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czwartek, 5 listopada 2009

Equivalence class and linear subspace

(Białynicki-Birula, Algebra liniowa z geometrią, str 55)

Assume that we have a linear space Z33 and let V be a set of three vectors: {(0,0,0),(1,0,1), (2,0,2)} .

Prove that:
1) V is a linear subspace of Z33
2) There are exactly 9 equivalence classes of V.

Z3 is a set {0, 1, 2} having operations defined in the set of integers modulo 3. Z3 is a field.

Z33 is a set of all 3-tuples (ordered) of Z3. Examples: (1,1,2), (0,1,2) etc. Z33 is a linear space over Z3.

Equivalence class:


It is very easy to prove 1).

Let's look at 2).

Equivalence class (Polish: warstwa) is a set of all vectors being the sum of any vector of Z33 minus V and all vectors belonging to V.
Consider the vector (1,0,0). The equivalence class defined by this vector (tuple) is: {(1,0,0), (2,0,1), (0,0,2)}.
Every equivalence class is defined by a vector (tuple). So what we have to do is to find the minimal set E (subset of Z33) that every vector of Z33 is the sum of one vector of E and one vector of V.

Let Z1 be a set of all vectors (a,b,c) where a != c. Forget for a moment about b.

Let V1 be a set {(0,0), (1,1), (2,2)}. It is clear that every vector (a,c) is the sum of one of the vector of V1 and (0,1) or (0,2).
For instance:
(2,0) = (2,2) + (0,1) (modulo 3 arithmetic) (0,2) = (0,0) + (0,2) etc.
So, with respect to b, we have a set E1 of six vectors = { (0,0,1),(0,0,2),(0,1,1),(0,1,2),(0,2,1),(0,2,2)} and it defines all vectors of Z1.

Let Z2 be a set of vectors (a,b,c) where a=c and vector does not belong to V . It is clear that a set E2 of three vector = { (0,1,0), (0,2,0), (2,0,2) } is enough to define every vector belonging to Z2.
For instance:
(2,1,2) = (2,0,2) + (0,1,0)
(1,0,1) = (1,0,1) + (2,0,2)
So set E = E1 + E2 (containing 9 vectors) defines Z1 + Z2.


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