Range, Null Space, Rank, and Nullity of a Linear Transformation from $\R^2$ to $\R^3$

Linear algebra problems and solutions

Problem 154

Define the map $T:\R^2 \to \R^3$ by $T \left ( \begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}\right )=\begin{bmatrix}
x_1-x_2 \\
x_1+x_2 \\
x_2
\end{bmatrix}$.

(a) Show that $T$ is a linear transformation.

(b) Find a matrix $A$ such that $T(\mathbf{x})=A\mathbf{x}$ for each $\mathbf{x} \in \R^2$.

(c) Describe the null space (kernel) and the range of $T$ and give the rank and the nullity of $T$.

 
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Proof.

(a) Show that $T$ is a linear transformation.

To show that $T: \R^2 \to \R^3$ is a linear transformation, the map $T$ needs to satisfy:

(i) $T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})$ for any $\mathbf{u}, \mathbf{v}\in \R^2$, and
(ii) $T(c\mathbf{v})=cT(\mathbf{v})$ for any $\mathbf{v} \in \R^2$ and $c\in \R$.

To check (i), let
\[\mathbf{u}=\begin{bmatrix}
u_1 \\
u_2
\end{bmatrix}, \mathbf{v}=\begin{bmatrix}
v_1 \\
v_2
\end{bmatrix}\in \R^2.\] We have
\begin{align*}
T(\mathbf{u}+\mathbf{v})&=T\left( \begin{bmatrix}
u_1+v_1 \\
u_2+v_2
\end{bmatrix} \right) =\begin{bmatrix}
(u_1+v_1)-(u_2+v_2) \\
(u_1+v_1)+(u_2+v_2) \\
u_2+v_2
\end{bmatrix}\\[6pt] &=
\begin{bmatrix}
(u_1-u_2)+(v_1-v_2) \\
(u_1+u_2)+(v_1+v_2) \\
u_2+v_2
\end{bmatrix}
=
\begin{bmatrix}
u_1-u_2\\
u_1+u_2 \\
u_2
\end{bmatrix}+
\begin{bmatrix}
v_1-v_2 \\
v_1+v_2 \\
v_2
\end{bmatrix}\\[6pt] &=T(\mathbf{u})+T(\mathbf{v}).
\end{align*}
Thus condition (i) holds.


To check (ii), consider any $\mathbf{v}=\begin{bmatrix}
v_1 \\
v_2
\end{bmatrix}$ and $c\in \R$.
Then we have
\begin{align*}
T(c\mathbf{v})=&T\left( \begin{bmatrix}
cv_1 \\
cv_2
\end{bmatrix}
\right)
= \begin{bmatrix}
cv_1-cv_2 \\
cv_1+cv_2 \\
cv_2
\end{bmatrix}\\[6pt] &= c\begin{bmatrix}
v_1-v_2 \\
v_1+v_2 \\
v_2
\end{bmatrix}
=cT(\mathbf{v}).
\end{align*}
Thus condition (ii) is met and the map $T$ is a linear transformation from $\R^2$ to $\R^3$.

(b) Find a matrix $A$ such that $T(\mathbf{x})=A\mathbf{x}$ for each $\mathbf{x} \in \R^2$.

(b) Let
\[\mathbf{e}_1=\begin{bmatrix}
1 \\
0
\end{bmatrix}, \mathbf{e}_2=\begin{bmatrix}
0 \\
1
\end{bmatrix}\] be the standard basis vectors of $\R^2$.
The matrix $A$ satisfying $T(\mathbf{x})=A\mathbf{x}$ can be obtained as
\[A=[T(\mathbf{e}_1)\quad T(\mathbf{e}_2)].\] Since we have
\[T(\mathbf{e}_1)=\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix} \text{ and } T(\mathbf{e}_2)=\begin{bmatrix}
-1 \\
1 \\
1
\end{bmatrix},\] we obtain the matrix
\[A=\begin{bmatrix}
1 & -1 \\
1 & 1 \\
0 &1
\end{bmatrix}.\]

(c) Describe the null space (kernel) and the range of $T$ and give the rank and the nullity of $T$

Null Space and Nullity

We fist find the null space of the linear transformation of $T$.
Note that the null space of $T$ is the same as the null space of the matrix $A$.

By definition, the null space is
\[ \calN(T)=\calN(A)=\{\mathbf{x} \in \R^2 \mid A\mathbf{x}=\mathbf{0}\}.\] So the null space is a set of all solutions for the system $A\mathbf{x}=\mathbf{0}$.

To find the solution of the system we consider the augmented matrix and reduce the matrix using the elementary row operations. (The Gauss-Jordan elimination method.)
We have
\begin{align*}
\left[\begin{array}{rr|r}
1 & -1 & 0 \\
1 &1 &0 \\
0 & 1 & 0
\end{array} \right] \xrightarrow{R_2-R_1}
\left[\begin{array}{rr|r}
1 & -1 & 0 \\
0 &2 &0 \\
0 & 1 & 0
\end{array} \right]\\[6pt] \xrightarrow[\text{then} R_2\leftrightarrow R_3] {\substack{R_1+R_3 \\ R_2-2R_1}}
\left[\begin{array}{rr|r}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 0
\end{array} \right].
\end{align*}

Thus the system has only zero solution
\[x_1=0, x_2=0.\] Therefore we obtained
\[\calN(T)=\calN(A)=\left \{ \begin{bmatrix}
0 \\
0
\end{bmatrix} \right \}.\]

Since the nullity is the dimension of the null space, we see that the nullity of $T$ is $0$ since the dimension of the zero vector space is $0$.

Range and Rank

Next, we find the range of $T$. Note that the range of the linear transformation $T$ is the same as the range of the matrix $A$.
We describe the range by giving its basis.

The range of $A$ is the columns space of $A$. Thus it is spanned by columns
\[\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-1 \\
1 \\
1
\end{bmatrix}.\]

From the above reduction of the augmented matrix, we see that these vectors are linearly independent, thus a basis for the range. (Basically, this is the leading 1 method.)
Hence we have
\[\calR(T)=\calR(A)=\Span \left\{\,\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-1 \\
1 \\
1
\end{bmatrix} \,\right\}\] and
\[\left\{\, \begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-1 \\
1 \\
1
\end{bmatrix} \,\right\}\] is a basis for $\calR(T)$.

The rank of $T$ is the dimension of the range $\calR(T)$.
Thus the rank of $T$ is $2$.

Remark that we obtained that the nullity of $T$ is $0$ and the rank of $T$ is $2$. This agrees with the rank-nullity theorem
\[(\text{rank of } T)+ (\text{nullity of } T)=2.\]

More Problems about Linear Transformations

Additional problems about linear transformations are collected on the following page:
Linear Transformation from $\R^n$ to $\R^m$


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