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Math Help - Adjoint relative to this Inner product... relative... what?

  1. #1
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    Adjoint relative to this Inner product... relative... what?

    I've been given this inner product and I am confused... because I don't know what my prof means by find the adjoint relative to this inner product... what the heck does he mean... relative to? Isn't the adjoint of this just itself? Am I reading this wrong, or did he word it horriby... or am i jsut stupid? (by the way I have the positive definate part... that was easy...)

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  2. #2
    Super Member Deadstar's Avatar
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    Right this is relative to what im doing now so ill maybe start things off...

    I think the matrix relative to the inner product is \left(\begin{array}{cc}2&1\\1&1\end{array}\right) with the row 1 being x_1, column 1 being y_1, etc.
    So the adjoint matrix A* would be \left(\begin{array}{cc}1&-1\\-1&2\end{array}\right).

    Yes? No? Im not at all sure about this...
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  3. #3
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    I believe the question is using this definition of the adjoint:

    Given a inner product space X,\langle\ \cdot\ ,\cdot\ \rangle and a linear transformation A:X\rightarrow X, we can define the adjoint of a matrix to be A^* such that \langle Ax, y \rangle=\langle x, A^*y\rangle. Then using the usual inner product in \mathbb{R}^2, we see that \langle Ax, y \rangle=y^{t}Ax=(Ay)^tx=\langle x, A^{t}y\rangle (where t stands for transpose) so relative to the usual inner product we see that A^*=A^t.

    So in the problem given, we instead have this new inner product which, as noted by Deadstar may be represented using the usual inner product as \langle x,y\rangle_{\rm new}=\langle Tx, y\rangle where T=\left(\begin{array}{cc}2&1\\1&1\end{array}\right  ).


    Then for a given A, we have

    \langle Ax,y\rangle_{\rm new}=\langle TAx, y\rangle=\langle x, A^{t}T^{t}y\rangle=\langle x, A^{t}Ty\rangle

    and since T is clearly invertible,
    A^tT=TT^{-1}AT, so

    \langle Ax,y\rangle_{\rm new}=\langle x, T(T^{-1}A^{t}T)y\rangle=\langle Tx, (T^{-1}A^{t}T)y\rangle=\langle x,T^{-1}A^{t}Ty\rangle_{\rm new} and thus we conclude that A^*=T^{-1}A^{t}T.
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