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Math Help - A square matrix A is called skew-symmetric

  1. #1
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    A square matrix A is called skew-symmetric

    A square matrix A is called skew-symmetric if A^T= A. Prove that if A is a skew-symmetric n x nmatrix and n is odd, then A is not inveritible.
    what if n is even ?

    Please help, I have no idea what is talking about.
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  2. #2
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    Quote Originally Posted by move123 View Post
    A square matrix A is called skew-symmetric if A^T= A. Prove that if A is a skew-symmetric n x nmatrix and n is odd, then A is not inveritible.
    what if n is even ?

    Please help, I have no idea what is talking about.
    Hello move123,

    First your definition is incorrect. A matrix is skew symmetric if

    A^T=-A

    First we know that a matrix and its transpose have the same determinant. This gives

    \det(A^T)=\det(A)

    Since the determinant function is multi-linear when we factor an number out of an n by n matrix it will have the power n e.g

    \det(cA)=c^n\det(A) when A is an n by n matrix.

    using this on the first equation gives

    \det(A)=\det(A^T)=\det(-A)=(-1)^n\det(A)

    This gives

    \det(A)-(-1)^n\det(A)=0 \iff [1+(-1)^{n+1}]\det(A)=0

    Now what can we say about the determinant when n is odd?
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by move123 View Post
    what if n is even ?

    It can be proved that if A=(a_{ij}) is a real skew symmetric matrix of even order then,


    \det (A)=(p_n(A))^n

    where p_n(A) is a polynomial of degree n/2 in the indeterminates a_{ij} . Besides, p_n(A) is determined up to a factor \pm1.

    For example:

    A=\begin{bmatrix}  \;\;0 &a\\-a & 0 \end{bmatrix}

    Then, \det (A)=a^2 where p_n(a)=a or p_n(a)=-a .
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    Quote Originally Posted by TheEmptySet View Post
    Hello move123,

    First your definition is incorrect. A matrix is skew symmetric if

    A^T=-A

    First we know that a matrix and its transpose have the same determinant. This gives

    \det(A^T)=\det(A)

    Since the determinant function is multi-linear when we factor an number out of an n by n matrix it will have the power n e.g

    \det(cA)=c^n\det(A) when A is an n by n matrix.

    using this on the first equation gives

    \det(A)=\det(A^T)=\det(-A)=(-1)^n\det(A)

    This gives

    \det(A)-(-1)^n\det(A)=0 \iff [1+(-1)^{n+1}]\det(A)=0

    Now what can we say about the determinant when n is odd?
    That's right. I was wrong for the A^T=-A
    So for n even. is it ivertible?
    thank you.
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by move123 View Post
    So for n even. is it ivertible?

    Hint

    \begin{bmatrix}0&0\\0&0\end{bmatrix},\quad \begin{bmatrix}\;\;0&1\\-1&0\end{bmatrix}

    are skew symmetric matrices.
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