Find a basis for the set of vectors in\({\mathbb{R}^{\bf{2}}}\)on the line\(y = {\bf{5}}x\).

The set of allhomogeneousequation solutions, \(A{\bf{x}} = 0\), is \(Nul A\). It is called null space of A.

The line equation \(y = 5x\)can also be written as\(y - 5x = 0\).

In the matrix form, it is\(A = \left( {\begin{array}{*{20}{c}}5&{ - 1}\end{array}} \right)\).

From the above equation, \(y\) corresponds to the pivot position. So, \(y\) is the basic variable, and \(x\) is the free variable.

Let \(x = a\).

Substitute the value \(x = a\) in the equation \(y = 5x\) to obtain the general solution

\(y = 5a\).

Obtain the vector in the parametric form using \(x = a\) and \(y = 5a\).

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}x\\y\end{array}} \right) = \left( {\begin{array}{*{20}{c}}a\\{5a}\end{array}} \right)\\ = a\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right)\end{array}\)

It can also be written as shown below:

\(\left( {\begin{array}{*{20}{c}}x\\y\end{array}} \right) = x\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right)\)

Thebasis of vectoris shown below:

\( \Rightarrow \left\{ {\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right)} \right\}\)

Thus, \({\bf{x}} = x\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right)\), and a basis for \(Nul A\) is \(\left\{ {\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right)} \right\}\).