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Questions about the concept of subspace of linear transformation

  1. Oct 22, 2007 #1
    Hi all,
    I have some questions about the concept of subspace of linear transformation and its dimension, when I try to prove following problems:
    Prove T is a finite dimensional subspace of L(V) and U is a finite dimensional subspace of V, then
    T(U) = {F(u) | F is in T, u is in U} is a subspace of V,
    Dim(T(U))<=(dim(T))(dim(U))

    What does “T is a finite dimensional subspace of L(V)” mean?
    L1(v)+L2(v) = (L1+L2)(v)
    aL1(v) = (aL1)(v) ?

    and what is dim(T) and dim(T(U))?

    Every Linear transformation has its Matrix format, dim(T) is dim(M(T))?
    I am a fish in Linear algebra. Hope I explain my questions clearly.
     
  2. jcsd
  3. Oct 22, 2007 #2

    radou

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    You are a fish in definitions, obviously. First of all, I assume that by L(V) you mean the vector space of all linear operators from some finite dimensional space V to V?
     
  4. Oct 22, 2007 #3
    Yeah, L(V) means the vector space of all linear operators from finite dimensional space V to V.
     
  5. Oct 23, 2007 #4

    radou

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    OK, so, first of all, I assume this is the problem statement:

    Prove that, if T is a finite dimensional subspace of L(V) and U is a finite dimensional subspace of V, then
    (i) T(U) = {F(u) | F is in T, u is in U} is a subspace of V,
    (ii) Dim(T(U))<=(dim(T))(dim(U)) .

    In general, if A : V --> W is a linear operator, and L is a subspace of V, then A(L) is a subspace of W. Do you know how to prove that?
     
  6. Oct 23, 2007 #5

    HallsofIvy

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    If T is a subspace of V, then what do you mean by T(U)? If F is in T, then F is a vector in V. What do you mean by F(u)?
     
  7. Oct 23, 2007 #6

    radou

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    He said T was a subspace of L(V), not V.
     
  8. Oct 23, 2007 #7
    yeah, I know how to prove it.
    I just has an idea of what does "T is a finite dimensional subspace of L(V)" mean.
    Does It mean that T is a subset of one kind of operator and this subset must be closed under addition and scalar multiplication.
    for example, the non-invertible operator is not a subspace of L(V), cause it is not closed under addition. [tex]\left(\begin{array}{cc}1&0\\0&0\end{array}\right)[/tex] and [tex]\left(\begin{array}{cc}0&0\\0&1\end{array}\right)[/tex] they are both non-invertible operator but their addition [tex]\left(\begin{array}{cc}1&0\\0&1\end{array}\right)[/tex] is invertible.
     
  9. Oct 23, 2007 #8

    radou

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    Yes, exactly.

    For example, take some subspace M of V. We say that M is invariant under the operator A, if A(M) is in M. Now, for some fixed subspace M, take the set of all operators such that M is invariant for these operators. Then this set is a subspace of L(V).

    Edit: perhaps a more instructive example of a subspace of L(V) would be the space of all continuous linear operators from V to V.
     
    Last edited: Oct 23, 2007
  10. Oct 23, 2007 #9
    Thank you radou.
    I think I've already figured out how to prove the problem.
     
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