I'm currently struggling to solve this question.
The first one $col(P_2) \subset col(P_1)^{\bot}$ is quite straightforward
($P_1$ is ($n \times r$) and $P_2$ is ($n \times (n-r)$))
$x \in col(P_2) \Rightarrow ^{\exists} y \in \mathbb R^{n-r} $ s.t $ x = P_2 y$
$z \in col(P_1) \Rightarrow ^{\exists} w \in \mathbb R^{r} $ s.t $ z = P_1 w$
Then, $x'z = y'{P_2}'P_1 w = y'Ow = 0$
This holds for all $z \in col(P_1) $, so $x \in col(P_1)^{\bot}$.
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But I can't think of the other way $col(P_1)^{\bot} \subset col(P_2)$.
For $a \in col(P_1)^{\bot}$ and $b = P_1 c \in col(P_1)$, $a'P_1c = 0$.
Then, I'm not sure how to associate this with $col(P_2)$.
Any comments with regard to this trial (including some errors or typos w.r.t this trial, if exist) is welcome. Thank you.
Let $P = [p_1|p_2| ... |p_n], p_i \in \mathbb R^n$
$a \in col(P_1)^{\bot}$ can be expressed as $a = \sum_{i = 1}^n a_i p_i$, $a_i \in \mathbb R$.
$a$ is orthogonal to all elements of $col(P_1)$ including $p_1, ..., p_r$.
Then, $a'p_i = a_i\Vert p_i \Vert ^2 = a_i = 0$ for $i = 1,...,r$.
So, $a = \sum_{i = r+1}^n a_i p_i$ and thus $a \in col(P_2)$.
$\Rightarrow$ Is this the correct method for the second direction?