# Find Matrix Representation of Linear Transformation From $\R^2$ to $\R^2$

## Problem 370

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

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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...

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