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Thread: Subspace Minimization

  1. #1
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    Question Subspace Minimization

    Let $\displaystyle f(x)$ be a $\displaystyle \mathcal{C}^1$ function and let $\displaystyle M$ be a subspace of $\displaystyle \mathbb{R}^n$. Suppose $\displaystyle x^\ast \in M$ minimizes $\displaystyle f(x)$ on $\displaystyle M$. Show $\displaystyle \nabla f(x^\ast) \in M^\perp$.

    My only idea is to take any $\displaystyle x \in M$ and try to show something with the function $\displaystyle \varphi(t) = f(x^\ast + tx)$, but I haven't been able to get anywhere with that. Any help is greatly appreciated!
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  2. #2
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    Quote Originally Posted by syme.gabriel View Post
    Let $\displaystyle f(x)$ be a $\displaystyle \mathcal{C}^1$ function and let $\displaystyle M$ be a subspace of $\displaystyle \mathbb{R}^n$. Suppose $\displaystyle x^\ast \in M$ minimizes $\displaystyle f(x)$ on $\displaystyle M$. Show $\displaystyle \nabla f(x^\ast) \in M^\perp$.

    My only idea is to take any $\displaystyle x \in M$ and try to show something with the function $\displaystyle \varphi(t) = f(x^\ast + tx)$, but I haven't been able to get anywhere with that. Any help is greatly appreciated!
    Suppose the gradient did not belong to the orthogonal complement of $\displaystyle M$.

    Then the gradient at $\displaystyle \bold{x}^*$ has a component $\displaystyle \bold{u}$ in $\displaystyle M$, so moving form $\displaystyle \bold{x}^*$ to $\displaystyle \bold{x}^*+\delta \bold{\hat{u}}$
    changes the value of the function to:

    $\displaystyle f(\bold{x}^*)+ \delta [\nabla f(\bold{x}^*).\bold{\hat{u}}]=f(\bold{x}^*)+ \delta |\bold{u}| $,

    so $\displaystyle \bold{x}^*$ will not be a mininmum in $\displaystyle M$.

    RonL
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