Conference rusure::math

1529.0. "question on linear mapping, how to find if 1:1?" by STAR::ABBASI () Mon Dec 09 1991 21:13

    hi ,
    
    for those who like functional analysis stuff, i can share this question with
    you.
    
    in my finals today we got this question. 
    
    Let V be some Hilbert space, let x1,x2 in V, be some two vectors. define
    a mapping T:V->V by Tx= <x,x1>x1 + <x,x2>x2
    (note that <.,.> is the inner product function, this is a function
     that takes VxV and returns real number., and has other 
     properties)
    
    a) show it is linear.
       ok , i showed that by showing that T(ax+by)= aTx+bTy, no problem
       here.
    b) is T continuouse? i struggled on this, i know the definition of
       continuouse, but iam not sure i convised the prof. with my argument.
    c) what is its NULL space?
       i knew that a NULL space is the set of vectors in V s.t. Tx->0
       so, any x vector that makes <x,x1>x1+<x,x2>x2 = 0 is in Null space.
       so i said <x,x1>x1 = - <x,x2>x2
       so here not knowing if x1,x2 are orthoginal or not (question did not
       say), i said the x is the vector that is orthogonal to both x1 and
       x2 ?
    d) under what conditions will T be 1:1 ?
       i knew that 1:1 means no two vectors in V will map to the same
       image point under T, but how to find if T is 1:1.
       i only knew if given Basis in V, i can build the Matrix
       representation of T induced on the the Basis, and can then try to
       find the determinant of [T] matrix and see if if singular or not.
       if det[T] = 0 , then it is not 1:1, if not zero, then it could be.
    
       any way, my question here to you guys, is how to show that a linear
       operator is 1:1 on some vector space, given T but we are not given the 
       Basis of the space? is there a *systematic* way of doing this?
       i have to show that no two vectors can have same image... humm
       may be i can assume there are two such vectors and show a
       contradiction, .. now i think about it..too late..
    
    c) what is the image of T, what is its rank, wnder what conditions will
       it be onto?
       i wrote some stuff herem hopefully somewhat convincing. 
       any way most of this questions is probably out of the window for me, 
       this one the easist question, so i feel doooommed, iam going to go 
       and cry a little and feel soory for myself now , hopefully i'll feel 
       better...i doubt it.
    
    than you.
    /nasser
       

What you missed is that you were given the basis: x1 and x2 are the basis by
construction.  Much of what you are looking for can be found by looking for
the implications of that fact.

Dick

    re -.1
    
    x1 and x2 are a basis for the image of T but V may be bigger.

    Rather than show a proof for 1-1, how about a hint?
    
    Consider 3-space with x1 not = x2.  Then you can form a basis 
    {x1, x2, x3}.  Note that the transformation T is then the coordinates
    of the vector x in the plane defined by x1 cross x2.  Therefore,  all
    vectors which share the same x1 and x2 coordinates will be mapped by
    T into the same vector!
    
    Think about this and you'll see what conditions are needed for T to be
    1-1.  
    
    Note:  In your proof, you'll have to consider a space with arbitrary
    dimension and also the case when x1 = scalar * x2.
    
    

    As for the image question, consider what vectors may be constructed by
    T  -  
    
    If x1 is not a scalar multiple of x2, then T defines the two
    dimensional space determined by the basis {x1,x2}.
    
    I.e.  The image is the x1,x2 plane.  
    
    Regarding the onto property, take a vector y which lies within the
    x1,x2 plane ie y = a x1 + b x2 and put it through the transformation T.
    What do you get?
    

                       <<< Note 1529.0 by STAR::ABBASI >>>
              -< question on linear mapping, how to find if 1:1? >-

>>    b) is T continuous? i struggled on this, i know the definition of
>>       continuous, but iam not sure i convinced the prof. with my argument.

It's easy to show that i) T is bounded and ii) any bounded linear mapping is
continuous.

    ref .5
    oh, i think that is what he wanted, he talked very little about
    bounded operators at the end of the course, so i did not study it 
    (also for lack of time), and so i did not know about this fact.
    
    ref all, thanks for input, i'll go over it , but if i do manage to get
    a B or better in that course iam gonna invite you guys to a pizza
    with all the toppings to celebrate. deal ?
    
    /nasser