---
title: Quiz1101#2
tags: Linear algebra
GA: G-77TT93X4N1
---

# 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 $\forall x\in\mathbb{R}^n$ we have $Ax=Bx$, then $A=B$.
* pf:
1. Choose $e_1 = (1, 0, \cdots, 0)^T$, then $Ae_1=Be_1$ 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 $\forall y \in\mathbb{R}^m$ we have $y^TA = y^TB$, 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 $A$.

* $P_cA=A$.
* 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 $x\in\mathbb{R}^n$, $Ax\in C(A)$, and then $P_cAx = P_c(Ax) = Ax$.
* Since $P_cAx = Ax$ for all $x$, we have $P_cA = A$.
* $P_r^TA^T=A^T$.
* pf:
* Since $P_r$ is the projection matrix onto the row space of $A$, and $P_r^T=P_r$, we have that $P_r^Ty = y$ for any $y\in C(A^T)$.
* Given $y\in\mathbb{R}^m$, $A^Ty\in C(A^T)$, and then $P_r^TA^Ty = P_r^T(A^Ty) = A^Ty$.
* Since $P_r^TA^Ty = A^Ty$ for all $y$, we have $P_r^TA^T = A^T$.
* Proof of theorem
* pf: $P_c A P_r = (P_c A) P_r = AP_r = A$.