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Math Help - Orthogonal matrix #2

  1. #1
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    Orthogonal matrix #2

    A is an orthogonal matrix, 0<a<1 is a scalar and (aA+(1-a)I) is also orthogonal. I need to find A.

    I'm trying to develop the expression (aA+(1-a)I)(aA+(1-a)I)^t = I, but that doesn't get me anywhere.
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  2. #2
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    Re: Orthogonal matrix #2

    Hmm...this is what I get:

    $(aA + (1-a)I)^T(aA + (1-a)I) = I$

    $(aA^T + (1-a)I)(aA + (1-a)I) = I$

    $a^2A^TA + (a - a^2)(A^T + A) + (1-a)^2I = I$

    $(a - a^2)(A^T + A) = I - a^2I - I + 2aI - a^2I = 2(a - a^2)I$.

    Since $0 < a < 1$, we have $a - a^2 \neq 0$, so that:

    $A^T + A = 2I$.

    What does this say about what the diagonal elements of $A$ must be?

    Finally, if $A_j$ is the $j$-th column of $A$, show that $\|A_j\| = 1$ (since $A$ is orthogonal). What does this force the off-diagonal elements to be?
    Last edited by Deveno; May 31st 2014 at 03:09 PM.
    Thanks from rebecca and LimpSpider
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  3. #3
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    Re: Orthogonal matrix #2

    Ones and zeroes, respectively; hence A = I.
    Thanks.
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