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Math Help - Linear Transformation - Questions reg Chr and Minimal Polynomial

  1. #1
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    Linear Transformation - Few quick questions

    Let V - finite dimensional vector space over field, F
    T,S be linear transformation from V into/onto V
    I be the identity tranformation

    Q1: m(x) is minimal polynomial of T
    p(x) is characteristic polynomial of T
    if q(x) is an irreducible polynomial over F and q(x)|p(x), prove that q(x)|m(x)
    as well.

    Q2: If ST=I => TS = I
    (basically I want to prove if a transformation is right or left invertible, it is invertible)

    Q3: T^n = 0 => T=0. I am not sure of this one, but couldn't find the reason/counter-example

    Any pointers would be welcome. I am just picking up the subject. So these might really be elementary question but will appreciate any help here.

    Thanks
    Last edited by aman_cc; September 18th 2009 at 01:24 AM. Reason: added more questions
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  2. #2
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    Quote Originally Posted by aman_cc View Post
    Q2: If ST=I => TS = I
    (basically I want to prove if a transformation is right or left invertible, it is invertible)
    Please validate if this argument is correct.
    ST = I => T(S(v)) = v for all v in V
    Thus T is onto and hence invertible. Let inverse of T be T'

    so TS = TS(TT') = T(ST)T' = TIT' = I (Hence Proved)

    Looks fine to me. Any better argument?
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  3. #3
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    Quote Originally Posted by aman_cc View Post
    Let V - finite dimensional vector space over field, F
    T,S be linear transformation from V into/onto V
    I be the identity tranformation

    Q1: m(x) is minimal polynomial of T
    p(x) is characteristic polynomial of T
    if q(x) is an irreducible polynomial over F and q(x)|p(x), prove that q(x)|m(x)
    as well.
    Hint: in general if \dim_F V = n, then p(x) \mid (m(x))^n.


    Q2: If ST=I => TS = I
    (basically I want to prove if a transformation is right or left invertible, it is invertible)
    S is 1-1 because it's onto. we also have STS(v)=S(v), for all v \in V, because ST=\text{id}. thus TS(v)=v because S is 1-1.


    Q3: T^n = 0 => T=0. I am not sure of this one, but couldn't find the reason/counter-example

    Any pointers would be welcome. I am just picking up the subject. So these might really be elementary question but will appreciate any help here.

    Thanks
    you forgot to mention what n is! whatever it is, the implication is false. for example, on V=\mathbb{R}^2, define T(x,y)=(y,0). then T^2=0 but T \neq 0.
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  4. #4
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    Thanks very much. I'll try your hint to Q1.

    Let T:V->V be a linear transformation. Am I correct in saying:
    T is onto IFF T is one-one

    Proof: Let v1,v2, v3...,vn be basis of V
    T is onto => T(v1), T(v2), ... T(vn) span V hence is a basis of V. So if T(v) = 0 => v = 0. Which further => T is one-one

    T is one-one. T(v1), T(v2)....T(vn) is basis for range of T which is a equal to/subspace of V. But any subspace of V with n basis = V itself hence T is onto.

    Is my reasoning correct?
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  5. #5
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    Quote Originally Posted by aman_cc View Post
    Thanks very much. I'll try your hint to Q1.

    Let T:V->V be a linear transformation. Am I correct in saying:
    T is onto IFF T is one-one

    Proof: Let v1,v2, v3...,vn be basis of V
    T is onto => T(v1), T(v2), ... T(vn) span V hence is a basis of V. So if T(v) = 0 => v = 0. Which further => T is one-one

    T is one-one. T(v1), T(v2)....T(vn) is basis for range of T which is a equal to/subspace of V. But any subspace of V with n basis = V itself hence T is onto.

    Is my reasoning correct?
    it's correct. you may also use the rank-nullity theorem for a faster proof.
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