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Math Help - Linear Algebra: Linear Transformations(1)

  1. #1
    Lord of certain Rings
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    Linear Algebra: Linear Transformations(1)

    Hello,

    I am teaching myself Linear Algebra following P.R.Halmos' "Finite Dimensional Vector Spaces".

    I am stuck at a couple of problems:
    -----------------------------------------------------------------------------------------------------------------------------

    Problem (1)


    Prove that corresponding to every linear transformation A on a finite dimensional vector space V, there exists an invertible linear transformation P such that APA = A.(Or equivalently prove that PA is a projection)

    -----------------------------------------------------------------------------------------------------------------------------

    Problem (2)

    If A,B,C are linear transformations on a finite dimensional vector space, then does (AB - BA)^2 commutes with C always?

    What happens when the dimension of the vector space is 2?

    -----------------------------------------------------------------------------------------------------------------------------

    Problem (3)

    From the basic definition of a determinant of a linear transformation(see below), prove that \text{det }(A \otimes B) = \text{det }(A) ^m \text{det }(B) ^n, where A and B are linear transformations on vector spaces of dimensions n and m respectively

    (I know the proof of a particular case when A and B are matrices, using eigenvalues. But I would love to see a proof using the below definition).

    -----------------------------------------------------------------------------------------------------------------------------

    Definition of Determinant of a linear transformation:

    If W is the space of all alternating n-linear forms on an n-dimensional vector space V,then we know that it is one-dimensional. Thus given a linear transformation A, we have a scalar \delta such that for every w in W, w(Ax_1,Ax_2,.....,Ax_n) = \delta w(x_1,x_2,....,x_n) \forall x_1,x_2,...x_n vectors in V. Then the det(A) is defined to be the scalar \delta.


    Thank you,
    Srikanth
    Last edited by Isomorphism; December 25th 2008 at 12:20 AM. Reason: Problem (3)
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  2. #2
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    Quote Originally Posted by Isomorphism View Post
    Hello,

    I am teaching myself Linear Algebra following P.R.Halmos' "Finite Dimensional Vector Spaces".

    I am stuck at a couple of problems:

    Problem (1)

    Prove that corresponding to every linear transformation A on a finite dimensional vector space V, there exists an invertible linear transformation P such that APA = A.(Or equivalently prove that PA is a projection)
    the proof of this is in the book and it's quite easy! see Theorem 3, page 94.



    Problem (2)

    If A,B,C are linear transformations on a finite dimensional vector space, then does (AB - BA)^2 commutes with C always?
    no. it's easier to work with matrices rather than transformations here. they're the same anyway: A=e_{12}, \ B=e_{21}, \ C=e_{13}. (here e_{ij} has 1 as its (i,j)-entry and 0 everywhere else.)



    What happens when the dimension of the vector space is 2?
    in this case, the claim is true: let AB-BA=X. clearly \text{tr}(X)=0. thus by Cayley Hamilton: X^2 = \alpha I, for some scalar \alpha. now it's obvious that X^2C=CX^2, for all C.
    Last edited by NonCommAlg; December 25th 2008 at 01:28 AM.
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  3. #3
    Lord of certain Rings
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    in this case, the claim is true: let clearly thus by Cayley Hamilton: for some scalar now it's obvious that for all C.
    Wow! This proof is fantastic .

    As of now, AB and BA are linear transformations and he has not defined trace or proved Cayley-Hamilton theorem. So probably there is a elementary proof of this result.But I know both the results for matrices.

    Nevertheless its a wonderful proof. Thank you NonCommAlg, you

    If you have any idea on how to do the third one, can you tell me?

    Thanks again,
    Srikanth
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  4. #4
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    here's a direct way for Problem 2, part 2: let \{e_1,e_2 \} be a basis for our vector space. let A(e_1)=ae_1 + be_2, \ A(e_2)=ce_1 + de_2, \ \ B(e_1)=a'e_1+b'e_2, \ B(e_2)=c'e_1 + d'e_2. see that:

    (AB-BA)(e_1)=re_2, \ \ (AB-BA)(e_2)=se_1, for some scalars r,s. thus: (AB-BA)^2(e_1)=rse_1, \ (AB-BA)^2(e_2)=rse_2. hence: (AB-BA)^2=rsI and the result follows.
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