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Thread: 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 $\displaystyle \left(\begin{array}{cc}2&1\\1&1\end{array}\right)$ with the row 1 being $\displaystyle x_1$, column 1 being $\displaystyle y_1$, etc.
    So the adjoint matrix A* would be $\displaystyle \left(\begin{array}{cc}1&-1\\-1&2\end{array}\right)$.

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

    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 $\displaystyle \langle x,y\rangle_{\rm new}=\langle Tx, y\rangle$ where $\displaystyle T=\left(\begin{array}{cc}2&1\\1&1\end{array}\right )$.


    Then for a given A, we have

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

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

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