Let $M$ be a subspace of $R^n$ and $z\notin M$. Show that the orthogonal proyection of $z$ in $M$ is $\bar x$ if and only if:

$(z-\bar x,x)=0, $ $ \forall x \in M$

How can i prove it? I know that for a convex, closed and not empty $C \subset R^n$, there is an only $\bar x \in C$ orthogonal projection of $z$ in $C$, that:

$(x-\bar x,\bar x-z)\geq 0, $ $ \forall x \in C$

But i dont know how can i used it for the proof.

  • $\begingroup$ Are you aware of the fact that z can be written as a sum of a vector in M and a vector in the orthogonal complement of M in a unique way? $\endgroup$ – Itamar Vigi Aug 16 at 18:42
  • $\begingroup$ No, because this is a proof by steps, and the orhogonal complement is defined later $\endgroup$ – JudeRyder Aug 16 at 18:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.