$W\subset U \subset V, u=proj_U(v), w=proj_W(u). \ \ w=proj_W(v)?$

Question

if $V$ is an inner product vector space, $W,U$ are subspaces S.T. $W\subset U, v\in V.\ $ $u$ is the orthogonal projection of $v$ onto $U$, $w$ is the orthogonal projection of $u$ onto $W$. The question is whether the orthogonal projection of $v$ onto $W$ is equal to $w$. I am almost certain this is not the case but failed to find a counter example. Any thoughts?

Answer 1

I will restrict this answer to the finite dimensional case.

Choose an orthonormal basis $\{w_1,\dots,w_r\}$ of $W$.

Extend to an orthonormal basis $\{w_1,\dots,w_r,u_1,\dots,u_m\}$ of $U$, the $u_i$'s coming from $W^{\perp}$.

Finally, extend to an orthonormal basis $\{w_1,\dots,w_r,u_1,\dots,u_m,v_1,\dots,v_l\}$ of $V$, the $v_i$'s coming from $U^{\perp}$.

Note that we have $U^{\perp}\subset W^{\perp}$.

Fix $v\in V$. We have $$\text{proj}_U(v)=\sum_{i=1}^{r}\langle v,w_i\rangle w_i+\sum_{j=1}^{m}\langle v,u_j\rangle u_j=u$$ $$\text{proj}_W(u)=\sum_{i=1}^{r}\langle v,w_i\rangle w_i=w\tag{$u_j$'s are in $W^{\perp}$}$$ $$\text{proj}_W(v)=\sum_{i=1}^{r}\langle v,w_i\rangle w_i=w\tag{$u_j$'s and $v_k$'s are in $W^{\perp}$}$$