Projection matrix and null space

Question

How can you show that the null space of a projection matrix is equal to the the column space of the identity matrix minus that matrix? $$N(A) = C(I - A)$$

Answer 1

The column space of a matrix is the same as the image of the transformation. (that's not very difficult to see but if you don't see it post a comment and I can give a proof)

Now for $v\in N(A)$, $Av=0$ Then $(I-A)v=Iv-Av=v-0=v$ hence $v$ is the image of $I-A$.

On the other hand if $v$ is the image of $I-A$, $v=(I-A)w$ for some vector $w$. Then $$ Av=A(I-A)w=Aw-A^2w=Aw-Aw=0 $$ where I used the fact $A^2=A$ ($A$ is projection). Then $v\in N(A)$.

Answer 2

Let

$\text{size}(A) = n; \tag 0$

since $A$ is a projection matrix,

$A^2 = A; \tag 1$

now if

$\vec x \in N(A), \tag 2$

then

$A \vec x = 0, \tag 3$

whence

$(I - A) \vec x = I \vec x - A \vec x = \vec x; \tag 4$

if the columns of $I - A$ are $\vec A_i$, $1 \le i \le n$, and

$\vec x = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}, \tag 5$

then

$\vec x = (I - A) \vec x = \displaystyle \sum_1^n x_i \vec A_i \in C(I - A), \tag 6$

which is easy to see since $\sum_1^n x_i \vec A_i$ is a linear combination of the $\vec A_i$, with coefficients $x_i$. Thus

$\vec x \in C(I - A), \tag 7$

and we have shown that

$N(A) \subset C(I - A); \tag 8$

if we now assume

$\vec x \in C(I - A), \tag 9$

there must exist $y_i$ with

$\vec x = \displaystyle \sum_1^n y_i \vec A_i; \tag{10}$

(10) may also be written

$\vec x = (I - A) \vec y, \tag{11}$

so

$A \vec x = A(I - A) \vec y = (A - A^2)\vec y = 0, \tag{12}$

by virtue of (1). We thus have

$C(I - A) \subset N(A), \tag{13}$

and conclude that

$N(A) = C(I - A) \tag{14}$

as was to be shown.