# Find Bases for the Null Space, Range, and the Row Space of a $5\times 4$ Matrix ## Problem 604

Let
$A=\begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 &1 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 2 & 2 & 2\\ 0 & 0 & 0 & 0 \end{bmatrix}.$

(a) Find a basis for the null space $\calN(A)$.

(b) Find a basis of the range $\calR(A)$.

(c) Find a basis of the row space for $A$.

(The Ohio State University, Linear Algebra Midterm) Add to solve later

## Solution.

### (a) Find a basis for the null space $\calN(A)$.

To find a basis for the null space $\calN(A)$, we first find an algebraic description of $\calN(A)$.
Recall that $\calN(A)$ consists of the solutions of the homogeneous system $A\mathbf{x}=\mathbf{0}$. Let us find the solution of this system by applying elementary row operations to the augmented matrix $[A\mid \mathbf{0}]$ as follows.
$\left[\begin{array}{rrrr|r} 1 & -1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 1 & -1 & 0 & 0 & 0 \\ 0 & 2 & 2 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right] \xrightarrow[R_4-2R_2]{R_3-R_1} \left[\begin{array}{rrrr|r} 1 & -1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right] \xrightarrow{R_1+R_2} \left[\begin{array}{rrrr|r} 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right].$ Hence the general solution satisfies
\begin{align*}
x_1&=-x_3-x_4\\
x_2&=-x_3-x_4
\end{align*}
and hence the elements in $\calN(A)$ are of the form
$\mathbf{x}=\begin{bmatrix} x_1\\ x_2\\ x_3 \\x_4 \end{bmatrix}=\begin{bmatrix} -x_3-x_4\\ -x_3-x_4\\ x_3 \\x_4 \end{bmatrix} =x_3 \begin{bmatrix} -1\\ -1\\ 1 \\0 \end{bmatrix} +x_4 \begin{bmatrix} -1\\ -1\\ 0 \\1 \end{bmatrix} .$

It follows that the set
$\left\{ \,\begin{bmatrix} -1\\ -1\\ 1 \\0 \end{bmatrix}, \, \begin{bmatrix} -1\\ -1\\ 0 \\1 \end{bmatrix} \, \right\}.$ is a spanning set for $\calN(A)$.
It is straightforward to see that these vectors are linearly independent.
Hence it is a basis for $\calN(A)$.

### (b) Find a basis of the range $\calR(A)$.

Note that we have already computed the reduced row echelon form matrix for $A$ in part (a) (just ignore the augmented part).
The first two columns contain leading 1’s. Hence by the leading 1 method, we see that
$\left\{ \,\begin{bmatrix} 1\\ 0\\ 1 \\0 \\0 \end{bmatrix}, \, \begin{bmatrix} -1\\ 1\\ -1 \\2\\0 \end{bmatrix} \, \right\}$ is a basis for the range $\calR(A)$.

### (c) Find a basis of the row space of $A$.

Again by the reduced row echelon form matrix in $A$, we see that the set of the nonzero rows
$\left\{ \,\begin{bmatrix} 1\\ 0\\ 1 \\1 \end{bmatrix}, \, \begin{bmatrix} 0\\ 1\\ 1 \\1 \end{bmatrix} \, \right\}$ is a basis for the row space of $A$ by the row space method.

## 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 Add to solve later

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