Interpreting & Solving Nullspace

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SUMMARY

The discussion focuses on interpreting the nullspace of the matrix $A = \left(\begin{matrix}2&-1&0\\1&0&0\\0&0&0\end{matrix}\right)$ using Gauss-Jordan elimination. The solution reveals that the first two variables, $x$ and $y$, are constrained to zero, while $z$ is a free variable. The parametric form of the solution is expressed as $x = 0$, $y = 0$, and $z = t$, where $t$ represents any real number. This illustrates the concept of nullspace in linear algebra, specifically for a $3 \times 3$ matrix.

PREREQUISITES
  • Understanding of linear algebra concepts, particularly nullspace and row reduction.
  • Familiarity with Gauss-Jordan elimination technique.
  • Knowledge of parametric equations in the context of linear systems.
  • Basic proficiency in matrix operations and vector spaces.
NEXT STEPS
  • Study the properties of nullspaces in linear transformations.
  • Learn about the implications of free variables in linear systems.
  • Explore advanced techniques in solving systems of equations, such as matrix rank and dimension.
  • Investigate the relationship between nullspace and column space in linear algebra.
USEFUL FOR

Students and professionals in mathematics, particularly those studying linear algebra, as well as educators teaching these concepts. This discussion is also beneficial for anyone looking to deepen their understanding of matrix theory and its applications.

cbarker1
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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
 
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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.
 
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.
 

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