# Rank and Nullity of Linear Transformation From $\R^3$ to $\R^2$

## Problem 369

Let $T:\R^3 \to \R^2$ be a linear transformation such that
$T(\mathbf{e}_1)=\begin{bmatrix} 1 \\ 0 \end{bmatrix}, T(\mathbf{e}_2)=\begin{bmatrix} 0 \\ 1 \end{bmatrix}, T(\mathbf{e}_3)=\begin{bmatrix} 1 \\ 0 \end{bmatrix},$ where $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$ are the standard basis of $\R^3$.
Then find the rank and the nullity of $T$.

(The Ohio State University, Linear Algebra Exam Problem)

## Solution.

The matrix representation of the linear transformation $T$ is given by
$A=[T(\mathbf{e}_1), T(\mathbf{e}_2), T(\mathbf{e}_3)]=\begin{bmatrix} 1 & 0 & 1 \\ 0 &1 &0 \end{bmatrix}.$

Note that the rank and nullity of $T$ are the same as the rank and nullity of $A$.
The matrix $A$ is already in reduced row echelon form.
Thus, the rank of $A$ is $2$ because there are two nonzero rows.

Another way to see this is to use the leading 1 method. It implies that the first two columns vectors form a basis of the range of $A$ because the first two columns contain the leading 1’s.
Thus, the rank (=the dimension of the range) is $2$.

The rank-nullity theorem says that
$\text{rank of A} + \text{ nullity of A}=3 \text{ (the number of columns of A)}.$ Hence the nullity of $A$ is $1$.

In summary, the rank of $T$ is $2$, and the nullity of $T$ is $1$.

## Linear Algebra Midterm Exam 2 Problems and Solutions

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##### Determine a Value of Linear Transformation From $\R^3$ to $\R^2$

Let $T$ be a linear transformation from $\R^3$ to $\R^2$ such that \[ T\left(\, \begin{bmatrix} 0 \\ 1 \\ 0...

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