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Thread: proof of similar matrices

  1. #1
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    proof of similar matrices

    A is a 2x2 matrices
    $\displaystyle A^2=-I$
    and there is a transformation for which $\displaystyle T^2=-I$
    there is a basis B={v,T(v)} for which
    $\displaystyle [T]_B$=
    (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
    $\displaystyle [T]_B$=
    (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?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: proof of similar matrices

    Quote Originally Posted by transgalactic View Post
    A is a 2x2 matrices
    $\displaystyle A^2=-I$
    and there is a transformation for which $\displaystyle T^2=-I$
    there is a basis B={v,T(v)} for which
    $\displaystyle [T]_B$=
    (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
    $\displaystyle [T]_B$=
    (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?
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    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 $\displaystyle B=\{v,T(v)\}$ is a basis of $\displaystyle \mathbb{R}^{2\times 2}$ and the coordinates on $\displaystyle B$ are expressed by rows, then $\displaystyle [T]_B=\begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}$ . So, it seems we are saying that $\displaystyle A$ is similar to $\displaystyle A$ .
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  4. #4
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    Re: proof of similar matrices

    Quote Originally Posted by FernandoRevilla View Post
    I agree. If $\displaystyle B=\{v,T(v)\}$ is a basis of $\displaystyle \mathbb{R}^{2\times 2}$ and the coordinates on $\displaystyle B$ are expressed by rows, then $\displaystyle [T]_B=\begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}$ . So, it seems we are saying that $\displaystyle A$ is similar to $\displaystyle A$ .
    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 $\displaystyle T^2=-I$
    i prooved that {T(v),v} basis $\displaystyle [T]_B=$
    (0 1)
    (-1 0)

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

    i am used to prove that if $\displaystyle B=P^-1AP$ 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?
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Re: proof of similar matrices

    Quote Originally Posted by transgalactic View Post
    in the third part i was asked:
    A is 2x2 matrices $\displaystyle A^2=-I$ prove that A is similar to
    (0 1)
    (-1 0)
    That is false. Choose for example $\displaystyle A=\begin{bmatrix}{i}&{0}\\{0}&{i}\end{bmatrix}$ , we have $\displaystyle A^2=-I$. However $\displaystyle \det A=-1 $ and $\displaystyle \det \begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}=1$ so, the matrices are not similar.
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  6. #6
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    Re: proof of similar matrices

    Quote Originally Posted by FernandoRevilla View Post
    That is false. Choose for example $\displaystyle A=\begin{bmatrix}{i}&{0}\\{0}&{i}\end{bmatrix}$ , we have $\displaystyle A^2=-I$. However $\displaystyle \det A=-1 $ and $\displaystyle \det \begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}=1$ 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 $\displaystyle T^2=-I$, 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 $\displaystyle \bigl[{\scriptstyle{0\atop-1}\:{1\atop0}}\bigr].$ So that matrix is similar to A.
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  7. #7
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    Re: proof of similar matrices

    why if $\displaystyle A^2=-I$ then $\displaystyle T^2=-I$
    ?
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    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 $\displaystyle T^2=-I$, 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 $\displaystyle \bigl[{\scriptstyle{0\atop-1}\:{1\atop0}}\bigr].$ 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 $\displaystyle B$ satisfying $\displaystyle B=\{v,T(v)\}$ the matrix of $\displaystyle T$ is $\displaystyle \begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}$ (row representation) or equivalently, for $\displaystyle B'=\{v,-T(v)\}$ is $\displaystyle \begin{bmatrix}{\;\;0}&{1}\\{-1}&{0}\end{bmatrix}$ (column representation). Now in the third part of the problem it seems we have a $\displaystyle 2\times 2$ matrix without a reference to the field $\displaystyle \mathbb{K}$ .
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  9. #9
    MHF Contributor FernandoRevilla's Avatar
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    Re: proof of similar matrices

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