Algebraic Methods
[20/45%]Recall that for a set one defines . The elements of this set are called vectors (or tuples, or points). If , then vectors can be added componentwise: . If , then a vector can be multiplied componentwise by a number : (this can also be done for , but it is not interesting).
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Among any tuples of length consisting of zeros and ones, there are several (a nonzero number of) tuples whose componentwise sum modulo two is the zero tuple.
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For any vectors there exist rational numbers , not all equal to zero, for which .
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The analog of theorem holds for the real numbers, the complex numbers, and the integers.
The tuples from Problems 7.1.1, are called linearly dependent — over and over , respectively. Linear independence is the negation of linear dependence. Linear (in)dependence of polynomials over and , respectively, is defined similarly.
Remark. Similar definitions can be given in a more general situation — this leads to the notions of a ring and of a module over it. An attempt to prove (and use!) an analog of the linear dependence theorem (Problems 7.1.1, ) leads to the notions of a field and of a linear space over it. (In the case of the integers, not all generalizations go through anymore.) Details can be found in a textbook on linear algebra.
The phrase $N$ elements'' (in particular, subsets) means pairwise distinct elements'' (in particular, subsets).
A linear subspace is a subset closed under vector addition and multiplication by rational numbers. A linear subspace is called -dimensional if there exist linearly independent vectors such that every vector is a linear combination of these vectors, i.e. there exist numbers for which . The number is called the dimension of the space .
Let be a family of subsets of the set .
If every subset in has an odd number of elements and the intersection of any two subsets in has an even number of elements, then .
Construct an example where this bound is attained.
If the intersection of any two subsets in has exactly elements and every subset in has more than elements, then .
If and the intersection of any two subsets in has exactly elements, then .
There exist subsets of a -element set such that each of them has an even number of elements and the intersection of any two of them has an even number of elements.
There cannot be more than subsets satisfying the conditions of part (1).
The distance between points of the space is defined by the formula
The largest number of points in with equal pairwise distances is .
Construct points in whose pairwise distances take only two distinct values.
For and points , , denote . If the pairwise distances between points are equal to , then the polynomials are linearly independent over .
If the pairwise distances between points in take only two distinct values, then .
For , denote
Among any pairwise intersecting -element subsets of a -element set, there are two subsets whose intersection has exactly or exactly elements.
Among any points in , there are two whose scalar product lies in .
For any , expand the product
where are variables. In each of the resulting monomials, perform the following operation: for each , if the monomial contains the factor , replace this factor with . Denote the resulting polynomial by . Prove that if the scalar product of no two vectors among is divisible by , then the polynomials are linearly independent over .
Specify polynomials whose linear combinations with rational coefficients yield every polynomial , .
Among any five-element subsets of a -element set, there are two subsets whose intersection has exactly elements.
The same for subsets.
The same for subsets.
It is impossible to color all five-element subsets of a -element set with colors so that any two five-element subsets intersecting in exactly two elements have different colors.
Cf.the remark in Problem 5.1.4. Here is an equivalent formulation. The vertices of the graph are all five-element subsets of a -element set. Its edges are the pairs of subsets intersecting in exactly two elements. Prove that this graph cannot be properly colored with colors.
For a prime and an integer , the number is divisible by if and only if is not divisible by .
Let be a prime and . Denote
Denote . For any , expand the brackets in the product , where are variables. In each of the resulting monomials, for each we replace by as long as possible. Denote the resulting polynomial by .
Prove that if no two vectors among have zero scalar product, then the polynomials are linearly independent over .
There exist and a bounded subset of that cannot be partitioned into nonempty parts of smaller diameter.
If is prime, then among any distinct subsets of an -element set, each of which has elements, there are two subsets whose intersection has a number of elements divisible by .
The same, except that an integer is given and is divisible by $t$'' is replaced by is congruent to modulo ''.
If the edge set of the graph is the union of the edge sets of edge-disjoint complete bipartite graphs, then .
Construct a family of bipartite graphs for which this bound is attained.
A more detailed and extended exposition can be found in [Mk, R1], and a more advanced one — in [BF].
If a matrix of size has all entries of absolute value at most , then .
A square matrix is called a Hadamard matrix if all its entries are equal to and , where is the order of the matrix and is the identity matrix.
Construct an Hadamard matrix for
;
;
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;
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Any two columns of a Hadamard matrix are orthogonal.
A matrix is a Hadamard matrix if and only if its entries are equal to and any two of its rows are orthogonal.
Hadamard matrices attain the upper bound in Hadamard's theorem 7.2.1. (Hadamard matrices got their name owing to this result.)
If there exists an Hadamard matrix and , then is divisible by .
Conjecture. An Hadamard matrix exists for every number divisible by .
The conjecture has not been proved even for some numbers less than , namely: for .
Solving this problem and the next one (7.2.5) requires the basic properties of quadratic residues; see [GIM, § 9], [Vi, § 5], [ZSS, Sec. 3.3].
For a prime number , denote (this is a sum of Legendre symbols).
Prove that does not depend on .
Find for each .
Solving this problem (along with the previous one, 7.2.4) requires the basic properties of quadratic residues; see [GIM, § 9], [Vi, § 5], [ZSS, Sec. 3.3].
Construct an Hadamard matrix for
if there exists an Hadamard matrix;
if there exist and Hadamard matrices;
, where is a prime number of the form ;
, where is a prime number of the form .
A Hadamard matrix is called normalized if its first row and first column consist entirely of ones.
Write out all normalized Hadamard matrices of orders .
The Hadamard property of a matrix is preserved under the following transformations:
multiplying a row or a column by ;
interchanging rows or columns.
Hadamard matrices obtained from each other by applying some number of transformations (1) and (2) are called equivalent.
Which of the matrices from Problem 7.2.6 are equivalent?
Every Hadamard matrix is equivalent to some normalized one.
The number of equivalence classes: for orders — , — , — , — , — , — more than a million.
Is it true that for any Hadamard matrices and the matrices and are equivalent? Here the tensor product is defined in the hint to Problem 7.2.5 (1, 2).
The Hadamard matrix built by means of the (Paley) construction from Problem 7.2.5 (4) is equivalent to the matrix .
Does there exist a Hadamard matrix that is not equivalent to the matrix ?