MHB Interpreting & Solving Nullspace

[cbarker1]

What is the interpretation of this nullspace? How to write the solution in parametric form if possible?

$N( \left(\begin{matrix}2&-1&0\\1&0&0\\0&0&0\end{matrix}\right))$

Using Gauss-Jordan Elimination

$\left(\begin{matrix}2&-1&0\\1&0&0\\0&0&0\end{matrix}\right)$ $\implies$ $\left(\begin{matrix}1&0&0\\0&1&0\\0&0&0\end{matrix}\right)$
Cbarker

 

[Euge]

Hi Cbarker1,

You have pivots in the first and second columns, so $x = 0$ and $y = 0$. The row of zeros indicate that $z$ is a free variable. You can write the solution parametrically as $x = 0$, $y = 0$, $z = t$, where $t$ is a parameter variable.

 

[Deveno]

Just for clarity's sake:

for a $m \times n$ matrix $A = (a_{ij})$ we can regard it as a function $\Bbb R^n \to \Bbb R^m$ that sends the vector:

$\begin{bmatrix}x_1\\ \vdots\\ x_n\end{bmatrix} \mapsto \begin{bmatrix}a_{11}&\cdots&a_{1n}\\ \vdots&\ddots&\vdots\\a_{m1}&\dots&a_{mn}\end{bmatrix}\begin{bmatrix}x_1\\ \vdots\\x_n\end{bmatrix}$

Then $N(A) = \left\{\begin{bmatrix}x_1\\ \vdots\\ x_n\end{bmatrix} \in \Bbb R^n: \begin{bmatrix}a_{11}&\cdots&a_{1n}\\ \vdots&\ddots&\vdots\\a_{m1}&\dots&a_{mn}\end{bmatrix}\begin{bmatrix}x_1\\ \vdots\\x_n\end{bmatrix} = \begin{bmatrix}0\\ \vdots\\0\end{bmatrix}\right\}$

In your case, then, we seek $(x,y,z)$ such that:

$\begin{bmatrix}2&-1&0\\1&0&0\\0&0&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}$

While Gaussian elimination (row-reduction) certainly works, it is possible to solve the related system of equations:

$2x - y +0z = 0$
$x + 0y + 0z = 0$
$0x + 0y + 0z = 0$

by *inspection*.

We can simplify this to:

$2x - y = 0$
$x = 0$

But I kept the $0$-coefficients in on purpose to show you that *any* $z$ will do.