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

## 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:

(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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