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I'm stuck in the following problem. I feel there may be an elegant solution that's avoiding me, as it's obvious choosing $a$ such that $x-ay$ orthogonal to y will minimize $x-ay$, but I'm getting stuck in the algebra. Here's the problem.

let $||\cdot||$ be a norm on $V$, derived from an inner product, let $x,y\in V$ and suppose $y\ne0$. Show that the scalar $a_0$ that minimizes $||x-ay||$ is $a_0=\langle x,y\rangle/||y||^2$, and that $x-a_0y$ and $y$ are orthogonal.

Once I have the first part, the second part is trivial to show.

I've tried expanding $\langle x-ay,x-ay\rangle$, but I haven't seen any obvious solution emerge, besides being rather tedious.

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Expanding $||x-ay||^2 = \langle x-ay, x-ay\rangle$ results in $$ ||x||^2 - 2a\langle x,y\rangle + a^2||y||^2, $$ which is a quadratic function of $a$. Since the coefficient of $a^2$ is positive, it has a minimum. The derivative with respect to $a$ is $$ 2a||y||^2 - 2\langle x,y\rangle, $$ which is zero when $a = \langle x,y\rangle/||y||^2$.

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  • $\begingroup$ Oh. I feel a bit silly then, thank you. $\endgroup$ Mar 29, 2019 at 19:56

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