Interpreting & Solving Nullspace

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In summary: So let $z = t$.In summary, the interpretation of the nullspace for the given matrix is that $x = 0$, $y = 0$, and $z = t$, where $t$ is a parameter variable. Using Gauss-Jordan Elimination, we can see that the first and second columns have pivots, indicating that $x = 0$ and $y = 0$. The row of zeros indicates that $z$ is a free variable. Therefore, the solution can be written in parametric form as $x = 0$, $y = 0$, $z = t$, where $t$ is a parameter variable.
  • #1
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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  • #2
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.
 
  • #3
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.
 

1. What is the nullspace of a matrix?

The nullspace of a matrix is the set of all vectors that, when multiplied by the matrix, result in a zero vector. In other words, it is the set of all solutions to the equation Ax = 0, where A is the given matrix.

2. How is the nullspace related to the linear independence of a matrix?

The linear independence of a matrix is determined by the dimension of its nullspace. If the nullspace has a dimension of 0, it means that the matrix is linearly independent. If the nullspace has a dimension greater than 0, it means that the matrix is linearly dependent.

3. How can the nullspace be used to interpret a system of equations?

The nullspace can be used to determine if a system of equations is consistent or inconsistent. If the nullspace has a dimension of 0, it means that the system has a unique solution and is consistent. If the nullspace has a dimension greater than 0, it means that the system has infinitely many solutions and is inconsistent.

4. Can the nullspace be used to solve a system of equations?

Yes, the nullspace can be used to solve a system of equations. By finding a basis for the nullspace, we can construct a general solution to the system of equations.

5. How does the nullspace relate to the rank of a matrix?

The rank of a matrix is equal to the number of pivot columns in its reduced row echelon form. The dimension of the nullspace is equal to the number of non-pivot columns in the reduced row echelon form. Therefore, the rank and the nullspace are complementary in determining the dimension of the original matrix.

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