Results 1 to 8 of 8

Math Help - clearification on Isomorphism question

  1. #1
    Senior Member
    Joined
    Jan 2008
    From
    Montreal
    Posts
    311
    Awards
    1

    clearification on Isomorphism question

    Let B be an n \times n invertible matrix. Define \Phi: M_{n \times n}(F) \ \mbox{by} \ \Phi (A) =B^{-1}AB Prove that \Phi is an isomorphism.

    From the answer provided, it shows:

    Let A =B^{-1}CB \Rightarrow \Phi A = \Phi B^{-1}CB \Rightarrow {\color{blue} B}(B^{-1}CB){\color{blue} B^{-1}} = C

    where do the values in blue come from?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by lllll View Post
    Let B be an n \times n invertible matrix. Define \Phi: M_{n \times n}(F) \ \mbox{by} \ \Phi (A) =B^{-1}AB Prove that \Phi is an isomorphism.

    From the answer provided, it shows:

    Let A =B^{-1}CB \Rightarrow \Phi A = \Phi B^{-1}CB \Rightarrow {\color{blue} B}(B^{-1}CB){\color{blue} B^{-1}} = C

    where do the values in blue come from?
    Hmm... I am stuck too...
    Can you write it completely? What are you trying to prove?

    Is \Phi (A) =B^{-1}AB or \Phi (A) =BAB^{-1}?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jan 2008
    From
    Montreal
    Posts
    311
    Awards
    1
    The first thing that was shown was \Phi was a linear

    \Phi was shown to be 1-1 by:

     \Phi (A) = 0 \therefore B^{-1}AB = 0 \Rightarrow N(\Phi) = 0

    \Phi was shown to be onto by:

    for any  C \in M_{n\times n}(\mathbb{R}) Find an A to show that \Phi (A) = C

    then A =B^{-1}CB \Rightarrow \Phi A = \Phi B^{-1}CB \Rightarrow {\color{blue} B}(B^{-1}CB){\color{blue} B^{-1}} = C

    That's everything I have shown.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by lllll View Post
    Let B be an n \times n invertible matrix. Define \Phi: M_{n \times n}(F) \ \mbox{by} \ \Phi (A) =B^{-1}AB Prove that \Phi is an isomorphism.

    From the answer provided, it shows:

    Let A =B^{-1}CB \Rightarrow \Phi A = \Phi B^{-1}CB \Rightarrow {\color{blue} B}(B^{-1}CB){\color{blue} B^{-1}} = C

    where do the values in blue come from?
    On second thought, I think I will just prove it my way. Probably the book has done something similar.

    Proving \Phi is a homomorphism is not hard. Proving 1-1 is even easier! So I dont understand why the textbook has done the mentioned manipulations...

    If \Phi(A) = \Phi(C) \Rightarrow B^{-1}AB = B^{-1}CB \Rightarrow A=C

    To prove homomorphism: Let W = XY, the let us prove \Phi(W) = \Phi(X)\Phi(Y).

    \Phi(W) = B^{-1}WB = B^{-1}(XY)B

     \Rightarrow B^{-1}(XY)B = B^{-1}(X{\color{blue}BB^{-1}}Y)B
    \Rightarrow(B^{-1}XB)(B^{-1}YB) = \Phi(X)\Phi(Y)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by lllll View Post
    The first thing that was shown was \Phi was a linear

    \Phi was shown to be 1-1 by:

     \Phi (A) = 0 \therefore B^{-1}AB = 0 \Rightarrow N(\Phi) = 0

    \Phi was shown to be onto by:

    for any  C \in M_{n\times n}(\mathbb{R}) Find an A to show that \Phi (A) = C

    then A =B^{-1}CB \Rightarrow \Phi A = \Phi B^{-1}CB \Rightarrow {\color{blue} B}(B^{-1}CB){\color{blue} B^{-1}} = C

    That's everything I have shown.
    I am sure they meant A =BCB^{-1}, its a typo
    OR they wanted to show \Phi (C) = A

    \Phi (A) = C \Rightarrow B^{-1}AB  = C \Rightarrow A  = BCB^{-1}
    Since C \in M_{n \times n} , \,det(C) \neq 0 and hence det(A) \neq 0, thus A \in M_{n \times n}

    So the map is onto.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Jan 2008
    From
    Montreal
    Posts
    311
    Awards
    1
    Is there a way that this can be shown without using determinants? Other then solving numerical problems, we haven't looked at theorems governing them.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by lllll View Post
    Is there a way that this can be shown without using determinants? Other then solving numerical problems, we haven't looked at theorems governing them.
    Well, than how will you establish that A is invertible? Without that you cannot be sure that A \in M_{n \times n}.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    There is another way to show the map is onto. A consequence of the rank nullity theorem says that if you have a one-to-one linear transformation f: V \mapsto U and the dimension of V and U are both equal and finite then the map is also onto. In this case V=U = M_{n\times n}(\mathbb{R}) which have the same finite dimension.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Clearification on Exponential Distribution
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: December 12th 2008, 12:38 PM
  2. clearification on Continuous Lemma
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 29th 2008, 03:00 PM
  3. clearification on probability
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: October 25th 2008, 01:49 PM
  4. clearification on series
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 12th 2008, 08:25 PM
  5. Clearification on Poisson Distribution
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: March 23rd 2008, 10:52 PM

Search Tags


/mathhelpforum @mathhelpforum