Projections in Linear Algebra I just started reading about Orthogonal Projections when I encountered this:

I'm fine with the definition. It seems natural. But I don't get the Note after it. Why is it that if we're given Vector spave $V$ and it's subspace $U$, then there are many projections $T$ ? Isn't the complementary subspace $W$ of $U$ uniquely determined by $V$ and $U$? If yes, hoow then can there be more than one projection $T$ ?
 A: No, for a subspace $U$, there can be more than one complementary subspace (and because of this, more than one projection). 
Think in $\mathbb R^2$. Let $U$ be the subspace $\{y=0\}=<(1,0)>$. Now, $W_1=\{x=0\}=<(0,1)>$ and $W_2=\{x-y=0\}=<(1,1)>$ are both complementary subspaces of $U$. In fact, every vector linearly independent with $(1,0)$ generates a complementary subspace of $U$.
Now, if you fix a suplementary $W$ for $U$, every vector of $V$ can be written $v=u+w$, with $u\in U$ and $w\in W$. In this case, the projection would be $T(v)=u$. Let's do it with the previous examples:


*

*If we let $U=<(1,0)>$ and $W=<(0,1)>$, then every vector $(x,y)\in \mathbb R^2$ is written $$(x,y)=(x,0)+(0,y), \ (x,0)\in U, \ (0,y)\in V$$
so the projection is $T((x,y))=(x,0)$.

*If we pick $W=(1,1)$, then $$(x,y)=(x-y,0)+(y,y)$$ so $T(x,y)=(x-y,0)$ in this case.
A: Perhaps you’re thinking of the orthogonal complement of $U$, which is indeed unique, but that’s not the only complementary subspace.
Thinking about what’s going on in $\mathbb R^2$ is illustrative. The non-trivial subspaces of $\mathbb R^2$ are the lines through the origin. A projection onto one of these lines $U$ “collapses” all of the vectors in some other subspace $W$—the complementary space—onto the origin. It also maps all points on a line parallel to $W$ onto the same point of $U$. 

Another way to put this is that the projection is parallel to $W$. If we pick some other line for $W$, we get a different projection:

Thus, projection onto $U$ is not unique and is determined by the choice of complementary subspace $W$. The situation is similar in higher-dimensional spaces, although the concept of “parallel to” an arbitrary subspace can become hard to visualize.
A: The complementary subspace is not uniquely determined by$V$.
One of the easier examples to see that is in $\Bbb R^2$. Let $U= \langle (1,0)\rangle$. Then one option for $W$ is $\langle (0,1)\rangle$ since then $V\oplus W = \langle (1,0), (0,1)\rangle = \Bbb R^2$. This is the standard basis. Note that we can also have $W=\langle (1,1)\rangle$, however, and still have $V\oplus W =\Bbb R^2$.
