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

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 ←The current problem
5. Matrix Representation, Rank, and Nullity of a Linear Transformation $T:\R^2\to \R^3$
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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