1. ## 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

2. ## Re: NullSpace

Write each equation on the form $\displaystyle$a_1x_1+a_2x_2+a_3x_3+a_4x_4$$where \displaystyle a_i$$ are scalars. A possible matrix will "appear".

3. ## Re: NullSpace

for the second equation, doesn't the x1 just cancel out?

4. ## Re: NullSpace

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

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

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

the fact that x1 "cancels out" is indeed relevant, it shows we are completely free to choose ANY value for it.

5. ## Re: NullSpace

Are you sure those are the right equations? $\displaystyle x_1= x_1+ x_3$ immediately gives $\displaystyle x_3= 0$. And $\displaystyle x_1= x_1- 2x_4$ immediately gives $\displaystyle x_4= 0$. So immediately, vectors in the solution space are of the form $\displaystyle <x_1, x_2, 0, 0>$. And your other equation, $\displaystyle x_2- 3x_3= -x_2- x_3- x_4$ becomes $\displaystyle x_2= -x_2$ which is equivalent to $\displaystyle 2x_2= 0$ or $\displaystyle x_2= 0$.

6. ## Re: NullSpace

it seems perfectly conceivable that A (the matrix in question) could have a 1-dimensional nullspace.