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Found in: Page 191

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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

A basis for $$Nul A$$ is $$\left\{ {\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right)} \right\}$$.

See the step by step solution

## Step 1: Define the null space of a matrix

The set of all homogeneous equation solutions, $$A{\bf{x}} = 0$$, is $$Nul A$$. It is called null space of A.

## Step 2: State the basis for the set of vectors

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:

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

The basis of vector is 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 $$\left\{ {\left( {\begin{array}{*{20}{c}}1\\5\end{array}} \right)} \right\}$$.