Abstract Algebra

I assume you mean covariance matrix and I assume that you are familiar with the definition:

C = E[(X-u)(X-u)T]

where X is a random vector and u = E(X) is the mean

The definition of non-negative definite is:

xTCx ≥ 0 for any vector x Є R

So is xTE[(X-u)(X-u)T]x ≥ 0?

Then, from one of the covariance properties:

E[(xT(X-u))((X-u)Tx)] = E[xT([(X-u)(X-u)T]x)] = E[((xTI)x)] = E[xTx]

Finally, since we've already defined x to have only real values, xTx is therefore non-negative definite by definition.

Assuming we're not dealing with complex numbers, the domain is:

R = {x Є R | x >= 0}, or equivalently, R0+, or [0,∞]

All three of the above terms say the same thing, the domain is all the real numbers greater than or equal to zero.

In computer science, relational algebra is an offshoot of first-order logic and of algebra of sets concerned with operations over finitary relations, usually made more convenient to work with by identifying the components of a tuple by a name (called attribute) rather than by a numeric column index, which is what is called a relation in database terminology.

The main application of relational algebra is providing a theoretical foundation for relational databases, particularly query languages for such databases, chiefly among which is SQLRelational algebra received little attention outside of pure mathematics until the publication of E.F. Codd's relational model of data in 1970. Codd proposed such an algebra as a basis for database query languages. (See section Implementations.)

Both a named and a unnamed perspective are possible for relational algebra, depending on whether the tuples are endowed with component names or not. In the unnamed perspective, a tuple is simply a member of a Cartesian product. In the named perspective, tuples are functions from a finite set U of attributes (of the relation) to a domain of values (assumed distinct from U).[1] The relational algebras obtained from the two perspectives are equivalent.[2] The typical undergraduate textbooks present only the named perspective though,[3][4] and this article follows suit.

Relational algebra is essentially equivalent in expressive power to relational calculus (and thus first-order logic); this result is known as Codd's theorem. One must be careful to avoid a mismatch that may arise between the two languages because negation, applied to a formula of the calculus, constructs a formula that may be true on an infinite set of possible tuples, while the difference operator of relational algebra always returns a finite result. To overcome these difficulties, Codd restricted the operands of relational algebra to finite relations only and also proposed restricted support for negation (NOT) and disjunction (OR). Analogous restrictions are found in many other logic-based computer languages. Codd defined the term relational completeness to refer to a language that is complete with respect to first-order predicate calculus apart from the restrictions he proposed. In practice the restrictions have no adverse effect on the applicability of his relational algebra for database purposes.

Furthermore, computing various functions on a column, like the summing up its elements, is also not possible using the relational algebra introduced insofar. There are five aggregate functions that are included with most relational database systems. These operations are Sum, Count, Average, Maximum and Minimum. In relational algebra the aggregation operation over a schema (A1, A2, ... An) is written as follows:

G1, G2, ..., Gm g f1(A1'), f2(A2'), ..., fk(Ak') (r)

where each Aj', 1 ≤ j ≤ k, is one of the original attributes Ai, 1 ≤ i ≤ n.

The attributes preceding the g are grouping attributes, which function like a "group by" clause in SQL. Then there are an arbitrary number of aggregation functions applied to individual attributes. The operation is applied to an arbitrary relation r. The grouping attributes are optional, and if they are not supplied, the aggregation functions are applied across the entire relation to which the operation is applied.

Let's assume that we have a table named Account with three columns, namely Account_Number, Branch_Name and Balance. We wish to find the maximum balance of each branch. This is accomplished by Branch_NameGMax(Balance)(Account). To find the highest balance of all accounts regardless of branch, we could simply write GMax(Balance)(Account).

There is no relational algebra expression E(R) taking R as a variable argument which produces R+. This can be proved using the fact that, given a relational expression E for which it is claimed that E(R) = R+, where R is a variable, we can always find an instance r of R (and a corresponding domain d) such that E(r) ≠r+.[15]

SQL however officially supports such fixpoint queries since 1999, and it had vendor-specific extensions in this direction well before that.

xy + x'z + yz ≡ xy + x'z

De Morgan's laws:

NOT (P OR Q) ≡ (NOT P) AND (NOT Q)

NOT (P AND Q) ≡ (NOT P) OR (NOT Q)

AKA:

(P+Q)'≡P'Q'

(PQ)'≡P'+Q'

AKA:

¬(P U Q)≡¬P ∩ ¬Q

¬(P ∩ Q)≡¬P U ¬Q

P must be true if there is a proposition Q such that the truth of P follows from the truth of "if Pthen Q". In particular, when Q is taken to be a false formula, the law says that if P must be true whenever it implies the false, then P is true.

Every Boolean algebra is isomorphic to a field of sets.

Source is linked

Any group must have an identity element e. As it has order 3, it must have two other elements, a and b. Now, clearly, ab = e, for if ab = b, then a = e:

abb-1 = bb-1, so ae = e, or a = e.

This contradicts the givens, so ab != b. Similarly, ab != a, leaving only possibility: ab = e. Multiplying by a-1, b = a-1. So our group has three elements: e, a, a-1.

What is a2? It cannot be a, because that would imply a = e, a contradiction of the givens. Nor can it be e, because then a = a-1, and these were shown to be distinct. One possibility remains: a2 = a-1.

That means that a3 = e, and the powers of a are: a0 = e, a, a2 = a-1, a3 = e, a4 = a, etc. Thus, the cyclic group generated by a is given by: = {e, a, a-1}.

QED.

Let g be any element other than the identity. Consider , the subgroup generated by g. By Lagrange's Theorem, the order of is either 1 or 3. Which is it? contains at least two distinct elements (e and a). Therefore it has 3 elements, and so is the whole group. In other words, g generates the group.

QED

In fact here is the proof that any group of order p where p is a prime number is cyclic. It follow precisely from the proof given for order 3.

Let p be any prime number and let the order of a group G be p. We denote this as

|G|=p. We know G has more than one element, so let g be an element of the group and g is not the identity element in G. We also know contains more than one element and ||<|G|, so by Lagrange (as above)

|| divided |G|. Therefore || divides a prime p=|G| which tells us

||=G from which it follows that G is cyclic.

QED

Notation...Order of an element is the number of elements in the subgroup generated by that element. It is also the min n>1 such that gn =1 if such an n exists.

Well, I would say in the range of $1000.

Unknown caliber. Contact Ruger