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Math Help - Block Matrices Positive Definiteness

  1. #1
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    Block Matrices Positive Definiteness

    Hello everyone,

    I got stuck in the following problem, any help would surely be appreciated:

    I have to prove that a block matrix of the form:

    \mathbf{R}=\left[\begin{array}{cc}\mathbf{B} & -\mathbf{C}\\\mathbf{C} & \mathbf{B}\end{array}\right]

    is positive definite, where \mathbf{B} is symmetric and positive definite, and \mathbf{C} is anti-symmetric, that is, \mathbf{C}^{T}=-\mathbf{C}.

    When I take the following nonzero vector:

    \mathbf{y}=\left[\begin{array}{c}<br />
\mathbf{y}_{1}\\<br />
\mathbf{y}_{2}\end{array}\right]

    and try to check the usual condition \mathbf{y}^{H}\mathbf{R}\mathbf{y}>0, I get the following:

    \mathbf{y}^{H}\mathbf{R}\mathbf{y}=\left[\begin{array}{cc}<br />
\mathbf{y}_{1}^{H} & \mathbf{y}_{2}^{H}\end{array}\right]\left[\begin{array}{cc}<br />
\mathbf{B} & \mathbf{C}^{T}\\<br />
\mathbf{C} & \mathbf{B}\end{array}\right]\left[\begin{array}{c}<br />
\mathbf{y}_{1}\\<br />
\mathbf{y}_{2}\end{array}\right]=

    =\mathbf{y}_{1}^{H}\mathbf{B}\mathbf{y}_{1}+\mathb  f{y}_{2}^{H}\mathbf{B}\mathbf{y}_{2}+\mathbf{y}_{1  }^{H}\mathbf{C}^{T}\mathbf{y}_{2}+\mathbf{y}_{2}^{  H}\mathbf{C}\mathbf{y}_{1}

    But:
    \mathbf{y}_{1}^{H}\mathbf{C}^{T}\mathbf{y}_{2}=(\m  athbf{y}_{2}^{H}\mathbf{C}\mathbf{y}_{1})^{H}

    And \mathbf{y}_{2}^{H}\mathbf{C}\mathbf{y}_{1} is a scalar, so:

    \mathbf{y}_{1}^{H}\mathbf{C}^{T}\mathbf{y}_{2}=(\m  athbf{y}_{2}^{H}\mathbf{C}\mathbf{y}_{1})^{*}

    Substituting above, we have:

    \mathbf{y}_{1}^{H}\mathbf{B}\mathbf{y}_{1}+\mathbf  {y}_{2}^{H}\mathbf{B}\mathbf{y}_{2}+(\mathbf{y}_{2  }^{H}\mathbf{C}\mathbf{y}_{1})^{*}+\mathbf{y}_{2}^  {H}\mathbf{C}\mathbf{y}_{1}=

    =\mathbf{y}_{1}^{H}\mathbf{B}\mathbf{y}_{1}+\mathb  f{y}_{2}^{H}\mathbf{B}\mathbf{y}_{2}+2\textrm{Re}\  {\mathbf{y}_{2}^{H}\mathbf{C}\mathbf{y}_{1}\}

    Now, how can I prove that the real number \mathbf{y}_{1}^{H}\mathbf{B}\mathbf{y}_{1}+\mathbf  {y}_{2}^{H}\mathbf{B}\mathbf{y}_{2}+2\textrm{Re}\{  \mathbf{y}_{2}^{H}\mathbf{C}\mathbf{y}_{1}\} is always positive?

    For sure, \mathbf{y}_{1}^{H}\mathbf{B}\mathbf{y}_{1} and \mathbf{y}_{2}^{H}\mathbf{B}\mathbf{y}_{2} are both positive, or at least one of them is positive and the other is zero, because \mathbf{B} is positive definite.

    Thanks,

    Henrique
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by thetouristbr View Post
    Hello everyone,

    I got stuck in the following problem, any help would surely be appreciated:

    I have to prove that a block matrix of the form:

    \mathbf{R}=\left[\begin{array}{cc}\mathbf{B} & -\mathbf{C}\\\mathbf{C} & \mathbf{B}\end{array}\right]

    is positive definite, where \mathbf{B} is symmetric and positive definite, and \mathbf{C} is anti-symmetric, that is, \mathbf{C}^{T}=-\mathbf{C}.

    When I take the following nonzero vector:

    \mathbf{y}=\left[\begin{array}{c}<br />
\mathbf{y}_{1}\\<br />
\mathbf{y}_{2}\end{array}\right]

    and try to check the usual condition \mathbf{y}^{H}\mathbf{R}\mathbf{y}>0, I get the following:

    \mathbf{y}^{H}\mathbf{R}\mathbf{y}=\left[\begin{array}{cc}<br />
\mathbf{y}_{1}^{H} & \mathbf{y}_{2}^{H}\end{array}\right]\left[\begin{array}{cc}<br />
\mathbf{B} & \mathbf{C}^{T}\\<br />
\mathbf{C} & \mathbf{B}\end{array}\right]\left[\begin{array}{c}<br />
\mathbf{y}_{1}\\<br />
\mathbf{y}_{2}\end{array}\right]=\ldots.
    That result is false. For example, take 22 matrices \mathbf{B} = \mathbf{0} and \mathbf{C} = \begin{bmatrix}0&1\\-1&0\end{bmatrix}. Let \mathbf{y}_{1} = \begin{bmatrix}1\\0\end{bmatrix} and \mathbf{y}_{2} = \begin{bmatrix}0\\1\end{bmatrix}. Then \mathbf{y}^{\textsc{h}}\mathbf{R}\mathbf{y} = -2.

    If you don't like taking \mathbf{B} = \mathbf{0} (because it's not strictly speaking positive definite) then you can replace it with a small positive multiple of the 22 identity matrix, and you will still find that \mathbf{y}^{\textsc{h}}\mathbf{R}\mathbf{y} is negative.
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  3. #3
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by Opalg View Post
    That result is false. For example, take 22 matrices \mathbf{B} = \mathbf{0} and \mathbf{C} = \begin{bmatrix}0&1\\-1&0\end{bmatrix}. Let \mathbf{y}_{1} = \begin{bmatrix}1\\0\end{bmatrix} and \mathbf{y}_{2} = \begin{bmatrix}0\\1\end{bmatrix}. Then \mathbf{y}^{\textsc{h}}\mathbf{R}\mathbf{y} = -2.

    If you don't like taking \mathbf{B} = \mathbf{0} (because it's not strictly speaking positive definite) then you can replace it with a small positive multiple of the 22 identity matrix, and you will still find that \mathbf{y}^{\textsc{h}}\mathbf{R}\mathbf{y} is negative.
    On the other hand, if the scalars are supposed to be real, and if \mathbf{R} = \begin{bmatrix}\mathbf{B}&\mathbf{C}\\\mathbf{C}&\  mathbf{B}\end{bmatrix} (without the minus sign in the top right corner), then you can check that the result holds. That is probably what was intended.
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