Orthonormal Basis of Null Space and Row Space

Ohio State University exam problems and solutions in mathematics

Problem 366

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

(a) Find an orthonormal basis of the null space of $A$.

(b) Find the rank of $A$.

(c) Find an orthonormal basis of the row space of $A$.

(The Ohio State University, Linear Algebra Exam Problem)
 
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Solution.

First of all, note that $A$ is already in reduced row echelon form.

(a) Find an orthonormal basis of the null space of $A$.

Let us find a basis of null space of $A$.
The null space consists of the solutions of $A\mathbf{x}=0$.
Since $A$ is in reduced row echelon form, the solutions $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}$ satisfy
\[x_1=-x_3 \text{ and } x_2=0,\] hence the general solution is
\[\mathbf{x}=\begin{bmatrix}
-x_3 \\
0 \\
x_3
\end{bmatrix}=x_3\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}.\] Therefore, the set
\[\left\{\, \begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix} \,\right\}\] is a basis of the null space of $A$.
Since the length of the basis vector is $\sqrt{(-1)^2+0^2+1^2}=\sqrt{2}$, it is not orthonormal basis.
Thus, we divide the vector by its length and obtain an orthonormal basis
\[\left\{\, \frac{1}{\sqrt{2}}\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix} \,\right\}.\]  

(b) Find the rank of $A$.

From part (a), we see that the nullity of $A$ is $1$. The rank-nullity theorem says that
\[\text{rank of $A$} + \text{ nullity of $A$}=3.\] Hence the rank of $A$ is $2$.

The second way to find the rank of $A$ is to use the definition of the rank: The rank of a matrix $B$ is the number of nonzero rows in a reduced row echelon matrix that is row equivalent to $B$.
Since $A$ is in echelon form and it has two nonzero rows, the rank is $2$.

The third way to find the rank is to use the leading 1 method. By the leading 1 method, we see that the first two columns form a basis of the range, hence the rank of $A$ is $2$.
 

(c) Find an orthonormal basis of the row space of $A$.

By the row space method, the nonzero rows in reduced row echelon form a basis of the row space of $A$. Thus
\[ \left\{\, \begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}, \begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix} \,\right\}\] is a basis of the row space of $A$.
Since the dot (inner) product of these two vectors is $0$, they are orthogonal.
The length of the vectors is $\sqrt{2}$ and $1$, respectively.
Hence an orthonormal basis of the row space of $A$ is
\[ \left\{\, \frac{1}{\sqrt{2}} \begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}, \begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix} \,\right\}\]  

Linear Algebra Midterm Exam 2 Problems and Solutions


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