# Find an explicit description of nul A by listing vectors that span the null space.

\begin{equation*} A = \begin{bmatrix} 1 & 2 & 3 & -7 \\ 0 & 1 & 4 & -6 \end{bmatrix} \end{equation*}

This problem aims to find the vectors in matrix A that span the null space. Null space of matrix A can be defined as the set of n column vectors x such that their multiplication of A and x produces a zero i.e. Ax = 0. These vectors will be the explicit description of null A.

Given Matrix:

$\begin{bmatrix} 1 & 2 & 3 & -7 & 0 \\ 0 & 1 & 4 & -6 & 0 \end{bmatrix}$

First thing to do is to find the parametric description for the homogeneous equation. To do that, we need to row reduce the homogeneous equation by some matrix $A$ times $x$ equals to $0$ vector, but we are going to convert it to its equivalent augmented matrix by row reduced echelon form.

Since the first pivot has a $0$ underneath it, we shall leave it as it is and operate the second pivot to eliminate the entry above $1$.

To make $0$ above $1$, we need to perform the following operation:

\begin{equation*} \begin{bmatrix} 1 & 2 & 3 & -7 & 0 \\ 0 & 1 & 4 & -6 & 0 \\ \end{bmatrix}R_1 \rightarrow R_1 – 2R_2    \begin{bmatrix} 1 & 0 & -5 & 5 & 0 \\ 0 & 1 & 4 & -6 & 0 \end{bmatrix} \end{equation*}

Now this row reduced echelon form is equivalent to the linear systems:

$x_1 – 5x_3 + 5x_4 = 0$

And the second row gives us:

$x_2 – 4x_3 + 6x_4 = 0$

$x_1$ and $x_2$ are our basic variables. Solving for these basic variables, we get the system as:

$x_1 = 5x_3 – 5x_4$

$x_2 = – 4x_3 + 6x_4$

Now $x_3$ and $x_4$ are free variables as they can be any real number. To find the spanning set, we rewrite this general solution as their parametric vector forms.

So the parametric Vector Form of $x$ is:

\begin{equation*} x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ \end{bmatrix} =  \begin{bmatrix} 5x_3 & -5x_4 \\ -4x_3 & 6x_4 \\ 1 & 0 \\ 0 & 1 \\  \end{bmatrix} \end{equation*}

where $x_3$ and $x_4$ are scalar quantities.

To find the spanning set of the null of matrix A, we need to see the column vectors.

So scalar multiples are the linear combination of the column vectors. Rewriting our answer gives us:

\begin{equation*} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ \end{bmatrix} =  x_3 \begin{bmatrix} 5 \\ -4 \\ 1 \\ 0 \\  \end{bmatrix} + x_4 \begin{bmatrix} -5 \\  6 \\  0 \\  1 \\  \end{bmatrix} \end{equation*}

### Numerical Results:

Spanning set for Null $A$ are these two vectors:

\begin{equation*} \left\{ \begin{bmatrix} 5 \\ -4 \\ 1 \\ 0 \\  \end{bmatrix} , \begin{bmatrix} -5 \\  6 \\  0 \\  1 \\  \end{bmatrix} \right\} \end{equation*}

• Note that every linear combination of these two column vectors is going to be an element of the null of $A$ because it solves the homogeneous equation.
• This means that the spanning set of Null($A$) is linearly independent, and $Ax=0$ has only the trivial solution.
• Also, when Null($A$) contains nonzero vectors, the number of vectors in the spanning set will be equal to the number of free variables in $Ax=0$.

### Example:

Find an explicit description of Null($A$) by listing vectors that span the null space.

\begin{equation*} A =\begin{bmatrix} 1 & 3 & -2 & -4 \\ 0 & 1 & 3 & -5 \end{bmatrix} \end{equation*}

Step 1 is to convert $A$ into Row Reduced Echelon Form to make $0$ above $1$ in second column. To do this, we need to perform the following operation:

\begin{equation*} \begin{bmatrix}1 & 3 & -2 & -4 & 0 \\ 0 & 1 & 3 & -5 & 0 \\ \end{bmatrix}R_1 \rightarrow R_1 – 3R_2    \begin{bmatrix} 1 & 0 & -11 & 19 & 0 \\ 0 & 1 & 3 & -5 & 0 \end{bmatrix} \end{equation*}

We first multiply the second row $R_2$ with $3$ and then subtract it from the first row $R_1$ to get a $0$ above $1$ in the second column.

Hence, $x_1$ and $x_2$ can then be found as:

$x_1 = 11x_3 – 19x_4$

$x_2 = – 3x_3 + 5x_4$

$x_1$ and $x_2$ are our basic variables.

Now $x_3$ and $x_4$ are free variables as they can be any real number. To find the spanning set, we rewrite this general solution as their parametric vector forms.

So the parametric Vector Form of $x$ is:

\begin{equation*} x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ \end{bmatrix} =  \begin{bmatrix} 11x_3 & -19x_4 \\ -3x_3 & 5x_4 \\ 1 & 0 \\ 0 & 1 \\  \end{bmatrix} \end{equation*}

\begin{equation*} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ \end{bmatrix} =  x_3 \begin{bmatrix} 11 \\ -3 \\ 1 \\ 0 \\  \end{bmatrix} + x_4 \begin{bmatrix} -19 \\  5 \\  0 \\  1 \\  \end{bmatrix} \end{equation*}

Spanning set for Null $A$ are these two vectors:

\begin{equation*} \left\{ \begin{bmatrix} 11 \\ -3 \\ 1 \\ 0 \\  \end{bmatrix} , \begin{bmatrix} -19 \\  5 \\  0 \\  1 \\  \end{bmatrix} \right\} \end{equation*}