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Thread: if det(aM + K) = det(bM + K) where a, b are scalars and M, K are matrices, is a = b?

  1. #1
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    if det(aM + K) = det(bM + K) where a, b are scalars and M, K are matrices, is a = b?

    hello,

    i have a problem. if i know that

    $\displaystyle
    \textnormal{det}(a \mathbf{M} + \mathbf{K}) = 0
    $

    $\displaystyle
    \textnormal{det}(b \mathbf{M} + \mathbf{K}) = 0
    $

    where $\displaystyle a$ and $\displaystyle b$ are scalars and $\displaystyle \mathbf{M}$ and $\displaystyle \mathbf{K}$ are matrices,
    does this mean that $\displaystyle a = b$?

    also, $\displaystyle \mathbf{M}$ is diagonal, if that helps.

    cheers for reading and sorry if this is a dumb question,
    jack.
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  2. #2
    MHF Contributor Swlabr's Avatar
    Joined
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    Quote Originally Posted by thepartymushroom View Post
    hello,

    i have a problem. if i know that

    $\displaystyle
    \textnormal{det}(a \mathbf{M} + \mathbf{K}) = 0
    $

    $\displaystyle
    \textnormal{det}(b \mathbf{M} + \mathbf{K}) = 0
    $

    where $\displaystyle a$ and $\displaystyle b$ are scalars and $\displaystyle \mathbf{M}$ and $\displaystyle \mathbf{K}$ are matrices,
    does this mean that $\displaystyle a = b$?

    also, $\displaystyle \mathbf{M}$ is diagonal, if that helps.

    cheers for reading and sorry if this is a dumb question,
    jack.
    Look at the 2x2 case. Assuming a=1 then we need to solve the equation
    (b-1)(n(p+bm+m)+sm) = 0 where p is the top left coordinate of the matrix K, s the bottom left. m, n are the coordinates of M. Assuming b is not 1 then we want to find n, m, b, p, s such that n(p+bm+m)+sm=0. So, let s=0=n.

    That is, $\displaystyle M =
    \left( \begin{array}{ccc}
    m & 0 \\
    0 & 0 \end{array} \right)$ and $\displaystyle K =
    \left( \begin{array}{ccc}
    p & q \\
    r & 0 \end{array} \right) $

    is a counter-example.
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  3. #3
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    Sep 2009
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    thanks, this helped get me thinking... what if the diagonal entries are positive? what if a and b are both negative? etc. seems like a bit of a long shot though.

    thanks again,
    jack.

    p.s. i assume you meant "s is the bottom right".
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