Construction of a projection operator in $\mathbb{R}^n$

Question

Is it possible to construct a projection operator (which is not the identity operator) for a given space? (e.g. $\mathbb{R}^4$)

Answer 1

Sure, if the vector space is finite dimensional. For let $V$ be a finite dimensional vector space over the field $\Bbb F$, and let $W \subset V$ be a subspace. Say $\dim V = n$ and $\dim W = m \le n$. Choose a basis $w_1, w_2, \ldots, w_m$ for $W$ and extend it to a basis for $V$ by adding $l = n - m$ linearly independent vectors $v_i$, $1 \le i \le l$, $v_i \notin W$ for all $i$. Define a map $P: V \to V$ by the rules

$Pw _i = w_i \tag{1}$

and

$Pv_i = 0; \tag{2}$

extend $P$ to all of $V$ by linearity; thus, since any $y \in V$ may be written

$y = \sum_1^m a_iw_i + \sum_1^l b_i v_i,\tag{3}$

where the $a_i, b_i \in \Bbb F$, we have

$Py = \sum_1^m a_i Pw_i + \sum_1^l b_i Pv_i = \sum_1^m a_iw_i \tag{4}$

by the definition of $P$. It follows from (3) and (4) that if $y \in W$, so that all the $b_i = 0$, that is

$y = \sum_1^m a_i w_i, \tag{5}$

that

$Py = \sum_1^m a_iw_i = y; \tag{6}$

furthermore, for any $y \in V$ as in (3) we find via (6) that

$P^2y = P(Py) = P(\sum_1^m a_iw_i) = Py, \tag{7}$

i.e.,

$P^2 = P, \tag{8}$

the definitive trait of projection operators.

The preceding discussion in fact provides a recipe for the construction of a projection onto a subspace $W$ of any finite dimensional vector space $V$ over any field $\Bbb F$. For example, if $V = \Bbb R^4$ and $W$ is he subspace spanned by the two vectors

$w_1 = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} \tag{9}$

and

$w_2 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}, \tag{10}$

we may take

$v_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} \tag{11}$

and

$v_2 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}; \tag{12}$

if we set

$Pw_i = w_i, i = 1, 2, \tag{13}$

and

$Pv_i = 0, i = 1, 2, \tag{14}$

it is easy to calculate for any vector

$y = \begin{pmatrix} y_1 \\ y_2 \\ y_3 \\ y_4 \end{pmatrix}, \tag{15}$

we have

$Py = \begin{pmatrix} 0 \\ y_2 \\ 0 \\ y_4 \end{pmatrix} \in W; \tag{16}$

one may also easily verify $P^2 = P$ for this $P$; it is indeed a projection operator.

Finally, it is worth noting that, though our example was taken with $V = \Bbb R^4$, a parallel result holds with $V = \Bbb F^4$, $\Bbb F$ any field; and similarly, more general examples may be had looking at $V = \Bbb F^n$ for any positive integer $n$; the reader may use our recipe to construct many projections.