# A Matrix Representation of a Linear Transformation and Related Subspaces

## Problem 164

Let $T:\R^4 \to \R^3$ be a linear transformation defined by
$T\left (\, \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} \,\right) = \begin{bmatrix} x_1+2x_2+3x_3-x_4 \\ 3x_1+5x_2+8x_3-2x_4 \\ x_1+x_2+2x_3 \end{bmatrix}.$

(a) Find a matrix $A$ such that $T(\mathbf{x})=A\mathbf{x}$.

(b) Find a basis for the null space of $T$.

(c) Find the rank of the linear transformation $T$.

(The Ohio State University Linear Algebra Exam Problem)

## Solution.

### (a) A matrix representation of a linear transformation

Let $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$, and $\mathbf{e}_4$ be the standard 4-dimensional unit basis vectors for $\R^4$.
Then the matrix representation for the linear transformation is given by the formula
$A:=\begin{bmatrix} T(\mathbf{e}_1) \quad T(\mathbf{e}_2) \quad T(\mathbf{e}_3)\quad T(\mathbf{e}_4) \end{bmatrix}.$ Then we calculate
\begin{align*}
T(\mathbf{e}_1)=T \left(\,\begin{bmatrix} 1\\0\\0\\0 \end{bmatrix} \, \right)
=\begin{bmatrix}
1\\3\\1
\end{bmatrix}.
\end{align*}

Similarly, we compute and obtain
$T(\mathbf{e}_2) =\begin{bmatrix} 2\\5\\1 \end{bmatrix}, T(\mathbf{e}_3)=\begin{bmatrix} 3\\8\\2 \end{bmatrix}, T(\mathbf{e}_4)=\begin{bmatrix} -1\\-2\\0 \end{bmatrix}.$

Therefore the matrix $A$ we are looking for is
$A=\begin{bmatrix} 1 & 2 & 3 & -1\\ 3 & 5 & 8 & -2\\ 1 & 1& 2 & 0 \end{bmatrix}.$

### (b) The null space of the linear transformation

First, note that the null space of a linear transformation $T$ is the same as the null space of the matrix $A$ that represents $T$.
Thus, we find a basis for the null space of the matrix $A$ we obtained in (a).

The null space of $A$ consists of the solutions of $A\mathbf{x}=\mathbf{0}$ and so we apply Gauss-Jordan elimination process to solve the linear equation as follows. We apply the elementary row operations to the augmented matrix $[A|\mathbf{0}]$ of $A$.
\begin{align*}
[A|\mathbf{0}]&= \left[\begin{array}{rrrr|r}
1 & 2 & 3 & -1 &0\\
3 & 5 & 8 & -2&0\\
1 & 1& 2 & 0 &0
\end{array} \right] \xrightarrow[R_3-R_1]{R_2-3R_1}
\left[\begin{array}{rrrr|r}
1 & 2 & 3 & -1 &0\\
0 & -1 & -1 & 1&0\\
0 & -1& -1 & 1 &0
\end{array} \right]\\
& \xrightarrow[\text{then } -R_2]{R_3-R_2}
\left[\begin{array}{rrrr|r}
1 & 2 & 3 & -1 &0\\
0 & 1 & 1 & -1&0\\
0 & 0 & 0 & 0 &0
\end{array} \right] \xrightarrow{R_1-2R_2}
\left[\begin{array}{rrrr|r}
1 & 0 & 1 & 1 &0\\
0 & 1 & 1 & -1&0\\
0 & 0 & 0 & 0 &0
\end{array} \right].
\end{align*}
Therefore the solutions are
\begin{align*}
x_1&=-x_3-x_4\\
x_2&=-x_3+x_4
\end{align*}
and thus the solution $\mathbf{x}$ of $A\mathbf{x}=\mathbf{0}$ is of the form
$\mathbf{x}=\begin{bmatrix} -x_3-x_4\\ -x_3+x_4\\ x_3\\ x_4 \end{bmatrix} =x_3 \begin{bmatrix} -1\\-1\\1\\0 \end{bmatrix}+x_4\begin{bmatrix} -1\\1\\0\\1 \end{bmatrix}.$ Hence we have
$\calN(T)=\calN(A)=\Span\left\{\, \begin{bmatrix} -1\\-1\\1\\0 \end{bmatrix}, \begin{bmatrix} -1\\1\\0\\1 \end{bmatrix} \, \right\}$ and the set
$B:=\left\{\, \begin{bmatrix} -1\\-1\\1\\0 \end{bmatrix}, \begin{bmatrix} -1\\1\\0\\1 \end{bmatrix}\, \right\}$ is a spanning set for the null space $\calN(T)$.

We claim that the set $B$ is linearly independent. Indeed, if we have a linear combination
$a_1\begin{bmatrix} -1\\-1\\1\\0 \end{bmatrix}+a_2\begin{bmatrix} -1\\1\\0\\1 \end{bmatrix}=\mathbf{0},$ then we have
$\begin{bmatrix} -a_1-a_2\\ -a_1+a_2\\ a_1\\ a_2 \end{bmatrix}=\mathbf{0}$ and thus $a_1=a_2=0$ and the set $B$ is linearly independent.

Since $B$ is a linearly independent spanning set for the null space $\calN(T)$, the set $B$ is a basis for $\calN(T)$.
In summary, we found a basis
$B:=\left\{\, \begin{bmatrix} -1\\-1\\1\\0 \end{bmatrix}, \begin{bmatrix} -1\\1\\0\\1 \end{bmatrix}\, \right\}.$

### (c) The rank of the linear transformation

We apply the rank-nullity theorem to find the rank of $T$.
From part (b), we see that nullity, the dimension of the null space $\calN(T)$, is $2$.

The linear transformation is a map from $\R^4$ to $\R^3$. Thus the rank-nullity theorem states that we have
$\text{rank of } T + \text{nullity of } T=4$ $\text{rank of } T + 2=4.$ Therefore the rank of the linear transformation $T$ is 2.

Let $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$ are vectors in $\R^n$. Suppose that vectors $\mathbf{u}_1$, $\mathbf{u}_2$ are orthogonal and the norm of $\mathbf{u}_2$...