Let $V$ be the vector space of $2 \times 2$ matrices with real entries, and $\mathrm{P}_3$ the vector space of real polynomials of degree 3 or less. Define the linear transformation $T : V \rightarrow \mathrm{P}_3$ by
\[T \left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = 2a + (b-d)x – (a+c)x^2 + (a+b-c-d)x^3.\]

Find the rank and nullity of $T$.

Solution.

To find the rank of $T$, we will use its matrix representation relative to the standard bases for $V$ and $\mathrm{P}_3$. Define the matrices
\[\mathbf{e}_1 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} , \, \mathbf{e}_2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} , \, \mathbf{e}_3 = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} , \, \mathbf{e}_4 = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\] and the polynomials
\[p_1(x) = 1 , \quad p_2(x) = x , \quad p_3(x) = x^2 , \quad p_4(x) = x^3.\]

Then the standard bases are
\[B = \{ \mathbf{e}_1 , \mathbf{e}_2 , \mathbf{e}_3 , \mathbf{e}_4 \} \subset V \mbox{ and } C = \{ p_1 , p_2 , p_3 , p_4 \} \subset \mathrm{P}_3.\]

We now find the matrix of representation of $T$ relative to these bases. This will be a $4 \times 4$ matrix whose $i$-th row is the column vector of $T(e_i)$ relative to the basis $C$. For example, we have
\[ T(e_1) = 2 – x^2 + x^3 = 2p_1 – p_3 + p_4 , \] and so the first column of the matrix is the column vector
\[ [T(e_1)]_{C} = \begin{bmatrix} 2 \\ 0 \\ -1 \\ 1 \end{bmatrix} . \]

This vector becomes the first column of the matrix. After performing this process for the other three basis elements $e_2 , e_3 , e_4$, we get the matrix
\[ [T]_{B}^{C} = \begin{bmatrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ -1 & 0 & -1 & 0 \\ 1 & 1 & -1 & -1 \end{bmatrix}.\]

To find the rank of $T$, we row-reduce this matrix:
\begin{align*}
\begin{bmatrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ -1 & 0 & -1 & 0 \\ 1 & 1 & -1 & -1 \end{bmatrix} \xrightarrow[-R_3]{ \frac{1}{2} R_1 } \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 1 & 0 & 1 & 0 \\ 1 & 1 & -1 & -1 \end{bmatrix} \xrightarrow{ R_3 – R_1 } \\[6pt] \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 1 & 1 & -1 & -1 \end{bmatrix}
\xrightarrow{R_4 – R_1 – R_2 + R_3} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}.
\end{align*}

This reduced matrix has three non-zero rows, and so has rank 3. Thus the orginal linear transformation $T$ also has rank 3. The nullity can be computed using the equation
\[ \mathtt{rank}(T) + \mathtt{nullity}(T) = \dim (V).\] Using the values $\mathtt{rank}(T) = 3$ and $\dim (V) = 4$, the nullity of $T$ must be 1.

The post The Rank and Nullity of a Linear Transformation from Vector Spaces of Matrices to Polynomials first appeared on Problems in Mathematics.

Let $U$ and $V$ be vector spaces over a scalar field $\F$.
Let $T: U \to V$ be a linear transformation.

Prove that $T$ is injective (one-to-one) if and only if the nullity of $T$ is zero.

Definition (Injective, One-to-One Linear Transformation).

A linear transformation is said to be injective or one-to-one if provided that for all $\mathbf{u}_1$ and $\mathbf{u}_1$ in $U$, whenever $T(\mathbf{u}_1)=T(\mathbf{u}_2)$, then we have $\mathbf{u}_1=\mathbf{u}_2$.

Proof.

$(\implies)$: If $T$ is injective, then the nullity is zero.

Suppose that $T$ is injective.
Our objective is to show that the null space $\calN(T)=\{\mathbf{0}_U\}$.

Since $T$ is a linear transformation, it sends the zero vector $\mathbf{0}_U$ of $U$ to the zero vector $\mathbf{0}_V$ of $V$.
In fact, we have
\begin{align*}
T(\mathbf{0}_U)&=T(\mathbf{0}_U-\mathbf{0}_U)\\
&=T(\mathbf{0}_U+(-1)\mathbf{0}_U)\\
&=T(\mathbf{0}_U)+(-1)T(\mathbf{0}_U) &&\text{by linearity of $T$}\\
&=T(\mathbf{0}_U)-T(\mathbf{0}_U)=\mathbf{0}_V.
\end{align*}
Hence $\mathbf{0}_U\in \calN(T)$.

On the other hand, if $\mathbf{u}\in \calN(T)$, then we have
\[T(\mathbf{u})=\mathbf{0}_V=T(\mathbf{0}_U).\] Since $T$ is injective, it yields that $\mathbf{u}=\mathbf{0}_U$.
Therefore we obtain $\calN(T)=\{\mathbf{0}_U\}$, and the nullity of $T$ is zero.
(Recall that the nullity of $T$ is the dimension of $\calN(T)$.)

$(\impliedby)$: If the nullity is zero, then $T$ is injective.

Next, suppose that the nullity of $T$ is zero.
This is equivalent to the condition $\calN(T)=\{\mathbf{0}_U\}$.
Our goal is to show that $T: U \to V$ is injective.

Suppose that $T(\mathbf{u}_1)=T(\mathbf{u}_2)$ for some $\mathbf{u}_1, \mathbf{u}_2\in U$.
Then we have
\begin{align*}
\mathbf{0}_V&=T(\mathbf{u}_1)-T(\mathbf{u}_2)\\
&=T(\mathbf{u}_1)+(-1)T(\mathbf{u}_2)\\
&=T(\mathbf{u}_1+(-1)\mathbf{u}_2) && \text{by linearity of $T$}\\
&=T(\mathbf{u}_1-\mathbf{u}_2)
\end{align*}
It follows that the vector $\mathbf{u}_1-\mathbf{u}_2$ is in the null space $\calN(T)=\{\mathbf{0}_U\}$.
Thus, we have $\mathbf{u}_1-\mathbf{u}_2=\mathbf{0}_U$, or $\mathbf{u}_1=\mathbf{u}_2$.
So the linear transformation $T$ is injective.

Related Question.

The proof of this problem is given in the post ↴
A Linear Transformation $T: U\to V$ cannot be Injective if $\dim(U) > \dim(V)$

The post A Linear Transformation is Injective (One-To-One) if and only if the Nullity is Zero first appeared on Problems in Mathematics.

Let $T: \R^2 \to \R^2$ be a linear transformation such that
\[T\left(\, \begin{bmatrix}
1 \\
1
\end{bmatrix} \,\right)=\begin{bmatrix}
4 \\
1
\end{bmatrix}, T\left(\, \begin{bmatrix}
0 \\
1
\end{bmatrix} \,\right)=\begin{bmatrix}
3 \\
2
\end{bmatrix}.\] Then find the matrix $A$ such that $T(\mathbf{x})=A\mathbf{x}$ for every $\mathbf{x}\in \R^2$, and find the rank and nullity of $T$.

(The Ohio State University, Linear Algebra Exam Problem)
 

Solution.

The matrix $A$ such that $T(\mathbf{x})=A\mathbf{x}$ is given by
\[A=[T(\mathbf{e}_1), T(\mathbf{e}_2)],\] where $\mathbf{e}_1=\begin{bmatrix}
1 \\
0
\end{bmatrix}, \mathbf{e}_2=\begin{bmatrix}
0 \\
1
\end{bmatrix}$ are standard basis of $\R^2$.
Since the vector $T(\mathbf{e}_2)$ is given, it remains to find $T(\mathbf{e}_1)$.
By inspection, we obtain the linear combination
\[\begin{bmatrix}
1 \\
0
\end{bmatrix}=\begin{bmatrix}
1 \\
1
\end{bmatrix}-\begin{bmatrix}
0 \\
1
\end{bmatrix}.\] Thus, we have
\begin{align*}
T(\mathbf{e}_1)&=T\left(\,\begin{bmatrix}
1 \\
1
\end{bmatrix}-\begin{bmatrix}
0 \\
1
\end{bmatrix} \,\right)\\
&=T\left(\, \begin{bmatrix}
1 \\
1
\end{bmatrix} \,\right)-T\left(\, \begin{bmatrix}
0 \\
1
\end{bmatrix} \,\right) && \text{by linearity of $T$}\\
&=\begin{bmatrix}
4 \\
1
\end{bmatrix}-\begin{bmatrix}
3 \\
2
\end{bmatrix}\\
&=\begin{bmatrix}
1 \\
-1
\end{bmatrix}.
\end{align*}
It follows that the matrix $A$ is given by
\[A=\begin{bmatrix}
1 & 3\\
-1& 2
\end{bmatrix}.\]

The rank and nullity of $T$ are the same as the rank and nullity of $A$.
We reduced the matrix $A$ by elementary row operations as follows:
\begin{align*}
A\xrightarrow{R_2+R_1} \begin{bmatrix}
1 & 3\\
0& 5
\end{bmatrix}
\xrightarrow{\frac{1}{5}R_2}
\begin{bmatrix}
1 & 3\\
0& 1
\end{bmatrix}
\xrightarrow{R_1-3R_2}
\begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix}.
\end{align*}
Hence the rank of $A$ is $2$ (because there are two non zero rows). The nullity of $A$ is determined by the rank nullity theorem
\[\text{rank of $A$} + \text{ nullity of $A$}=2 \text{ (the number of columns of $A$)}.\] Hence the nullity of $A$ is $0$.

In a nutshell, the rank of $T$ is $2$, and the nullity of $T$ is $0$.

Linear Algebra Midterm Exam 2 Problems and Solutions

• True of False Problems and Solutions: True or False problems of vector spaces and linear transformations
• Problem 1 and its solution: See (7) in the post “10 examples of subsets that are not subspaces of vector spaces”
• Problem 2 and its solution: Determine whether trigonometry functions $\sin^2(x), \cos^2(x), 1$ are linearly independent or dependent
• Problem 3 and its solution: Orthonormal basis of null space and row space
• Problem 4 and its solution: Basis of span in vector space of polynomials of degree 2 or less
• Problem 5 and its solution: Determine value of linear transformation from $R^3$ to $R^2$
• Problem 6 and its solution: Rank and nullity of linear transformation from $R^3$ to $R^2$
• Problem 7 and its solution (current problem): Find matrix representation of linear transformation from $R^2$ to $R^2$
• Problem 8 and its solution: Hyperplane through origin is subspace of 4-dimensional vector space

The post Find Matrix Representation of Linear Transformation From $\R^2$ to $\R^2$ first appeared on Problems in Mathematics.

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

• True of False Problems and Solutions: True or False problems of vector spaces and linear transformations
• Problem 1 and its solution: See (7) in the post “10 examples of subsets that are not subspaces of vector spaces”
• Problem 2 and its solution: Determine whether trigonometry functions $\sin^2(x), \cos^2(x), 1$ are linearly independent or dependent
• Problem 3 and its solution: Orthonormal basis of null space and row space
• Problem 4 and its solution: Basis of span in vector space of polynomials of degree 2 or less
• Problem 5 and its solution: Determine value of linear transformation from $R^3$ to $R^2$
• Problem 6 and its solution (current problem): Rank and nullity of linear transformation from $R^3$ to $R^2$
• Problem 7 and its solution: Find matrix representation of linear transformation from $R^2$ to $R^2$
• Problem 8 and its solution: Hyperplane through origin is subspace of 4-dimensional vector space

The post Rank and Nullity of Linear Transformation From $\R^3$ to $\R^2$ first appeared on Problems in Mathematics.

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 $T$.

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:

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 $\R^m$

The post Range, Null Space, Rank, and Nullity of a Linear Transformation from $\R^2$ to $\R^3$ first appeared on Problems in Mathematics.