thus $M$ is representable over a field of characteristic $p$ if and only if $\mathcal{T}\cup\{S_{p}\}$ can be satisfied. Assume that $M$ is not representable over a field with characteristic zero. Then the collection $\mathcal{T}\cup\{\neg S_{p}\colon p\ \text{is a prime}\}$ is inconsistent: it cannot be satisfied by any model. The Compactness Theorem tells us that there is a finite subcollection which is inconsistent. Therefore we can assume that $\mathcal{T}\cup\{\neg S_{p} \colon p\ \text{is a prime and}\ p\leq N\}$ has no model, for some integer $N$. This implies that there can be no representation of $M$ over a field with characteristic greater than $N$, so we have proved the contrapositive of Theorem 1. $\square$

The next proof appeared in 1989, and comes courtesy of Imre Leader [3]. I think it might be my favourite out of the four. It uses the notion of an ultraproduct.

Proof 3. Let $F_{1},F_{2},F_{3},\ldots$ be an infinite sequence of fields such that $M$ is representable over each $F_{i}$ and $\operatorname{char}(F_{i+1})>\operatorname{char}(F_{i})$. Let $\mathcal{U}$ be a non-principal ultrafilter of the positive integers. This means that $\mathcal{U}$ is a collection of subsets of positive integers, and $\mathcal{U}$ satisfies: (1) any superset of a member of $\mathcal{U}$ is also in $\mathcal{U}$; (2) if $U,V\in\mathcal{U}$, then $U\cap V\in \mathcal{U}$; and (3) if $U$ is any subset of $\mathbb{Z}^{+}$, then exactly one of $U$ or $\mathbb{Z}^{+}-U$ is in $\mathcal{U}$. Non-principal ultrafilters cannot be described explicitly, but we know that they exist by an application of Zorn’s Lemma. If $U$ is a finite subset of $\mathbb{Z}^{+}$, then $\mathbb{Z}^{+}-U$ is in $\mathcal{U}$. We consider sequences $(x_{1},x_{2},x_{3},\ldots)$, where each $x_{i}$ is an element of $F_{i}$. We declare $(x_{1},x_{2},x_{3},\ldots)$ and $(y_{1},y_{2},y_{3},\ldots)$ to be equivalent if $\{i\in\mathbb{Z}^{+}\colon x_{i}=y_{i}\}$ is a member of $\mathcal{U}$. We perform componentwise addition and multiplication on sequences. It is not too difficult to see that these operations respect our notion of equivalence, and that the equivalence classes of sequences form a field, $F$, under componentwise addition and multiplication. Moreover, the characteristic of $F$ is zero. We consider a sequence of representations of $M$ over the fields $F_{1},F_{2},F_{3},\ldots$. This leads to a representation of $M$ over $F$ in the obvious way: we simply “stack” the infinite sequence of matrices, and produce a matrix whose entries are infinite sequences of numbers. $\square$

Finally, we come full circle! In 2003, Vámos (along with Rosemary Baines) published another proof of Theorem 1 [2]. The proof is actually an easy consequence of the algorithm that Stefan mentioned in a previous post, and the main ingredient is the use of Gröbner bases. I will tweak the proof slightly, for the sake of simplicity.

Proof 4. Let $f$ be the product of all polynomials in $\mathcal{B}$. We introduce the new, “dummy”, variable, $y$. From now on, we will operate in the ring $\mathbb{Z}[X\cup\{y\}]$. Note that an evaluation of the variables in $X\cup\{y\}$ will make $yf-1$ zero if and only if it makes each polynomial in $\mathcal{B}$ non-zero. Let $I$ be the ideal of $\mathbb{Z}[X\cup\{y\}]$ that is generated by $\mathcal{N}\cup\{yf-1\}$. Thus a representation of $M$ is really a ring homomorphism from $\mathbb{Z}[X\cup\{y\}]$ to a field that takes every polynomial in $I$ to zero.

Let $G$ be a strong Gröbner basis of $I$ (see [1]). Thus $I$ is the ideal generated by $G$. In addition, we have an ordering on products of variables (the lexicographic order will suffice). The leading term of a polynomial is the highest non-zero term under this ordering. For $G$ to be a strong Gröbner basis, we require that if $h$ is in $I$, then there is a polynomial $g\in G$ such that the leading term of $g$ divides the leading term of $h$. Assume that $G$ contains an integer, $N$. If $\phi$ is a homomorphism from $\mathbb{Z}[X\cup\{y\}]$ to a field, $F$, such that $\ker(\phi)$ contains $I$, then $N$ is taken to zero by $\phi$. This means that the characteristic of $F$ is at most $N$, so $M$ is not representable over infinitely many prime characteristics. Therefore we will now assume that $G$ contains no integer.

Let $I_{\mathbb{Q}}$ be the ideal of $\mathbb{Q}[X\cup\{y\}]$ generated by $\mathcal{N}\cup\{yf-1\}$. Assume that $1\in I_{\mathbb{Q}}$. Then $1$ can be expressed as a polynomial combination of the polynomials in $\mathcal{N}\cup\{yf-1\}$. By multiplying this combination through by the denominators of the coefficients, we see that $I$ contains an integer, $N$. Thus $G$ contains a polynomial whose leading term divides the leading term of $N$, which is $N$ itself. But in the lexicographic ordering, a constant polynomial is below all non-constant polynomials, from which we deduce that $G$ contains a constant polynomial, contradicting our conclusion from the previous paragraph. Therefore $1\notin I_{\mathbb{Q}}$, which means that $I_{\mathbb{Q}}$ is contained in a maximal ideal, $P$. Note that $P$ contains no constant polynomial. Now $\mathbb{Q}[X\cup\{y\}]/P$ is a field, and the natural homomorphism from $\mathbb{Q}[X\cup\{y\}]$ to this field leads to a representation of $M$ over a field of characteristic zero. $\square$


[1] W. W. Adams and P. Loustaunau. An introduction to Gröbner bases. Graduate Studies in Mathematics, 3. American Mathematical Society, Providence, RI, 1994.

[2] R. Baines and P. Vámos. An algorithm to compute the set of characteristics of a system of polynomial equations over the integers. J. Symbolic Comput. 35 (2003), no. 3, 269–279.

[3] I. Leader. A short proof of a theorem of Vámos on matroid representations. Discrete Math. 75 (1989), no. 1-3, 315–317.

[4] R. Rado. Note on independence functions. Proc. London Math. Soc. (3) 7 (1957), 300–320.

[5] P. Vámos. A necessary and sufficient condition for a matroid to be linear. Möbius algebras (Proc. Conf., Univ. Waterloo, Waterloo, Ont., 1971), pp. 162–169. Univ. Waterloo, Waterloo, Ont., 1971.

[6] P. Vámos. Linearity of matroids over division rings. Notes by G. Roulet. Möbius algebras (Proc. Conf., Univ. Waterloo, Waterloo, Ont., 1971), pp. 170–174. Univ. Waterloo, Waterloo, Ont., 1971.

[7] S. S. Wagstaff Jr. Infinite matroids. Trans. Amer. Math. Soc. 175 (1973), 141–153.

Partial Fields, III. Universal partial fields

I’ve talked about partial fields before here and, earlier, here. A quick refresher:

Definition 1. partial field is a pair $\mathbb{P} = (R,G)$ of a commutative ring $R$ and a subgroup $G$ of the invertible elements of $R$ such that $-1 \in G$.

Definition 2. A matrix $A$ over $R$ is a (strong) $\mathbb{P}$-matrix if every square submatrix has a determinant in $G\cup \{0\}$.

Definition 3. An $r\times n$ matrix $A$ over $R$ is a weak $\mathbb{P}$-matrix if every $r\times r$ square submatrix has a determinant in $G\cup \{0\}$, and at least one such matrix has nonzero determinant.

One can check that, if $D$ is a submatrix with nonzero determinant as in the latter definition, then $D^{-1}A$ represents the same matroid, and is a strong $\mathbb{P}$-matrix. The following was Theorem 10 in Part I:

Proposition 4. If $A$ is a weak $\mathbb{P}$-matrix with columns labeled by a set $E$, then $\mathcal{B} = \{ B \subseteq E : |B| = r, \det(A[B]) \neq 0\}$ is the set of bases of a matroid $M$, denoted $M = M[A]$.

Today, we are interested in the following question: given a matroid $M$, can we find a partial field $\mathbb{P}_M$ that is a “best fit” for $M$? The qualities we are looking for are:

  • $M$ is representable over $\mathbb{P}_M$;
  • We can find (information about) other representations of $M$;
  • We can compute (with) $\mathbb{P}_M$;
  • The representation of $M$ over $\mathbb{P}_M$ is unique.

The fourth property turns out to be impractical, but we can get the first three. This post is based on Section 4 of [PvZ10]; see also Section 3.3 of my thesis [vZ09].

Constructing the universal partial field

Let $M$ be a rank-$r$ matroid with ground set $E$. Fix a basis $B$ of $M$. First, let $D^\#$ be the $B$-fundamental-circuit incidence-matrix (see [Oxl11, p.182]). Recall that any representation of $M$ of the form $[I\ D]$ will have the same zero/nonzero pattern in $D$ as in $D^\#$, and that we can scale rows and columns of $D$ to make certain nonzero entries equal to $1$. Let $T$ be the set of coordinates of a maximal set of coordinates that we can scale to be $1$ (see [Oxl11, Theorem 6.4.7] for details).

Next, for each $x \in B$ and $y \in E-B$ introduce a variable $a_{xy}$; also introduce variables $i_{B’}$ for every basis $B’$ of $M$. Let $\mathcal{Y}$ be the set of all these variables, and consider the ring of polynomials $\mathbb{Z}[\mathcal{Y}]$. Let $\hat D$ be the $B\times (E-B)$ matrix with entries $a_{xy}$, and let $\hat A = [I\ \hat D]$.

Now consider the ideal $I_{M,B,T}$ in $\mathbb{Z}[\mathcal{Y}]$ generated by the following relations:

  • $\det(\hat A[Z])$ for all $r$-subsets $Z\subseteq E$ that are nonbases of $M$;
  • $\det(\hat A[Z])i_Z – 1$ for all bases $Z$ of $M$;
  •  $a_{xy} – 1$ for all $xy \in T$.

Finally, we set $\mathbb{B}_M = \mathbb{Z}[\mathcal{Y}] / I_{M,B,T}$ and the partial field

$$\mathbb{P}_M = (\mathbb{B}_M, \langle \{-1\}\cup \{ i_{B’} : B’ \text{ basis of } M\}\rangle),$$

where $\langle\cdot\rangle$ denotes “multiplicative group generated by”. Note that this formalism is nothing other than introducing notation for the steps you would already take to produce a representation of a given abstract matroid. We have the following nice properties (which I won’t prove):

Proposition 5. Let $M$ be a matroid and $\mathbb{P}_M$, $\hat A$, etc. as above.

  1. The partial field $\mathbb{P}_M$ does not depend on the choice of $B$ or $T$.
  2. If $\mathbb{P}_M$ is not the trivial partial field (that is, if $1 \neq 0$), then $M$ is represented over $\mathbb{P}_M$ by the image $A$ of $\hat A$ under the obvious map from $\mathbb{Z}[\mathcal{Y}] \to \mathbb{Z}[\mathcal{Y}] / I_{M,B,T}$.
  3. If $M$ is represented over a (partial) field $\mathbb{P}$ by a matrix $A’$, then there is a partial field homomorphism $\phi:\mathbb{P}_M\to\mathbb{P}$ with $\phi(A) = A’$.

Here a partial field homomorphism $\phi: (R_1, G_1) \to (R_2, G_2)$ is a ring homomorphism $\phi:R_1 \to R_2$ such that $\phi(G_1) \subseteq G_2$. Such homomorphisms preserve matroids, i.e. $M[A] = M[\phi(A)]$ if $A$ is a $\mathbb{P}$-matrix. This third property earns $\mathbb{P}_M$ the name universal partial field: every representation of $M$ can be obtained from it!

Computation: Baines and Vámos and the set of characteristics

The set of characteristics of a matroid is the subset of $\{p: p = 0$ or $p$ is prime $\}$ such that $M$ has a representation over some field of that characteristic. It is known that the set of characteristics can be any finite subset not containing 0, and any infinite subset containing 0 (see [Oxl11, Section 6.8]). In [BV03], Baines and Vámos gave an algorithm to compute the set of characteristics from the ideal $I_{M,B,T}$ defined above. A brief summary is as follows:

  1. Compute a Gröbner basis $G$ over the integers for $I_{M,B,T}$.
  2. If the $G$ contains 1, then $M$ is not representable.
  3. If the $G$ contains a constant $k > 1$, then the prime factors of $k$ are exactly the characteristics over which $M$ is representable.
  4. Otherwise, let $\gamma$ be the least common multiple of the leading coefficients of the polynomials in $G$. Compute the Gröbner basis for the ideal generated by $G \cup \{\gamma\}$, and let $k$ be its constant member. Then $M$ is representable over characteristic 0 and all prime characteristics, except those dividing $\gamma$ but not $k$.

Note that Gröbner basis computations are notoriously difficult to compute and can take up huge amounts of memory. But for small examples this works well. The Gröbner basis generates the same ideal as the one we started with, and these computations often give a simpler representation of the partial field.


The universal partial field of $\text{PG}(2,q)$ is $\text{GF}(q)$. The non-Fano matroid and $P_8$ both have the dyadic partial field as their universal partial field, as do the ternary Dowling geometries. The Betsy Ross matroid can be shown to yield the Golden Ratio partial field (that is, it’s representable over $\text{GF}(4)$ and $\text{GF}(5)$), and the matroid represented by the following diagram is representable over the partial field $\mathbb{P}_4$, and is the source of Conjecture 15 in Part I. The matroid was obtained from Gordon Royle, out of his database of matroids with up to 9 elements, through a certain query regarding matroids with only 1 representation over $\text{GF}(5)$.


Related results and concepts

Many people have devoted attention to the theory of generic representations of a matroid. In [PvZ10] we discuss a construction of the universal partial field that does not rely on the choice of a special basis and scaling set. This construction is built on the idea of a bracket ring, introduced by White [Whi75], where we introduce a variable for each $r$-tuple of elements of $E$, followed by relations that make these $r$-tuples behave like determinants (alternating, 0 for repeated columns, 0 for nonbases, and relations encoding basis exchange). In addition to White’s construction we introduce symbols to make the bracket of each basis invertible.

Dress and Wenzel [DW89] introduce the Tutte Group, which again introduces a symbol for each basis, but only takes multiplicative relations into account. This group has received a considerable amount of attention. In particular the torsion of this group can give information on characteristics over which $M$ is not representable. Dress and Wenzel give a number of constructions of their group, based on various matroid axiomatizations.


I’ll conclude with some (computational) questions I’ve recently been asking myself.

  • Can we employ scaling techniques as in the definition of $\mathbb{P}_M$ to obtain a (quotient of the) Tutte-group that is faster to compute?
  • Can we combine computations of the Tutte-group with Gröbner basis techniques for a faster computation of the set of characteristics, or to determine whether $M$ is representable over any given finite field?

Unfortunately, $M$ is not always uniquely representable over $\mathbb{P}_M$ (up to partial field automorphisms, row- and column-scaling). The only obstacles I know involve partial fields $\mathbb{P}_M$ with an infinite set of cross ratios, such as the “near-regular-mod-2” partial field. Perhaps unique representability is recovered when the set of cross ratios is finite?


[BV03] R. Baines, P. Vámos, An algorithm to compute the set of characteristics of a system of polynomial equations over the integers. J. Symbolic Computat. 35, pp. 269-279 (2003).

[DW89] A.W.M. Dress, W. Wenzel, Geometric Algebra for Combinatorial Geometries. Adv. in Math. 77, pp. 1-36 (1989).

[Oxl11] J. Oxley, Matroid Theory. Second Edition. Oxford University Press (2011).

[PvZ10] R.A. Pendavingh, S.H.M. van Zwam, Confinement of matroid representations to subsets of partial fields. J. Comb. Th. Ser. B, vol. 100, pp. 510-545 (2010).

[Whi75] N.L. White. The bracket ring of a combinatorial geometry. I. Trans. Amer. Math. Soc. 214, pp. 233-248 (1975).

[vZ09] S.H.M. van Zwam, Partial Fields in Matroid TheoryPhD Thesis, Eindhoven University of Technology (2009).