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

Vector Space Problems and Solutions

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 $A^{\trans}A$.

 
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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)
=
\left\{\mathbf{x}\in \R^3 \quad\middle|\quad \mathbf{x}=x_{3}
\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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