Quiz1101#2

Theorem: Given a matrix
$A \in M_{m \times n}$ , let
$P_{c}$ and
$P_{r}$ be the projection matrices onto the column space and row space of
$A$ , respectively. Show that
$P_{c} A P_{r} = A$ .

Preliminary

Lemma: If
$\forall x \in R^{n}$ we have
$A x = B x$ , then
$A = B$ .
- pf:
  1. Choose
    $e_{1} = (1, 0, \dots, 0)^{T}$ , then
    $A e_{1} = B e_{1}$ says that the first column of
    $A$ and
    $B$ are exactly the same.
  2. Follow the same idea, choose
    $e_{k}$ that is
    $1$ in k-th element and otherwise
    $0$ . We have that the
    $k$ -th column of
    $A$ and
    $B$ are exactly the same, for any chosen
    $k$ .
  3. Therefore,
    $A = B$ .
Lemma: If
$\forall y \in R^{m}$ we have
$y^{T} A = y^{T} B$ , then
$A = B$ .
- pf: Choose the elementary row vector
  $e_{k}^{T}$ and follow the same proof as in the previous lemma.

Proof

Notation:

$C (A)$ : column space of
$A$ .

$C (A^{T})$ : row space of
$A$ .

$P_{c} A = A$ .
- pf:
  - Since
    $P_{c}$ is the projection matrix onto the column space of
    $A$ ,
    $P_{c} x = x$ for any
    $x \in C (A)$ .
  - Given
    $x \in R^{n}$ ,
    $A x \in C (A)$ , and then
    $P_{c} A x = P_{c} (A x) = A x$ .
  - Since
    $P_{c} A x = A x$ for all
    $x$ , we have
    $P_{c} A = A$ .
$P_{r}^{T} A^{T} = A^{T}$ .
- pf:
  - Since
    $P_{r}$ is the projection matrix onto the row space of
    $A$ , and
    $P_{r}^{T} = P_{r}$ , we have that
    $P_{r}^{T} y = y$ for any
    $y \in C (A^{T})$ .
  - Given
    $y \in R^{m}$ ,
    $A^{T} y \in C (A^{T})$ , and then
    $P_{r}^{T} A^{T} y = P_{r}^{T} (A^{T} y) = A^{T} y$ .
  - Since
    $P_{r}^{T} A^{T} y = A^{T} y$ for all
    $y$ , we have
    $P_{r}^{T} A^{T} = A^{T}$ .
Proof of theorem
- pf:
  $P_{c} A P_{r} = (P_{c} A) P_{r} = A P_{r} = A$ .