# NullSpace

• Mar 14th 2012, 04:47 AM
LettyWong
NullSpace
Hi, can anyone help me with this question?
Let S be the subspace of R4 given by the solution set of the equations
x2 - 3 x3 = -x2 - x3 - x4 and x1 = x1 + x3 = x1 - 2 x4

Find An example of a matrix for which S is the nullspace
• Mar 14th 2012, 08:09 AM
girdav
Re: NullSpace
Write each equation on the form $a_1x_1+a_2x_2+a_3x_3+a_4x_4$ where $a_i$ are scalars. A possible matrix will "appear".
• Mar 14th 2012, 03:59 PM
LettyWong
Re: NullSpace
for the second equation, doesn't the x1 just cancel out?
• Mar 14th 2012, 08:33 PM
Deveno
Re: NullSpace

$2x_2 - 2x_3 + x_4 = 0$
$x_3 + 2x_4 = 0$
$x_3 = 0$

we can model this system of equations with the matrix equation:

$\begin{bmatrix}0&2&-2&1\\0&0&1&2\\0&0&1&0 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\\x_4 \end{bmatrix} = \begin{bmatrix}0\\0\\0 \end{bmatrix}$

the fact that x1 "cancels out" is indeed relevant, it shows we are completely free to choose ANY value for it.
• Mar 16th 2012, 07:26 AM
HallsofIvy
Re: NullSpace
Are you sure those are the right equations? $x_1= x_1+ x_3$ immediately gives $x_3= 0$. And $x_1= x_1- 2x_4$ immediately gives $x_4= 0$. So immediately, vectors in the solution space are of the form $$. And your other equation, $x_2- 3x_3= -x_2- x_3- x_4$ becomes $x_2= -x_2$ which is equivalent to $2x_2= 0$ or $x_2= 0$.
• Mar 16th 2012, 09:55 AM
Deveno
Re: NullSpace
it seems perfectly conceivable that A (the matrix in question) could have a 1-dimensional nullspace.