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)
 
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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

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