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## Problem

The rank-nullity theorem states:

Let $V$ and $W$ be vector spaces, where $V$ is finite in dimension. Let $T \colon V \to W$ be a linear transformation. Then,

$\operatorname{Rank}(T) + \operatorname{Nullity}(T) = \dim V$.

Determine the the rank and nullity of the linear transformation

$\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$

## Solution

Row reduce the matrix until the pivots are one.

$\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$

Step 1: Subtract row 2 by four times row 1

$\begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \ \end{bmatrix}$

Step 2: Divide row 2 by minus 3

$\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \end{bmatrix}$

The row-reduced matrix illustrates the system

\begin{align} x + 2y + 3z &= 0 \\ y + 2z &= 0 \end{align}

Let $z = t$, where $t \in \mathbb{R}$.

Then $y = -2z$ and $x = -3z - 2y$.

Subsequently, we find what mathematicians call the kernel (or nullspace).

$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = t \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}$

Since the kernel has one vector, the nullity is 1.

Since $\operatorname{Rank}(T) = \dim V - \operatorname{Nullity}(T)$, the rank of this linear transformation is

$\operatorname{Rank}(T) = 3 - 1 = 2$.
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