# 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*)

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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

- True of False Problems and Solutions: True or False problems of vector spaces and linear transformations
- Problem 1 and its solution: See (7) in the post “10 examples of subsets that are not subspaces of vector spaces”
- Problem 2 and its solution: Determine whether trigonometry functions $\sin^2(x), \cos^2(x), 1$ are linearly independent or dependent
- Problem 3 and its solution (current problem): Orthonormal basis of null space and row space
- Problem 4 and its solution: Basis of span in vector space of polynomials of degree 2 or less
- Problem 5 and its solution: Determine value of linear transformation from $R^3$ to $R^2$
- Problem 6 and its solution: Rank and nullity of linear transformation from $R^3$ to $R^2$
- Problem 7 and its solution: Find matrix representation of linear transformation from $R^2$ to $R^2$
- Problem 8 and its solution: Hyperplane through origin is subspace of 4-dimensional vector space

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