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Linear Algebra : Proving that Every map is an identity operator

  1. Nov 3, 2012 #1
    Suppose T belongs to L(V,V) where L(A,W) denotes the set of linear mappings from Vector spaces A to W, is such that every subspace of V with dimension dim V - 1 is invariant under T. Prove that T is a scalar multiple of the identity operator.

    My attempt : Let U be one of the sub spaces of V which has dimension = dim V -1.
    Let a basis to U be defined as (V1, ....,Vj-1,Vj+1,.......Vn) where j ranges from 1 to n

    Now, this basis can be extended to form a basis of V as (V1,......,Vn)
    => v = a1V1+....+aj-1Vj-1+aj+1Vj+1.......+anVn+ajVj
    =>Tv =a1T(V1) + ....... + an T(Vn) + ajT(Vj)
    ===================
    A

    now since T belongs to L(V,V), => T(Vi) belongs to V
    or A belongs to V -------------(1)

    However, by the problem statement:

    subspace of dimension dim V - 1 is invariant under T
    =>A belongs to U
    =>T(Vi) (i≠j) belongs to span ( Basis of U )
    Since, j ranges from 1 to n
    => T(Vi) = λiVi where λ is a scalar.

    Now, having proved that every basis vector is an eigen vector, I am stuck at trying to prove that T is a scalar multiple of the identity operator for any vector V
     
  2. jcsd
  3. Nov 3, 2012 #2

    micromass

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    I have moved this to homework.

    Here are some remarks. Some remarks are very minor, and some are more vital:

    First, using [itex]V_i[/itex] as elements of the basis is quite confusing. We don't want to confuse this with the space V!
    Furthermore, you seem to leave out the vector Vj. This indicates to me that j is a fixed number. Saying that j ranges from 1 to n makes no sense.
    For example: let j=2. Then your basis U is defined as (V1,V3,V4,...,Vn). Then you say that 2 ranges from 1 to n. This clearly makes little sense.
    Finally, I don't see why you just don't take a basis of U to be defined as (V1,...,V_(n-1)).

    These remarks are very minor of course, but I'm telling you anyway.

    What is v?? Why do we care about Tv?

    What is A? I can't really follow the rest of your proof until you say what A is. In the very beginning you say something about L(A,W). This seems to indicate that A is a space? How can a space belong to V?

    Isn't it obvious that T(vi) belongs to V as T is a linear map with codomain V? Is this what you meant?
     
  4. Nov 3, 2012 #3
    Hi
    I did not intend to portray j as a fixed number . but you are right regarding taking basis of U to be defined as (V_1...,V_(n-1)).
    I actually meant that if V has the basis as V1,.......Vn . then there can be C(N,1) subspaces which have dimension = dim V - 1.

    Sorry about not being able to use the subscripts. For some reason, my browser does not support them.

    v is a vector in the span of (V1,.......Vn). I thought Tv can be used to conclude some results because it is given that T maps from V to V

    I did actually mark A . I meant A to be the span of the basis vectors of a particular subspace U which is different from that given in the question. I should have marked it as something different. Lets take it as B.

    Thanks
     
  5. Nov 3, 2012 #4

    micromass

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    C(N,1)??

    Wait, so A is the span of the basis vectors of U?? Then isn't A=U??
    And what do you mean with "A belongs to U" or "A belongs to V" then?? You mean subset?
     
  6. Nov 4, 2012 #5
    So, there can be many subspaces with dimension = dim V - 1 . they equal =NC1 . dim V = N
    *where NCr = N!/(r!)(N-r)! .*

    A = T ( a1 V_1 +aj V_j-1 +aj+1 V_j+1...+an V_n)

    Since U is invariant under T, A has to belong to the subspace U.
     
    Last edited: Nov 4, 2012
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