The AMS has a new book out, Jiri Matousek, Thirty-three Miniatures: Mathematical and Algorithmic Applications of Linear Algebra. Info at http://www.ams.org/bookstore-getitem/item=STML-53

"This volume contains a collection of clever mathematical applications of linear algebra, mainly in combinatorics, geometry, and algorithms."

Hoffman and Singleton proved that a regular graph with girth 5 and diameter 2 has to have degree 2, 3, 7, or 57. If I recall correctly, the proof used spectral properties of the adjacency matrix to produce some non-polynomial equation for which these were the integer solutions.

There are unique examples of the first three cases: degree 2 is a pentagon, degree 3 is the Petersen graph, and degree 7 is the Hoffman-Singleton graph. The existence of the degree 57 graph is still open (as far as I know).

The Lindstrom-Gessel-Viennot Lemma uses the reflection principle on $S_n$ to say that the number of nonintersecting families of lattice paths in the plane equals the determinant of a matrix so that the $i,j$-th entry is the number of paths from the $i$th source to the $j$th sink.

This was not a linear algebra proof. However, this determinant can be used to enumerate plane partitions inside an $a\times b \times c~$ box, to $q$-enumerate plane partitions by weight, and to count domino tilings of an Aztec diamond. The resulting determinants can be manipulated and evaluated in ways which are natural in linear algebra, but not as clear on the objects, such as factoring the matrices. These enumerations can be viewed as applications of simple results in linear algebra.

Notes:

Lattice paths are defined and the sources and sinks are restricted so that any nonintersecting family must be an even permutation from source indices to sink indices, usually the identity.

Others independently discovered this result, e.g., Karlin and McGregor.

The same idea applies to Brownian motion.

Here is a link to a Tricki article that has some further examples.

http://www.tricki.org/article/Dimension_arguments_in_combinatorics

These references may be more shallow than you desired, but they are both fun and lucid.

1) Noga Alon's Tools From Higher Algebra contains many things (or at least references to those things) that only require linear algebra at heart, such as Rayleigh's Principle.

2) A Course in Combinatorics by van Lint and Wilson is laced with gems in self-contained sections, such that each page is an adventure. You'll find Lots of techniques here that only require linear algebra, including the awkward-looking "interlacing property" of eigenvalues that have popped up way too much for me to ignore by now.

My favorite is actually the aforementioned Babai/Frankl manuscript, which is still very readable and useful.

-Yan

This is a crosspost from Why linear algebra is fun!(or ?), suggested by Kevin O'Bryant. I think it's relevant here. Everything below is verbatim from the earlier post.

My favorite application of linear algebra, as introduced to me by Fan Chung, is Oddtown (which I learned about from a manuscript of Lovasz, but may not be due to him).

The $n$ residents of Oddtown love to form clubs; call the family of these $\mathcal{F}$. If $F_1$ and $F_2$ are in $\mathcal{F}$, then $|F_1|$ must be odd (this is Oddtown!) and $|F_1 \cap F_2|$ must be even unless $F_1 = F_2$ ($\scriptsize{go\;Oddtown?}$). The question is, how many clubs may these $n$ people form?

The answer (taken from Tibor Szabó's lecture notes) is this:

Let $\mathcal{F} = \{F_1,\ldots,F_m\} \subseteq 2^{[n]}$ be a set of clubs in Oddtown. Let $\mathbf{v}_i \in \{0,1\}^n$ be the characteristic vector of $F_i$; the $j$th coordinate is 1 iff $j \in F_i$.

Note that $\mathbf{v}_i^T \mathbf{v}_j = |F_i \cap F_j|$.

Now, $\mathbf{v}_1,\ldots,\mathbf{v}_m$ is independent over $\mathbb{F}^n_2$: if $\lambda_1\mathbf{v}_1 + \cdots + \lambda_m\mathbf{v}_m = 0$, then for each $i$ we have $$0 \;=\; (\lambda_1\mathbf{v}_1 + \cdots + \lambda_m\mathbf{v}_m)^T\mathbf{v}_i \;=\; \lambda_1\mathbf{v}_1^T\mathbf{v}_i + \cdots + \lambda_i\mathbf{v}_i^T\mathbf{v}_i + \ldots + \lambda_m\mathbf{v}_m^T\mathbf{v}_i \;=\; \lambda_i $$

Since $\mathbf{v}_1,\ldots,\mathbf{v}_m$ are linearly independent vectors over $\mathbb{F}^n_2$, $m \leq n$, and Oddtown can have at most $n$ clubs.

It's not quite what you have asked for, but very close:

Some facts - and proofs! - in combinatorics can be interpreted as linear algebra over the "field with one element". In this very nicely written article Henry Cohn gives a concrete meaning to this and shows how to make a proof from linear algebra into a proof about a combinatorical statement by rephrasing it into axiomatic projective geometry.

(by the way: Lior's answer is an instance of linear algebra over field with one element)

Here is an example I learned about this month: The edges of the complete graph cannot be partitioned into fewer than $n-1$ complete bipartite graphs. Apparently the only known proofs involve linear algebra.

The polynomial method in combinatorial incidence geometry relies crucially on linear algebra to locate a non-trivial polynomial of controlled degree that vanishes at a specified set of points. A good example of the method in action is Dvir's proof of the finite field Kakeya conjecture, see e.g. http://terrytao.wordpress.com/2008/03/24/dvirs-proof-of-the-finite-field-kakeya-conjecture/ .

There is nice linear-algebra proof of the following result in discrete geometry:

Any $n$ lines in general position cut from the plane at least $n-2$ triangles.

See А. Я. Белов Об одной задаче комбинаторной геометрии (thanks to Arseny and Garry). You will find there more examples of such problems.

The proof that every $n\times n$ semi-magic square can be written as an integer linear combination of $n^2-2n+2$ permutation matrices.

The following is a good illustration:

Let $P$ be a finite set ("points"), and let $L\subset 2^P$ ("lines") be such that distinct lines intersect in at most one point and any two distinct points are contained in a line. Let $V$ be the real vector space with basis $P$, $W$ the vector space with basis $L$. There are natural linear maps $T\colon V\to W$ and $S\colon W\to V$ mapping every point to the sum of the lines containing it, and every line to the sum of the points in it. Then $ST = J+D-I$ where $J$ is the all-ones matrix (through every two distinct points there is a unique line), $I$ the identity matrix and $D$ is diagonal with entries counting the lines through each point.

Assume that not all points are collinear. Then all the diagonal entries of $D-I$ are at least one; it is then easy to verify that the determinant of $ST$ is positive, and conclude that $|L| \geq |P|$.

Linear algebra is also useful for proving lower bounds in extremal "bootstrap percolation" type problems. For example, Alon and Kalai used linear algebra (actually, exterior algebra) to (independently) answer the following question, first considered by Bollobás:

What is the minimum number of edges in a graph $G$ on $n$ vertices such that the non-edges of $G$ can be added, one edge at a time, so that every edge completes a $k$-clique when it is added?

Some more recent applications of linear algebra in bootstrap percolation can be found in a paper of Balogh, Bollobás, Morris and Riordan. Specifically, they use linear algebra to prove Lemma 3, which is the main tool of the paper.

I can personally tell you that this lemma can be quite useful. It is used in two (joint) papers of mine: see here and here. In particular, in the second paper, we use this linear algebraic lemma to prove a conjecture of Balogh and Bollobás on $r$-neighbour bootstrap percolation in the hypercube.

Here's a surprising use of inner product spaces in algebraic combinatorics. For $S,T\subseteq[n-1]$, let $$ \beta(S,T) := \#\{w\in \mathfrak{S}_n\colon D(w)=S, D(w^{-1})=T\},$$ where $D(w)$ is the descent set of the permutation $w$. Set $f(n):=\mathrm{max}_{S,T\subseteq[n-1]}\beta(S,T)$.

Claim: There is some $S\subseteq[n-1]$ for which $f(n)=\beta(S,S)$.

Proof: It follows from the RSK correspondence and some basic theory of symmetric/quasi-symmetric functions that for any $S,T\subseteq[n-1]$, $$ \beta(S,T) = \langle s_{B_S}, s_{B_T}\rangle $$ where here $s_{B_S}$ is the (skew) Schur function associated to the border strip shape $B_S$ determined by the subset $S$, and $\langle \cdot , \cdot \rangle$ is the canonical inner product on the space of symmetric functions (where usual Schur functions give an orthonormal basis): see e.g. Stanley's Enumerative Combinatorics, Vol. 2, Corollary 7.23.8. So the Cauchy-Schwarz inequality for inner product spaces tells us that $$ \beta(S,T)^2 \leq \beta(S,S) \beta(T,T),$$ and thus one of $\beta(S,S)$ or $\beta(T,T)$ must be greater than or equal to $\beta(S,T)$, from which the claim immediately follows. $\square$

I must mention the critical lemma proving the sensitivity conjecture:

Every induced subgraph of the hypercube graph $Q_n$ with more than $2^{n-1}$ vertices has a vertex with degree at least $\sqrt n$.

Which is proved using some basic linear algebra here.

There is also a book in Russian, Линейно-алгебраический метод в комбинаторике by Raygorodsky, that deals with this.

http://www.ozon.ru/context/detail/id/3625051/

Variants of the EKR Theorem offer a wide class of examples. This page has a nice list going, by the way.

A friend of mine once made the outrageous claim -- but hear me out -- that most "linear algebra proofs" in combinatorics are not truly using linear algebra. I think he was getting at such arguments' use of a preferential basis (think positive or $\{0,1\}$-matrices) when linear algebra should, in its purest sense, be basis-independent. In the standard proof of Fisher's inequality, you set up and compute a determinant, get that some matrix has full rank, and then conclude the inequality. But there are no linear transformations (debateable). He conceded that, for instance, Perron-Frobenius belongs inside "matrix analysis", but not "linear algebra"!

I argued a little, but then eventually kind of saw his point. I guess I now really appreciate the obvious (but fortunate) fact that inner products count something!

Here are some examples where the dimension of a vector space of polynomials is used to solve a combinatorial problem.

Theorem 1 There are at most $n(n+1)/2$ equiangular lines in $\mathbb{R}^n$.

Proof. [Koornwinder] Let $L_1, \dots, L_m$ be $m$ lines passing through origin in $\mathbb{R}^n$ with angle $\arccos(\alpha)$ between every pair of them. Pick unit vectors $u_1, \dots, u_m$ on each line. Then we have $\langle u_i, u_j \rangle ^2 = \alpha^2$ for all $i \neq j$, and $\langle u_i, u_i \rangle = 1$ for all $i$. Define polynomials $P_1, \dots, P_m$ with $P_i(x) = \langle u_i, x \rangle^2 - \alpha^2 \langle x, x \rangle$. Then we have $P_i(u_j) = (1-\alpha^2)\delta_{i, j}$, and therefore, these $m$ polynomials are linearly independent (unless $\alpha^2 = 1$, but this case is easy since $\mathbb{R}^n$ has no more than $n$ mutually orthogonal nonzero vectors). The space of $n$-variable homogenous polynomials with degree at most $2$ has dimension ${n + 1\choose 2}$, and therefore $m \leq n(n+1)/2$.

Theorem 2 [Larman, Rogers, Seidel] A two-distance set in $\mathbb R^n$ has cardinality at most $(n+4)(n+1)/2$.

Proof. Let $u_1, \dots, u_m$ be $m$ points in $R^n$ and $a$, $b$ be the two non-zero real number such that $\|u_i - u_j\| \in \{a, b\}$. Define $P_i(x) = (\|u_i - x\|^2 - a^2)(\|u_i - x\|^2 - b^2)$. Then, $P_i(u_j) = a^2b^2\delta_{i, j}$. Therefore, the polynomials $P_i$'s are linearly independent. Moreover, these polynomials lie in the vector space spanned by polynomials of the type $$\left(\sum_{i = 1}^nx_i^2\right)^2, \left(\sum_{i = 1}^n x_i^2\right)x_j, x_ix_j, x_i, 1.$$ The number of such polynomials is $1 + n + n(n+1)/2 + n + 1 = (n+4)(n+1)/2$, which gives us the bound.

For more such examples see "Linear Algebra Methods in Combinatorics" by Babai and Frankl, linked in Stanley's answer.

The friendship theorem of Erdős, Rényi, and Sós is a classical example of a combinatorial result with a linear-algebraic proof: if a group of people has the property that every pair of people has exactly one friend in common, then there must be one person who is a friend to all the others. (There now exist combinatorial proofs, but the original question did not insist that no combinatorial proof exist.) More generally, one can find other examples of this type in books on spectral graph theory. I vaguely recall that there was another MO question about results in spectral graph theory that either have no combinatorial proof or that have a much more complicated combinatorial proof, but I can't find it.

Counting the number of even and odd sized subsets of a set $A$ of size $n$:

• $P(A)$ has a structure of vector space over $\mathbb F_2$ with the symmetric difference. Equivalently, one just takes the characteristic vectors in $\mathbb F_2^n$ with the usual addition.

• The even sized subsets form a subspace $W$.

• The odd sized is then the “affine” subspace $W+{a}$ for some $a\in A$.

• so both should have the size $\frac{2^n}{2}=2^{n-1}$.

(I find the other counting proofs (at least for even $n$) less satisfying).

Using the natural inner product, it follows quickly that the character table of a finite group in characteristic 0 is square - that is, there are as many conjugacy classes as irreducible representations.

One of the references I gave in the comments ( http://www.mathlinks.ro/viewtopic.php?t=290708 ) has disappeared from AoPS, so let me repost my answer there here. Sorry for my sophomoric writing (I was an undergrad back then) and for the sloppy LaTeX. Note that Theorem 2 (b) has since become Exercise 10.5.1 (b) in my Math 235 Fall 2023, Worksheet 10: Applications of linear algebra, and Theorem 2 (a) follows easily from part (b) (nice exercise!), but there is Theorem 2' in this answer, and anyway I am angry and want to make a point about bit rot on AoPS. So here it comes:

mildly edited posts from 2009 follow:

Problem (12.24 in the old Problems from the Book subforum).

Let $ A_1,A_2,\dots ,A_m$ and $ B_1,B_2,\dots ,B_p$ be subsets of $\left\{1,2,\dots,n\right\}$ such that $\left|A_i\cap B_j\right|$ is an odd number for all $i$ and $j$. Prove that $mp\le 2^{n-1}$.

Solution.

Let $ \mathbb{F}_2$ be the field with $ 2$ elements. We denote the zero and the one of the field $ \mathbb{F}_2$ by $ 0_{\mathbb{F}_2}$ and $ 1_{\mathbb{F}_2}$. Sometimes we will abbreviate $ 0_{\mathbb{F}_2}$ as $ 0$, since this cannot cause confusion.

Let $ \pi: \mathbb{Z}\to\mathbb{F}_2$ be the canonical surjection which maps every even integer to $ 0_{\mathbb{F}_2}$ and every odd integer to $ 1_{\mathbb{F}_2}$. Then, $ \pi\left(n\right) = n\cdot 1_{\mathbb{F}_2}$ for every $ n\in\mathbb{Z}$.

Let $ n\in\mathbb{N}$. For any vector $ v\in\mathbb{F}_2^n$, and any $ i\in\left\{1,2,...,n\right\}$, we denote by $ v_i$ the $ i$-th coordinate of the vector $ v$. Also, for any $ k\in\left\{1,2,...,n\right\}$, define a vector $ e_k\in\mathbb{F}_2^n$ by $ \left(e_k\right)_i = \left[i = k\right]\cdot 1_{\mathbb{F}_2}$ for every $ i\in\left\{1,2,...,n\right\}$, where $\left[i = k\right]$ denotes the truth value of $i = k$ (that is, the number $1$ if $i = k$ and the number $0$ otherwise). Then, $ \left(e_1,e_2,...,e_n\right)$ is a basis of the $ \mathbb{F}_2$-vector space $ \mathbb{F}_2^n$, and every vector $ v\in\mathbb{F}_2^n$ satisfies $ v = \sum_{i = 1}^n v_ie_i$.

Define a bilinear form $ \left < \cdot ,\cdot\right >$ on the $ \mathbb{F}_2$-vector space $ \mathbb{F}_2^n$ by $ \left < x,y\right > = x^Ty = \sum_{i = 1}^nx_iy_i$ for any $ x\in\mathbb{F}_2^n$ and $ y\in\mathbb{F}_2^n$. This bilinear form $ \left < \cdot ,\cdot\right >$ is symmetric and non-degenerate.

For any subspace $ U$ of the $ \mathbb{F}_2$-vector space $ \mathbb{F}_2^n$, define a subspace $ U^{\perp}$ of the $ \mathbb{F}_2$-vector space $ \mathbb{F}_2^n$ by

$ U^{\perp} = \left\{x\in\mathbb{F}_2^n\mid \left < x,u\right > = 0\text{ for all }u\in U\right\}$.

Then, according to a known fact, $ \dim U^{\perp} = n - \dim U$ (since the bilinear form $ \left < \cdot,\cdot\right >$ is non-degenerate).

We will use a famous fact:

Theorem 1. Define a map $ f: \mathcal{P}\left(\left\{1,2,...,n\right\}\right)\to\mathbb{F}_2^n$ by $ f\left(S\right) = \sum_{i\in S}e_i$ for every $ S\in\mathcal{P}\left(\left\{1,2,...,n\right\}\right)$.

(a) The map $ f: \mathcal{P}\left(\left\{1,2,...,n\right\}\right)\to\mathbb{F}_2^n$ is a bijection.

(b) For any two subsets $ A$ and $ B$ of $ \left\{1,2,...,n\right\}$, we have $ \pi\left(\left|A\cap B\right|\right) = \left < f\left(A\right),f\left(B\right)\right >$.

Now let's solve our problem and a bit more:

Theorem 2. Let $ U$ and $ V$ be two subsets of the set $ \mathcal{P}\left(\left\{1,2,...,n\right\}\right)$.

(a) If $ \left|A\cap B\right|$ is even for every $ A\in U$ and every $ B\in V$, then $ \left|U\right|\cdot\left|V\right|\leq 2^n$.

(b) If $ \left|A\cap B\right|$ is odd for every $ A\in U$ and every $ B\in V$, then $ \left|U\right|\cdot\left|V\right|\leq 2^{n - 1}$.

Proof of Theorem 2.

(a) Assume that $ \left|A\cap B\right|$ is even for every $ A\in U$ and every $ B\in V$. Then, for every $ A\in U$ and every $ B\in V$, we have

$ \left < f\left(A\right),f\left(B\right)\right > = \pi\left(\left|A\cap B\right|\right)$ (by Theorem 1 (b))

$ = 0$

(since $ \left|A\cap B\right|$ is even). Hence, for any $ v\in \operatorname{span}\left( f\left(U\right)\right )$ and for any $ w\in \operatorname{span}\left( f\left(V\right)\right )$, we have $ \left < v,w\right > = 0$. In other words, $ \operatorname{span}\left( f\left(U\right)\right ) \subseteq\left( \operatorname{span}\left( f\left(V\right)\right ) \right)^{\perp}$. Thus,

$ \dim \operatorname{span}\left( f\left(U\right)\right ) \leq\dim\left( \operatorname{span}\left( f\left(V\right)\right ) \right)^{\perp} = n - \dim \operatorname{span}\left( f\left(V\right)\right )$,

so that $ \dim \operatorname{span}\left( f\left(U\right)\right ) + \dim \operatorname{span}\left( f\left(V\right)\right ) \leq n$.

Now,

$ \left|U\right| = \left|f\left(U\right)\right|$ (since $ f$ is a bijection)

$ \leq\left| \operatorname{span}\left( f\left(U\right)\right ) \right|$ (since $ f\left(U\right)\subseteq \operatorname{span}\left( f\left(U\right)\right )$)

$ = \left|\mathbb{F}_2\right|^{\dim \operatorname{span}\left( f\left(U\right)\right ) }$ (since $ \operatorname{span}\left( f\left(U\right)\right )$ is a $ \mathbb{F}_2$-vector space)

$ = 2^{\dim \operatorname{span}\left( f\left(U\right)\right ) }$

and similarly $ \left|V\right|\leq 2^{\dim \operatorname{span}\left( f\left(V\right)\right ) }$.

Thus,

$ \left|U\right|\cdot\left|V\right|\leq 2^{\dim \operatorname{span}\left( f\left(U\right)\right ) }\cdot 2^{\dim \operatorname{span}\left( f\left(V\right)\right ) } = 2^{\dim \operatorname{span}\left( f\left(U\right)\right ) + \dim \operatorname{span}\left( f\left(V\right)\right ) }$

$ \leq 2^n$ (since $ \dim \operatorname{span}\left( f\left(U\right)\right ) + \dim \operatorname{span}\left( f\left(V\right)\right ) \leq n$).

Thus, Theorem 2 (a) is proven.

(b) If $ U = \emptyset$, then Theorem 2 (b) becomes trivial. Hence, we can assume that $ U\neq\emptyset$. Then, the set $ U$ has an element $ A_0$. Similarly, we can assume that $ V\neq\emptyset$. Then, the set $ V$ has an element $ B_0$.

Define a map $ \rho: \mathbb{F}_2^n\to\mathbb{F}_2^n$ by $ \rho\left(v\right) = v + f\left(A_0\right)$ for every $ v\in\mathbb{F}_2^n$. Obviously, the map $ \rho$ is bijective.

Define a map $ \sigma: \mathbb{F}_2^n\to\mathbb{F}_2^n$ by $ \sigma\left(v\right) = v + f\left(B_0\right)$ for every $ v\in\mathbb{F}_2^n$. Obviously, the map $ \sigma$ is bijective.

Assume that $ \left|A\cap B\right|$ is odd for every $ A\in U$ and every $ B\in V$. Then, for every $ A\in U$ and every $ B\in V$, we have

(1) $ \left < f\left(A\right),f\left(B\right)\right > = \pi\left(\left|A\cap B\right|\right)$ (by Theorem 1 (b))

$ = 1_{\mathbb{F}_2}$

(since $ \left|A\cap B\right|$ is odd). Applying this to $ A = A_0$, we obtain

(2) $ \left < f\left(A_0\right),f\left(B\right)\right > = 1_{\mathbb{F}_2}$.

Thus, for every $ A\in U$ and every $ B\in V$, we have

$ \left < \rho\left(f\left(A\right)\right),f\left(B\right)\right > = \left < f\left(A\right) + f\left(A_0\right),f\left(B\right)\right > = \underbrace{\left < f\left(A\right),f\left(B\right)\right > }_{ = 1_{\mathbb{F}_2}} + \underbrace{\left < f\left(A_0\right),f\left(B\right)\right > }_{ = 1_{\mathbb{F}_2}}$

$ = 1_{\mathbb{F}_2} + 1_{\mathbb{F}_2} = 0_{\mathbb{F}_2} = 0$.

Hence, for any $ v\in \operatorname{span}\left( \rho\left(f\left(U\right)\right)\right )$ and for any $ w\in \operatorname{span}\left( f\left(V\right)\right )$, we have $ \left < v,w\right > = 0$. In other words, $ \operatorname{span}\left( \rho\left(f\left(U\right)\right)\right ) \subseteq\left( \operatorname{span}\left( f\left(V\right)\right ) \right)^{\perp}$. Thus,

$ \dim \operatorname{span}\left( \rho\left(f\left(U\right)\right)\right ) \leq\dim\left( \operatorname{span}\left( f\left(V\right)\right ) \right)^{\perp} = n - \dim \operatorname{span}\left( f\left(V\right)\right )$,

so that $ \dim \operatorname{span}\left( \rho\left(f\left(U\right)\right)\right ) + \dim \operatorname{span}\left( f\left(V\right)\right ) \leq n$.

Now, $ \sigma\left(f\left(V\right)\right)\subseteq \operatorname{span}\left( f\left(V\right)\right )$ (because $ \sigma\left(v\right) = \underbrace{v}_{\in \operatorname{span}\left( f\left(V\right)\right ) } + \underbrace{f\left(B_0\right)}_{\in \operatorname{span}\left( f\left(V\right)\right ) }\in \operatorname{span}\left( f\left(V\right)\right )$ for every $ v\in f\left(V\right)$) and $ f\left(V\right)\subseteq \operatorname{span}\left( f\left(V\right)\right )$ (obviously). Thus, $ \sigma\left(f\left(V\right)\right)\cup f\left(V\right)\subseteq \operatorname{span}\left( f\left(V\right)\right )$.

Applying (2) to $ B = B_0$, we obtain

(3) $ \left < f\left(A_0\right),f\left(B_0\right)\right > = 1_{\mathbb{F}_2}$.

For every $ B\in V$, we have

$ \left < f\left(A_0\right),\sigma\left(f\left(B\right)\right)\right > = \left < f\left(A_0\right),f\left(B\right) + f\left(B_0\right)\right >$

$ = \left < f\left(A_0\right),f\left(B\right)\right > + \left < f\left(A_0\right),f\left(B_0\right)\right >$

$ = 1_{\mathbb{F}_2} + 1_{\mathbb{F}_2}$ (by (2) and (3))

$ = 0_{\mathbb{F}_2}\neq 1_{\mathbb{F}_2}$.

In other words, for every $ v\in \sigma\left(f\left(V\right)\right)$, we have

(4) $ \left < f\left(A_0\right),v\right > \neq 1_{\mathbb{F}_2}$.

On the other hand, for every $ v\in f\left(V\right)$, we have

(5) $ \left < f\left(A_0\right),v\right > = 1_{\mathbb{F}_2}$

(by (2)). Comparison of (4) and (5) shows that $ \sigma\left(f\left(V\right)\right)\cap f\left(V\right) = \emptyset$.

Now,

$ \left|U\right| = \left|\rho\left(f\left(U\right)\right)\right|$ (since $ \rho$ and $ f$ are bijections)

$ \leq\left| \operatorname{span}\left( \rho\left(f\left(U\right)\right)\right ) \right|$ (since $ \rho\left(f\left(U\right)\right)\subseteq \operatorname{span}\left( \rho\left(f\left(U\right)\right)\right )$)

$ = \left|\mathbb{F}_2\right|^{\dim \operatorname{span}\left( \rho\left(f\left(U\right)\right)\right ) }$ (since $ \operatorname{span}\left( \rho\left(f\left(U\right)\right)\right )$ is a $ \mathbb{F}_2$-vector space)

$ = 2^{\dim \operatorname{span}\left( \rho\left(f\left(U\right)\right)\right ) }$

and

$ 2\left|V\right| = 2\left|f\left(V\right)\right|$ (since $ \left|V\right| = \left|f\left(V\right)\right|$, since $ f$ is a bijection)

$ = \left|f\left(V\right)\right| + \left|f\left(V\right)\right| = \left|\sigma\left(f\left(V\right)\right)\right| + \left|f\left(V\right)\right|$ (since $ \left|f\left(V\right)\right| = \left|\sigma\left(f\left(V\right)\right)\right|$, since $ \sigma$ is a bijection)

$ = \left|\sigma\left(f\left(V\right)\right)\cup f\left(V\right)\right|$ (since $ \sigma\left(f\left(V\right)\right)\cap f\left(V\right) = \emptyset$)

$ \leq\left| \operatorname{span}\left( f\left(V\right)\right ) \right|$ (since $ \sigma\left(f\left(V\right)\right)\cup f\left(V\right)\subseteq \operatorname{span}\left( f\left(V\right)\right )$)

$ = \left|\mathbb{F}_2\right|^{\dim \operatorname{span}\left( f\left(V\right)\right ) }$ (since $ \operatorname{span}\left( f\left(V\right)\right )$ is a $ \mathbb{F}_2$-vector space)

$ = 2^{\dim \operatorname{span}\left( f\left(V\right)\right ) }$.

Hence,

$ 2\left|U\right|\left|V\right| = \left|U\right|\cdot 2\left|V\right|\leq 2^{\dim \operatorname{span}\left( \rho\left(f\left(U\right)\right)\right ) }\cdot 2^{\dim \operatorname{span}\left( f\left(V\right)\right ) }$

$ = 2^{\dim \operatorname{span}\left( \rho\left(f\left(U\right)\right)\right ) + \dim \operatorname{span}\left( f\left(V\right)\right ) }\leq 2^n$ (since $ \dim \operatorname{span}\left( \rho\left(f\left(U\right)\right)\right ) + \dim \operatorname{span}\left( f\left(V\right)\right ) \leq n$).

Division by $ 2$ transforms this into

$ \left|U\right|\left|V\right|\leq 2^{n - 1}$.

Thus, Theorem 2 (b) is proven. $\blacksquare$

Remark. In http://www.mathlinks.ro/viewtopic.php?t=281597 (now https://artofproblemsolving.com/community/h281597 ), Allnames asked for an elementary solution to this problem. This is not hard to do and not particularly interesting (you can rewrite every step of the above solution in sets and subsets; the worst linear algebra that it uses is $ \dim U^{\perp} = n - \dim U$, which, in the worst case, can be proven using Gaussian elimination - and that can be rewritten in terms of sets and subsets just like everything else in the proof). But the interesting question is whether there is a substantially different proof.

Addendum

Let's slightly sharpen the inequality:

Theorem 2'. Let $ U$ and $ V$ be two subsets of the set $ \mathcal{P}\left(\left\{1,2,...,n\right\}\right)$.

(a) If $ \left|A\cap B\right|$ is even for every $ A\in U$ and every $ B\in V$, then $ \left\lceil\log_2\left|U\right|\right\rceil + \left\lceil\log_2\left|V\right|\right\rceil\leq n$.

(b) If $ \left|A\cap B\right|$ is odd for every $ A\in U$ and every $ B\in V$, then $ \left\lceil\log_2\left|U\right|\right\rceil + \left\lceil\log_2\left|V\right|\right\rceil\leq n - 1$.

Proof of Theorem 2'.

(a) Continue at the end of the proof of Theorem 2 (a). The estimate $ \left|U\right|\leq 2^{\dim \operatorname{span}\left( f\left(U\right)\right ) }$ yields $ \log_2\left|U\right|\leq \dim \operatorname{span}\left( f\left(U\right)\right )$. Hence, $ \dim \operatorname{span}\left( f\left(U\right)\right )$ is an integer which is greater or equal to $ \log_2\left|U\right|$. But $ \left\lceil\log_2\left|U\right|\right\rceil$ is the smallest integer which is greater or equal to $ \log_2\left|U\right|$. Thus, $ \left\lceil\log_2\left|U\right|\right\rceil \leq\dim \operatorname{span}\left( f\left(U\right)\right )$. Similarly, $ \left\lceil\log_2\left|V\right|\right\rceil \leq\dim \operatorname{span}\left( f\left(V\right)\right )$. Thus,

$ \underbrace{\left\lceil\log_2\left|U\right|\right\rceil}_{\leq \dim \operatorname{span}\left( f\left(U\right)\right ) } + \underbrace{\left\lceil\log_2\left|V\right|\right\rceil}_{\leq \dim \operatorname{span}\left( f\left(V\right)\right ) }\leq \dim \operatorname{span}\left( f\left(U\right)\right ) + \dim \operatorname{span}\left( f\left(V\right)\right ) \leq n$.

Thus, Theorem 2' (a) is proven.

(b) Continue at the end of the proof of Theorem 2 (b). The estimate $ \left|U\right|\leq 2^{\dim \operatorname{span}\left( \rho\left(f\left(U\right)\right)\right ) }$ yields $ \log_2\left|U\right|\leq \dim \operatorname{span}\left( \rho\left(f\left(U\right)\right)\right )$. Hence, $ \dim \operatorname{span}\left( \rho\left(f\left(U\right)\right)\right )$ is an integer which is greater or equal to $ \log_2\left|U\right|$. But $ \left\lceil\log_2\left|U\right|\right\rceil$ is the smallest integer which is greater or equal to $ \log_2\left|U\right|$. Thus, $ \left\lceil\log_2\left|U\right|\right\rceil \leq\dim \operatorname{span}\left( \rho\left(f\left(U\right)\right)\right )$.

Also, $ 2\left|V\right|\leq 2^{\dim \operatorname{span}\left( f\left(V\right)\right ) }$ yields $ \left|V\right|\leq \frac12\cdot 2^{\dim \operatorname{span}\left( f\left(V\right)\right ) } = 2^{\dim \operatorname{span}\left( f\left(V\right)\right ) - 1}$, so that $ \log_2\left|V\right|\leq \dim \operatorname{span}\left( f\left(V\right)\right ) - 1$. Hence, $ \dim \operatorname{span}\left( f\left(V\right)\right ) - 1$ is an integer which is greater or equal to $ \log_2\left|V\right|$. But $ \left\lceil\log_2\left|V\right|\right\rceil$ is the smallest integer which is greater or equal to $ \log_2\left|V\right|$. Thus, $ \left\lceil\log_2\left|V\right|\right\rceil \leq\dim \operatorname{span}\left( f\left(V\right)\right ) - 1$.

Thus,

$ \underbrace{\left\lceil\log_2\left|U\right|\right\rceil}_{\leq \dim \operatorname{span}\left( \rho\left(f\left(U\right)\right)\right ) } + \underbrace{\left\lceil\log_2\left|V\right|\right\rceil}_{\leq \dim \operatorname{span}\left( f\left(V\right)\right ) - 1}\leq \underbrace{\dim \operatorname{span}\left( \rho\left(f\left(U\right)\right)\right ) + \dim \operatorname{span}\left( f\left(V\right)\right ) }_{\leq n} - 1\leq n - 1$.

Thus, Theorem 2' (b) is proven. $\blacksquare$

Note that Theorem 2' generalizes the problems PP and II in Pierre Bornsztein, Xavier Caruso, Des formes bilinéaires en combinatoire.