# Basis and dimension of the solution space

## Homework Statement

Find (a) a basis for and (b) the dimension of the solution space of the homogeneous system of equations.

x - 2y + z = 0
y - z + w = 0
x - y + w = 0

## The Attempt at a Solution

(a)
[1 -2 1 0] => [1 0 -1 2]
[0 1 -1 1] => [0 1 -1 1]
[1 -1 0 1] => [0 0 0 0]
basis = {<1,0,-1,2>, <0,1,-1,1>}

(b) for b i make
x = z - 2w
y = z - w

would i set w = s and z = t?
if so.
[x] = [ t - 2s] =  + [-2]
[y] = [ t - s] = t + s[-1]
[z] = [ t ] =  + 
[w] = [ s ] =  + 

so the dimension would be the number of vectors
so dimension = 2?

im uncertain about that dimension part

LCKurtz
Homework Helper
Gold Member
I didn't check your arithmetic, but assuming it is OK try writing your solution like this:

$$\left ( \begin{array}{c}x\\y\\z\\w\end{array}\right) = t\left ( \begin{array}{c}1\\1\\1\\0\end{array}\right) +s\left ( \begin{array}{c}-2\\-1\\0\\1\end{array}\right)$$
and the answer should become apparent.

vela
Staff Emeritus
Homework Helper
You found that the solution of the system is given by

$$\begin{bmatrix}x\\y\\z\\w\end{bmatrix} = t\begin{bmatrix}1\\1\\1\\0\end{bmatrix}+s\begin{bmatrix}-2\\-1\\0\\1\end{bmatrix}$$

The vectors multiplying t and s are a basis for the solution space. Your answer to (a) is the basis for the row space of the matrix, not the solution space.

The dimension is just the number of vectors in the basis, so in this case, it's 2.

oh so for (a) i got it mixed up with the basis for the row space
the correct answer is { <1,1,1,0>, < -2,-1,0,1>}