# Determine Bases for Nullspaces $\calN(A)$ and $\calN(A^{T}A)$

## Problem 713

Determine bases for $\calN(A)$ and $\calN(A^{T}A)$ when
$A= \begin{bmatrix} 1 & 2 & 1 \\ 1 & 1 & 3 \\ 0 & 0 & 0 \end{bmatrix} .$ Then, determine the ranks and nullities of the matrices $A$ and $A^{\trans}A$.

## Solution.

We will first compute
\begin{align*}
A^{T}
=&
\begin{bmatrix}
1 & 1 & 0 \\
2 & 1 & 0 \\
1 & 3 & 0
\end{bmatrix}
\;\text{and}\\
A^{T}A
=&
\begin{bmatrix}
1 & 1 & 0 \\
2 & 1 & 0 \\
1 & 3 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 2 & 1 \\
1 & 1 & 3 \\
0 & 0 & 0
\end{bmatrix}
=
\begin{bmatrix}
1+1 & 2+1 & 1+3 \\
2+1 & 4+1 & 2+3 \\
1+3 & 2+3 & 1+9
\end{bmatrix}
=
\begin{bmatrix}
2 & 3 & 4 \\
3 & 5 & 5 \\
4 & 5 & 10
\end{bmatrix}
.
\end{align*}

Next, we will find $\calN(A)$ by row reducing $[A\mid\mathbf{0}]$:
$\left[\begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ 1 & 1 & 3 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \xrightarrow{R_{2}-R_{1}} \left[\begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ 0 & -1 & 2 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \xrightarrow{-R_{2}} \left[\begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]$ $\xrightarrow{R_{1}-2R_{2}} \left[\begin{array}{ccc|c} 1 & 0 & 5 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] .$ Thus the solution to $A\mathbf{x}=\mathbf{0}$ is given by
$\mathbf{x} = \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} -5x_{3} \\ 2x_{3} \\ x_{3} \end{bmatrix} = x_{3} \begin{bmatrix} -5 \\ 2 \\ 1 \end{bmatrix} .$ Thus
\begin{align*}
\calN(A)
=
\begin{bmatrix}
-5 \\ 2 \\ 1
\end{bmatrix} \text{ for any } x_{3}\in\R
\right\}
=\Span\left\{
\begin{bmatrix}
-5 \\ 2 \\ 1
\end{bmatrix}
\right\}
.
\end{align*}
Therefore,
$\left\{ \begin{bmatrix} -5 \\ 2 \\ 1 \end{bmatrix} \right\}$ is a basis for $\calN(A)$.

Similarly, we will compute $\calN(A^{T}A)$ by row reducing $[A^{T}A\mid\mathbf{0}]$:
$\left[\begin{array}{ccc|c} 2 & 3 & 4 & 0 \\ 3 & 5 & 5 & 0 \\ 4 & 5 & 10 & 0 \end{array}\right] \xrightarrow[R_{3}-2R_{1}]{R_{2}-R_{1}} \left[\begin{array}{ccc|c} 2 & 3 & 4 & 0 \\ 1 & 2 & 1 & 0 \\ 0 & -1 & 2 & 0 \end{array}\right] \xrightarrow{R_{1}-2R_{2}} \left[\begin{array}{ccc|c} 0 & -1 & 2 & 0 \\ 1 & 2 & 1 & 0 \\ 0 & -1 & 2 & 0 \end{array}\right]$ $\xrightarrow{R_{3}-R_{1}} \left[\begin{array}{ccc|c} 0 & -1 & 2 & 0 \\ 1 & 2 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \xrightarrow{R_{1}\leftrightarrow R_{2}} \left[\begin{array}{ccc|c} 1 & 2 & 1 & 0 \\ 0 & -1 & 2 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]$ $\xrightarrow[\text{then}\;-R_{2}]{R_{1}+2R_{2}} \left[\begin{array}{ccc|c} 1 & 0 & 5 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] .$ Since the row reduced matrices for $[A\mid\mathbf{0}]$ and $[A^{T}A\mid\mathbf{0}]$ are identical, we can immediately conclude that
$\calN\left(A^{T}A\right) = \Span\left\{ \begin{bmatrix} -5 \\ 2 \\ 1 \end{bmatrix} \right\} ,$ and that
$\left\{ \begin{bmatrix} -5 \\ 2 \\ 1 \end{bmatrix} \right\}$ is a basis for $\calN(A^{T}A)$.

It follows that the nullities of $A$ and $A^{\trans}A$ are both $1$.
The rank-nullity theorem tells
$\text{rank of A} + \text{nullity of A} =3.$ Hence the rank of $A$ is $2$. Similarly, the rank of $A^{\trans}A$ is $2$.

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##### In which $\R^k$, are the Nullspace and Range Subspaces?

Let $A$ be an $m \times n$ matrix. Suppose that the nullspace of $A$ is a plane in $\R^3$ and...

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