About $\Bbb {proj}_U \Bbb v$

Question

Let $V$ be an inner product space. Let $U \subseteq V$ and $U^{\perp} \subseteq V$.

Then I know that for any $\Bbb v \in V$, we can write a uniq form of $\Bbb v = u_1 + u_2$,

such that $u_1 \in U$ and $u_2 \in U^{\perp}$, while we know that $u_1 = \Bbb {proj}_U \Bbb v$.

So $\Bbb v = \Bbb {proj}_U \Bbb v +(\Bbb v - \Bbb {proj}_U \Bbb v)$

My question is:

Is that correct to write $\Bbb {proj}_{U^{\perp}} \Bbb v = \Bbb v - \Bbb {proj}_U \Bbb v$

So $\Bbb v = \Bbb {proj}_U \Bbb v + \Bbb {proj}_{U^{\perp}} \Bbb v~~$ ?

Thanks a lot!

Answer 1

Yes. By the way that you have defined $\operatorname{proj}_U(v)$, you have$$\operatorname{proj}_{U^\perp}(v)=u_2=v-u_1=v-\operatorname{proj}_U(v).$$And therefore you have indeed that$$v=\operatorname{proj}_U(v)+\operatorname{proj}_{U^\perp}(v).$$