Subspace Minimization

**syme.gabriel** · March 21st 2008, 03:50 PM

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

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

**CaptainBlack** · March 23rd 2008, 12:55 AM

Originally Posted by syme.gabriel

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

My only idea is to take any $x \in M$ and try to show something with the function $\varphi(t) = f(x^\ast + tx)$ , 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 $M$ .

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

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

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

RonL

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