Results 1 to 9 of 9

Math Help - proof of similar matrices

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    proof of similar matrices

    A is a 2x2 matrices
    A^2=-I
    and there is a transformation for which T^2=-I
    there is a basis B={v,T(v)} for which
    [T]_B=
    (0 1)
    (-1 0)

    prove that A is similar to
    (0 1)
    (-1 0)
    matrices.

    how i tried:

    i got T being represented by base B
    [T]_B=
    (0 1)
    (-1 0)

    i got T being represented by the stansdart base which is
    A matrices.
    what so A is similar to
    (0 1)
    (-1 0)
    ??

    bacause of what?
    is it correct?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21

    Re: proof of similar matrices

    Quote Originally Posted by transgalactic View Post
    A is a 2x2 matrices
    A^2=-I
    and there is a transformation for which T^2=-I
    there is a basis B={v,T(v)} for which
    [T]_B=
    (0 1)
    (-1 0)

    prove that A is similar to
    (0 1)
    (-1 0)
    matrices.

    how i tried:

    i got T being represented by base B
    [T]_B=
    (0 1)
    (-1 0)

    i got T being represented by the stansdart base which is
    A matrices.
    what so A is similar to
    (0 1)
    (-1 0)
    ??

    bacause of what?
    is it correct?
    Sadly, this question makes no sense. Is there some way you can reword it?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    44

    Re: proof of similar matrices

    Quote Originally Posted by Drexel28 View Post
    Sadly, this question makes no sense. Is there some way you can reword it?
    I agree. If B=\{v,T(v)\} is a basis of \mathbb{R}^{2\times 2} and the coordinates on B are expressed by rows, then [T]_B=\begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix} . So, it seems we are saying that A is similar to A .
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    Re: proof of similar matrices

    Quote Originally Posted by FernandoRevilla View Post
    I agree. If B=\{v,T(v)\} is a basis of \mathbb{R}^{2\times 2} and the coordinates on B are expressed by rows, then [T]_B=\begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix} . So, it seems we are saying that A is similar to A .
    this question is the third part of some bigger question in which i solved the first two parts.

    V is lenear space dimV=n n>1
    T:V->V is a transformation for wheach T^2=-I
    i prooved that {T(v),v} basis [T]_B=
    (0 1)
    (-1 0)

    in the third part i was asked:
    A is 2x2 matrices
    A^2=-I
    prove that A is similar to
    (0 1)
    (-1 0)

    i am used to prove that if B=P^-1AP then they are similar

    but here in the book they some thing liketwo representation in diffeerent basis
    so they are similar
    i cant understand this thing
    why is that?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    44

    Re: proof of similar matrices

    Quote Originally Posted by transgalactic View Post
    in the third part i was asked:
    A is 2x2 matrices A^2=-I prove that A is similar to
    (0 1)
    (-1 0)
    That is false. Choose for example A=\begin{bmatrix}{i}&{0}\\{0}&{i}\end{bmatrix} , we have A^2=-I. However \det A=-1 and \det \begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}=1 so, the matrices are not similar.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7

    Re: proof of similar matrices

    Quote Originally Posted by FernandoRevilla View Post
    That is false. Choose for example A=\begin{bmatrix}{i}&{0}\\{0}&{i}\end{bmatrix} , we have A^2=-I. However \det A=-1 and \det \begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}=1 so, the matrices are not similar.
    But the result is true for matrices with real entries. Let T be the linear transformation whose matrix (with respect to the standard basis) is A. Then T^2=-I, so T has no real eigenvalues and hence no (real) eigenvectors. It follows that if x is any nonzero vector then the vectors x and Tx are linearly independent. If B is the basis consisting of those two vectors then the matrix of T with respect to B is \bigl[{\scriptstyle{0\atop-1}\:{1\atop0}}\bigr]. So that matrix is similar to A.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    Re: proof of similar matrices

    why if A^2=-I then T^2=-I
    ?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    44

    Re: proof of similar matrices

    Quote Originally Posted by Opalg View Post
    But the result is true for matrices with real entries. Let T be the linear transformation whose matrix (with respect to the standard basis) is A. Then T^2=-I, so T has no real eigenvalues and hence no (real) eigenvectors. It follows that if x is any nonzero vector then the vectors x and Tx are linearly independent. If B is the basis consisting of those two vectors then the matrix of T with respect to B is \bigl[{\scriptstyle{0\atop-1}\:{1\atop0}}\bigr]. So that matrix is similar to A.
    Of course. These things happen when the formulation of the problem is not clear. For example in the answer #2 we already commented that for a basis B satisfying B=\{v,T(v)\} the matrix of T is \begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix} (row representation) or equivalently, for B'=\{v,-T(v)\} is \begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix} (column representation). Now in the third part of the problem it seems we have a 2\times 2 matrix without a reference to the field \mathbb{K} .
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    44

    Re: proof of similar matrices

    Quote Originally Posted by transgalactic View Post
    why if A^2=-I then T^2=-I ?
    If A is the matrix of T with respect to a determined basis B and we write the coordinates of the vectors in columns then, by a well known result, Y=AX where X are the coordinates of x\in V with respect to B and Y the coordinates of T(x)\in V with respect to B . Conclude.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Proof of two similar matrices have same eigenvalue and a question
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 3rd 2011, 03:58 PM
  2. Another similar matrices Q
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: July 31st 2011, 01:47 AM
  3. Similar matrices Q
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: July 30th 2011, 03:41 AM
  4. Similar matrices
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: December 12th 2009, 01:48 AM
  5. Similar Matrices
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: February 8th 2007, 11:24 AM

Search Tags


/mathhelpforum @mathhelpforum