NullSpace

**LettyWong** · March 14th 2012, 04:47 AM

Hi, can anyone help me with this question?
Let S be the subspace of R⁴given by the solution set of the equations
x₂ - 3 x₃ = -x₂ - x₃ - x₄ and x₁ = x₁ + x₃ = x₁ - 2 x_{4

Find An example of a matrix for which S is the nullspace}

**girdav** · March 14th 2012, 08:09 AM

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".

**LettyWong** · March 14th 2012, 03:59 PM

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

**Deveno** · March 14th 2012, 08:33 PM

your equations are equivalent to:

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

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.

**HallsofIvy** · March 16th 2012, 07:26 AM

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 $<x_1, x_2, 0, 0>$ . 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 $x_2= 0$ .

**Deveno** · March 16th 2012, 09:55 AM

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

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