# Orthonormal Basis of Null Space and Row Space

## 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)

## 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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