
NullSpace
Hi, can anyone help me with this question?
Let S be the subspace of R^{4 }given by the solution set of the equations
x_{2}  3 x_{3} = x_{2}  x_{3}  x_{4} and x_{1} = x_{1} + x_{3} = x_{1}  2 x_{4
Find An example of a matrix for which S is the nullspace }

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".

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

Re: NullSpace
your equations are equivalent to:
$\displaystyle 2x_2  2x_3 + x_4 = 0$
$\displaystyle x_3 + 2x_4 = 0$
$\displaystyle x_3 = 0$
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 x_{1} "cancels out" is indeed relevant, it shows we are completely free to choose ANY value for it.

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 $\displaystyle x_2= 0$.

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