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Problem

Rank-nullity.svg

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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