---
tags: Linear Algebra, 線代啟示錄, ccjou
---

# $\text{RREF}$ and Relationships bewteen Column Vectors

## Observation

Suppose we have matrix $A_{ij}$ of which $\text{RREF}$ is $R = \begin{bmatrix}
1 & a & 0 & b\\
0 & 0 & 1 & c\\
0 & 0 & 0 & 0
\end{bmatrix}$
Then we have $\bar{a_2} = a\cdot\bar{a_1}$ and $\bar{a_4} = b\cdot\bar{a_1} + c\cdot\bar{a_3}$, where $\bar{a_j}$ denotes column vector of $A$.

## Illustration

### Method 1 System of Equations

Suppose we're solving $A\bar{x} = \bar{0}$, equivalently $R\bar{x} = \bar{0}$
Then we have
$\bar{x} =
\begin{bmatrix}
x \\ y \\ z \\ w
\end{bmatrix} =
\left( y\cdot \begin{bmatrix}
-a \\ 1 \\ 0 \\ 0
\end{bmatrix} \right) +
\left( w \cdot \begin{bmatrix}
-b \\ 0 \\ -c \\ 1
\end{bmatrix} \right)$
Let $y=0$ and $w=1$, we have $(-b) \times a_{i,1} + (-c) \times a_{i,3} + a_{i,4} = 0$
Which means $a_{i,4} = b \cdot a_{i,1} + c \cdot a_{i,3} ~\forall i\in\{1,2,3\}$, that is, $\bar{a_4} = b\cdot\bar{a_1} + c\cdot\bar{a_3}$
We conclude that $\text{RREF}$ do give us relationships between column vectors.<div style="text-align: right"> $\small\square$ </div>

### Method 2 Elementary Matrix

Let $B$ be [elementary matrix](https://en.wikipedia.org/wiki/Elementary_matrix) s.t. $BA=R=\text{RREF of }A$
Since elementary square matrices are invertible, we have $B^{-1}R = A$
Which means each column of $A$ could be written as *linear combination* of columns of $B^{-1}$, where the coefficients are exactly elements in row vectors of $R$ <div style="text-align: right"> $\small\square$ </div>
Notice $B^{-1}$ and $B$ are full rank, equivalently all the column vectors of the square matrix $B^{-1}$ are linear independent, which guarantees the unique representation.

# Reference

I learnt this from a superb linear algebra site maintained by [ccjou](https://ccjou.wordpress.com/2009/03/25/%e7%94%b1%e7%b0%a1%e7%b4%84%e5%88%97%e6%a2%af%e5%bd%a2%e5%bc%8f%e5%88%a4%e6%96%b7%e8%a1%8c%e7%a9%ba%e9%96%93%e5%9f%ba%e5%ba%95/ "線代啟示錄 由簡約列梯形式判斷行空間基底")