Linear Algebra

Quiz 7. Find a Basis of the Range, Rank, and Nullity of a Matrix

Introduction to Linear Algebra at the Ohio State University quiz problems and solutions

Problem 320

(a) Let $A=\begin{bmatrix}
1 & 3 & 0 & 0 \\
1 &3 & 1 & 2 \\
1 & 3 & 1 & 2
\end{bmatrix}$.
Find a basis for the range $\calR(A)$ of $A$ that consists of columns of $A$.

(b) Find the rank and nullity of the matrix $A$ in part (a).

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Solution.

(a) Find a basis for the range $\calR(A)$

Note that the range $\calR(A)$ is the column space of $A$. Thus,
\[\calR(A)=\Span \{A_1, A_2, A_3, A_4\},\] where $A_i$ is the $i$-th column vector of $A$.

We use the “leading 1” method.
We reduce the matrix $A$ by elementary row operations as follows.
We have
\begin{align*}
A\xrightarrow{R_3-R_2} \begin{bmatrix}
1 & 3 & 0 & 0 \\
1 &3 & 1 & 2 \\
0 & 0 & 0 & 0
\end{bmatrix} \xrightarrow{R_2-R_1}
\begin{bmatrix}
1 & 3 & 0 & 0 \\
0 &0 & 1 & 2 \\
0 & 0 & 0 & 0
\end{bmatrix}.
\end{align*}

The first and the third column contain the leading 1’s.
Thus, the set $\{A_1, A_3\}$ is a basis of the range $\calR(A)$, which consists of the first and the third column vectors of $A$.

(b) Find the rank and nullity of the matrix $A$

By part (a), we know that $\{A_1, A_3\}$ is a basis of the range of $A$.
Hence the dimension of the range is $2$.
Thus the rank of $A$, which is the dimension of the range $\calR(A)$, is $2$

Recall the rank-nullity theorem. Since $A$ is a $3\times 4$ matrix, we have
\[\text{rank of } A + \text{nullity of } A=4.\] Since we know that the rank of $A$ is $2$, it follows from the rank-nullity theorem that the nullity of $A$ is $2$.

Comment.

These are Quiz 7 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017.

List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017

There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions.

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