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

Ohio State University exam problems and solutions in mathematics

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)
 
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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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  1. 04/06/2017

    […] Problem 6 and its solution: Rank and nullity of linear transformation from $R^3$ to $R^2$ […]

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