# 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)*

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## 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

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

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