Projection definition in Linear Algebra

Question

Why is a mathematical projection defined as $P^2 = P$? I understand that it might be because once a vector has been projected onto a space, if projected again, it should give the same thing. Is there anything more to it?

Answer 1

There's little more. If $P$ is a projection and $w\in\operatorname{Im}P$, we want to have $P(w)=w$. This is equivalent to the assertion:$$(\forall v\in V):P\bigl(P(v)\bigr)=P(v).$$And this, in turn, is equivalent to $P^2=P$. $\phantom{}$

Answer 2

If, $ E^n = V \oplus W $, then,

$Px=x \;\;\forall x\in V $

$Px=0 \;\;\forall x \in W$

If dimensionality of $V$ is $r$, you will find that, there exists matrices such that,

$P=T\Delta_rT^{-1}$ where $\Delta_r$ is a diagonal matrix with first $r$ diagonals equal to $1$ and rest other are $0$.

So what we get,

$P^2=(T\Delta_rT^{-1})(T\Delta_rT^{-1})=T\Delta_rT^{-1}=P$

Answer 3

It is useful to understand the equivalence between the two definitions mentioned in the existing answers:

1. $P$ is a projection (or projector) iff $P^2 = P$.
2. $P$ is a projection iff $Px \mapsto x_S, \forall x \in V$, where $x = x_S + x_T$ is the representation of $x$ corresponding to the direct sum of linear subspaces $V=S\oplus T$, for some $S, T$.

This equivalence is proved in many linear algebra textbooks, together with the criterion:

3. $P$ is a projection iff $I-P$ is a projection. $I-P$ is called the complementary projection to $P$.

In my opinion, definition (2) expresses the gist of the concept: there is a correspondence between complementary pairs of subspaces $S\oplus T = V$ and projection operators.

Answer 4

It is useful to remark that one has $X^2-X=X(X-1)$, so this polynomial has simple roots $0$ and $1$. For a projection $p$ onto a subspace $V$ and parallel to a complementary subspace $W$, you want that $p$ acts by the scalar$~0$ on $W$ (i.e., $W\subseteq\ker(p)$) and that $p$ acts scalar$~1$ on $V$ (i.e., $p(v)=v$ for all $v\in V$, which can also be expressed as $V\subseteq\ker(p-I)$); since by assumption of complementarity every vector can be written as sum of a vector in $V$ and one in $W$, one easily sees that both inclusions will actually be equalities of subspaces. So for any such $p$, the polynomial $p^2-p$ will vanish on $W$ (since $0$ is a root of $X^2-x$) and also on $V$ (since $1$ is a root of $X^2-X$), and therefore on the whole space.

Conversely if $p$ is any linear operator that satisfies $p^2-p=0$, then every vector $x$ can be written as the sum of the vector $x-p(x)$, which by assumption is a vector is in the eigenspace $W=\ker(p)$ of $p$ for $\lambda=0$, and of the vector $p(x)$, which by assumption is a vector is in the eigenspace $V=\ker(p-I)$ of $p$ for $\lambda=1$. Hence the whole space is the direct sum of these two eigenspaces, and $p$ is the projection onto $W$parallel to$~V$.