\HHT linear functions h\\B#aaddcc

 
  \fts3\bLinear functions \  \_\_

  V and v denotes vectors space and its element on this page.\_\_

  \t#aaffddp4RD  
  Even if one consider functions from any set M to vector space V, this functions very
  natively allow to define operations \ArFrom_m_to_v.htm"  "+" and "." on them \ . The set of
  all this functions FF comprises even vector space itself. 
  This fact relates to how functions "interact with themselves."
  t\
  \_
  
  In this page, we will discuss how linear functions "act on vector spaces".
  Suppose we have two vector spaces V and V'.
  By definition, LL = { f | f: V --> V'; f-linear } 
  - set of all linear functions from V to V'.\_
  "f - linear" means that 
  for every a,b \^.be "  R, u,v \^.be "  V,
  f(au+bv) = af(u) + bf(v).
  \_\_
  
  Linear function reveal themselves even from very beginning of definition of vector space.
  For every number b, multiplication vector with this number give one linear function 
  f\[b  \ : V --> V; f\[b \ (v) = bv. If we choose b=0, then f will be 0-element in FF.\_
  Actually, LL is subspace in FF.\_\_

  For every f from LL and every subspace S, f(S) is subspace in V'. \_
  For every f from LL and every subspace S' from V' f\]-1 \ (S') is subspace in V.\_
  Subspace f\]-1 \ ({0}} we will call kernel or kern f. \_
  
  For every element v' belonging V' and every element v such as f(v)=v',\_
  f\]-1 \ (v') = (v united with kern f). So, any "preimage" of any vector v' has the 
  same card as kern f.\_\_

  \t#aaffddp4RD  
  This is interesting that all sets f\]-1 \ (v') comprise a partition of V into disjoint 
  subsets, and
  all such subsets 1-1 mapped into kern f. Let us call a set of this subsets a factor set and denote
  it V/V'. In other words, V/V' = { f\]-1 \ (v')| v' belongs V' }. f\]-1 \ (v') 1-1 maps V' into V/V'.
  Moreover, if to define H+G and aH for elements of V/V' using arbitrary vectors from H and G, then\_\_
  1) such definition does not depend on the choice of arbitraty vectiors; \_
  2) V/V' comprises a vector space; \_
  3) function f\]-1 \ (v') establishes an isomorphism between f(V) and V/V'.\_\_
  t\\_

  It is easy to see now that the following sentences are equivalent: \_\_
 
  kern f = {0} \_
  function f is 1-1 \_
  function f is isomorphism between V and f(V)\_
  there is an element v' that its preimage has only one element\_\_ 

  Because of vital importance of isomorphism between vector spaces, 
  let us to give defintion of this term. \b1-1 \ function I from vector space V \bon \
  vector space V' is an isomorphism 
  if I "preserves operations" "+" and ".". "Preserves" means
  that: \_\_

  
  1) if u,v,g belong V, and u + v = g, then I(u) + I(v) = I(g) \_
  2) if b is a number; u,v belong to V, and bv = u, then  I(bv) = I(u)
  1') if u',v',g' belong to V', and u' + v' = g', then J(u') + J(v') = J(g') \_
  2') if b is a number; u',v' belong to V', and bv' = u', then  J(bv') = J(u') \_\_

  where J is a map reversed to I. Such map exists because I is 1-1 function. Practically,
  isomorphism means that "applying" operations to elements V is "interchangable" with
  applying operations to elements V', or vector spaces V and V' completely "identical" as
  vector spaces. If one defined additional "structures" on V and V', V and V' may be not
  "identical" in respect to this additional structures. For example, V can be topological
  space, and V' can be set of continuous functions. From context, it must be clear
  which properties of V and V' does isomorphism identify.\_\_


 \_\_

 \Arvector_space.asp"    \fs1#ff4400  GO TO PREVIOUS a\ \-\-\-
 \Arlinear_algebra.asp"  \fs1#ff4400  GO TO TOP a\ \-\-\-
 \Arbasis.asp"           \fs1#ff4400  GO TO NEXT a\ \-\-\-

 \Arconsistence.asp"  \fs1#884400  LOGICAL STRICTNESS a\
 \-\-\-\-\-
 \Ardisclaimer.asp"  \fs1#884400  DISCLAIMER a\    


h\