# Matrix Representation, Rank, and Nullity of a Linear Transformation $T:\R^2\to \R^3$

## Problem 605

Let $T:\R^2 \to \R^3$ be a linear transformation such that
$T\left(\, \begin{bmatrix} 3 \\ 2 \end{bmatrix} \,\right) =\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \text{ and } T\left(\, \begin{bmatrix} 4\\ 3 \end{bmatrix} \,\right) =\begin{bmatrix} 0 \\ -5 \\ 1 \end{bmatrix}.$

(a) Find the matrix representation of $T$ (with respect to the standard basis for $\R^2$).

(b) Determine the rank and nullity of $T$.

(The Ohio State University, Linear Algebra Midterm)

## Solution.

### (a) Find the matrix representation of $T$.

The matrix representation $A$ of the linear transformation 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 the standard basis vectors for $\R^2$.

To determine $T(\mathbf{e}_1)$, we first express $\mathbf{e}_1$ as a linear combination of $\begin{bmatrix} 3 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 4 \\ 3 \end{bmatrix}$ as follows.
Let
$\mathbf{e}_1=\begin{bmatrix} 1 \\ 0 \end{bmatrix}=a\begin{bmatrix} 3 \\ 2 \end{bmatrix}+b\begin{bmatrix} 4 \\ 3 \end{bmatrix}.$ This can be written as
$\begin{bmatrix} 3 & 4\\ 2& 3 \end{bmatrix}\begin{bmatrix} a \\ b \end{bmatrix}=\begin{bmatrix} 1 \\ 0 \end{bmatrix}.$ The determinant of the coefficient matrix $\begin{bmatrix} 3 & 4\\ 2& 3 \end{bmatrix}$ is $3\cdot 3-4\cdot 2=1\neq 0$ and thus it is invertible.
Hence we have
$\begin{bmatrix} a \\ b \end{bmatrix} =\begin{bmatrix} 3 & 4\\ 2& 3 \end{bmatrix}^{-1}\begin{bmatrix} 1 \\ 0 \end{bmatrix} =\begin{bmatrix} 3 & -4\\ -2& 3 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix}=\begin{bmatrix} 3 \\ -2 \end{bmatrix}.$ Hence $a=3$ and $b=-2$.
It yields that
$\mathbf{e}_1=3\begin{bmatrix} 3 \\ 2 \end{bmatrix}-2\begin{bmatrix} 4 \\ 3 \end{bmatrix}.$

It follows that
\begin{align*}
T(\mathbf{e}_1)&=T\left(\,3\begin{bmatrix}
3 \\
2
\end{bmatrix}-2\begin{bmatrix}
4 \\
3
\end{bmatrix} \,\right)\6pt] &=3T\left(\, \begin{bmatrix} 3 \\ 2 \end{bmatrix}\,\right)-2T\left(\, \begin{bmatrix} 4 \\ 3 \end{bmatrix} \,\right) &&\text{by linearity of T}\\[6pt] &=3\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}-2\begin{bmatrix} 0 \\ -5 \\ 1 \end{bmatrix}=\begin{bmatrix} 3 \\ 16 \\ 7 \end{bmatrix}. \end{align*} Similarly, we compute T(\mathbf{e}_2) as follows. Let \[\mathbf{e}_2=\begin{bmatrix} 0 \\ 1 \end{bmatrix}=c\begin{bmatrix} 3 \\ 2 \end{bmatrix}+d\begin{bmatrix} 4 \\ 3 \end{bmatrix}. This can be written as
$\begin{bmatrix} 3 & 4\\ 2& 3 \end{bmatrix}\begin{bmatrix} a \\ b \end{bmatrix}=\begin{bmatrix} 0 \\ 1 \end{bmatrix}.$ Hence, we obtain
$\begin{bmatrix} c \\ d \end{bmatrix}=\begin{bmatrix} 3 & 4\\ 2& 3 \end{bmatrix}^{-1}\begin{bmatrix} 0 \\ 1 \end{bmatrix}=\begin{bmatrix} 3 & -4\\ -2& 3 \end{bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix}=\begin{bmatrix} -4 \\ 3 \end{bmatrix},$ and $c=-4, d=3$.

Hence
$\mathbf{e}_2=\begin{bmatrix} 0 \\ 1 \end{bmatrix}=-4\begin{bmatrix} 3 \\ 2 \end{bmatrix}+3\begin{bmatrix} 4 \\ 3 \end{bmatrix}$ and we have
\begin{align*}
T(\mathbf{e}_2)&=T\left(\,-4\begin{bmatrix}
3 \\
2
\end{bmatrix}+3\begin{bmatrix}
4 \\
3
\end{bmatrix} \,\right)\6pt] &=-4T\left(\,\begin{bmatrix} 3 \\ 2 \end{bmatrix}\,\right)+3T\left(\,\begin{bmatrix} 4 \\ 3 \end{bmatrix}\,\right) &&\text{by linearity of T}\\[6pt] &=-4\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}+3\begin{bmatrix} 0 \\ -5 \\ 1 \end{bmatrix}=\begin{bmatrix} -4 \\ -23 \\ -9 \end{bmatrix}. \end{align*} Therefore the matrix representation A of T is \[A=[T(\mathbf{e}_1), T(\mathbf{e}_2)]=\begin{bmatrix} 3 & -4 \\ 16 & -23 \\ 7 &-9 \end{bmatrix}.

### (b) Determine the rank and nullity of $T$.

Note that the rank and nullity of $T$ are the same as the those of the matrix representation $A$ of $T$.
In part (a), we obtained
$A=\begin{bmatrix} 3 & -4 \\ 16 & -23 \\ 7 &-9 \end{bmatrix}.$ Let us first determine the rank of $A$.
We have
\begin{align*}
A=\begin{bmatrix}
3 & -4 \\
16 & -23 \\
7 &-9
\end{bmatrix}
\xrightarrow{R_3-2R_1}
\begin{bmatrix}
3 & -4 \\
16 & -23 \\
1 &-1
\end{bmatrix}
\xrightarrow{R_1\leftrightarrow R_3}
\begin{bmatrix}
1 & -1 \\
16 & -23 \\
3 &-4
\end{bmatrix}\6pt] \xrightarrow[R_3-3R_1]{R_2-16R_1} \begin{bmatrix} 1 & -1 \\ 0 & -7 \\ 0 &-1 \end{bmatrix} \xrightarrow{-R_3} \begin{bmatrix} 1 & -1 \\ 0 & -7 \\ 0 &1 \end{bmatrix} \xrightarrow[R_2+7R_3]{R_1+R_3} \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ 0 &1 \end{bmatrix} \xrightarrow{R_2\leftrightarrow R_3} \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 &0 \end{bmatrix}. \end{align*} Hence, the reduced row echelon form matrix of A has two nonzero rows. So the rank of A is 2. By the rank-nullity theorem, we know that \[\text{(rank of A)+(nullity of A)}=2. As the rank of $A$ is $2$, we see that the nullity of $A$ is $0$.

## Comment.

This is one of the midterm 2 exam problems for Linear Algebra (Math 2568) in Autumn 2017.

## List of Midterm 2 Problems for Linear Algebra (Math 2568) in Autumn 2017

1. Vector Space of 2 by 2 Traceless Matrices
2. Find an Orthonormal Basis of the Given Two Dimensional Vector Space
3. Are the Trigonometric Functions $\sin^2(x)$ and $\cos^2(x)$ Linearly Independent?
4. Find Bases for the Null Space, Range, and the Row Space of a $5\times 4$ Matrix
5. Matrix Representation, Rank, and Nullity of a Linear Transformation $T:\R^2\to \R^3$ ←The current problem
6. Determine the Dimension of a Mysterious Vector Space From Coordinate Vectors
7. Find a Basis of the Subspace Spanned by Four Polynomials of Degree 3 or Less