Let $u_1$ and $u_2$ be orthonormal vectors in an inner product space $U=(U,\langle\cdot,\cdot\rangle)$ over $F$ and let $a$ be in $F$. Show that the linear transformation $P_u = \langle u,u_1 + au_2\rangle u_1$ is a projection but not orthogonal unless $a = 0$.

I know that orthonormal implies the inner product of u1 and u2 is zero, and that each has unit length. I need to show P^2 = P, and I know this should be trivial but I don't understand properties of the inner product. I have [u + u1,u2] = [u,u2] + [u1,u2] and other relations from the definition of inner product, but I can't seem to string them together to show Pu = P(Pu).

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    $\begingroup$ The "conventional" symbols here are written in $\LaTeX$. $\endgroup$ – Sujaan Kunalan Oct 22 '13 at 16:05
  • $\begingroup$ Thank you, I don't know how to use latex. I'll learn before I post again! $\endgroup$ – Protagger Oct 22 '13 at 17:14

It's a projection because $P^2u = Pu$: you can check this easily: just apply the formula for $Pu$ twice. Namely:

$$ P^2 u = P(\langle u, u_1 + au_2\rangle u_1) = \langle u, u_1 + au_2 \rangle P(u_1) = \dots $$

And you can continue from here easily (oh, you should prove indeed that $P$ is linear in order to justify the second equality, of course). :-)

As for being orthogonal, this means that $u - Pu$ should be orthogonal to $u_1$ for all $u$. So let's impose this:

$$ 0 = \langle u-Pu , u_1 \rangle = \langle u , u_1 \rangle - \langle Pu, u_1 \rangle = \langle u , u_1 \rangle - \langle u,u_1 \rangle \langle u_1, u_1 \rangle - a\langle u, u_2 \rangle \langle u_1, u_1\rangle \ . $$

Now, use the fact that vector $u_1$ is unitary and that this equality must hold for all $u$. (You don't need $u_1, u_2$ to be orthonormal: just $u_1$ unitary and $u_2 \neq 0$.)

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  • $\begingroup$ Thank you for this answer, I see the approach for orthogonality now. I think I am having difficulties with properties of the inner product. Would you mind showing why P^2 = P? $\endgroup$ – Protagger Oct 22 '13 at 16:49
  • $\begingroup$ I've added some hints to my answer. $\endgroup$ – Agustí Roig Oct 22 '13 at 16:55
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    $\begingroup$ Aha! I was not thinking about the fact that the inner product results in some scalar. Thank you so much I can get it now. $\endgroup$ – Protagger Oct 22 '13 at 17:13

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