# Algebra, the basis of a solution space

1. Oct 22, 2011

### FunkReverend

1. The problem statement, all variables and given/known data
Find the basis of the solution space W $\subset$ $\Re^{4}$
of the system of linear equations

$2x_{1}$ + $1x_{2}$ + $2x_{3}$ +$3x_{4}$ =0
$_{ }$
$1x_{1}$ + $1x_{2}$ + $3x_{3}$ = 0

2. Relevant equations
The basis must span W and be independent.

3. The attempt at a solution
Solving the above system, I get
$x_{2}$ = $-x_{1}$ - $x_{3}$
$x_{4}$ = $\frac{x_{3}-x_{1}}{3}$

With 2 degrees of freedom, $x_{1}$ and $x_{3}$,
so I must need a 2D basis. I separately fixed $x_{1}$ and $x_{3}$ to 1 and the other to zero and got the following vectors:
[1, -1, 0, -1/3] and [0, -3, 1, 1/3]
I feel like this is right, as I've been looking up some examples, but I'm not sure this spans all the solutions.

Am I on the right track?

2. Oct 22, 2011

### Staff: Mentor

Yes, your vectors span W.

A more systematic way to do things is the row-reduce your matrix, which gives this matrix:
$$\begin{bmatrix}1&0&-1&3\\0&1&4&-3 \end{bmatrix}$$

From this matrix you can read off your solutions as
x1 = x3 - 3x4
x2 = -4x3 + 3x4
x3 = x3
x4 = ........ x4

From this you might be able to see that any vector x in the solution space is a linear combination of these two vectors: <1, -4, 1, 0>T and <-3, 3, 0, 1>T.

3. Oct 22, 2011

### HallsofIvy

Staff Emeritus
To add just a little bit, you are saying that for any vector in the solution space, $<x_1, x_2, x_3, x_4>$, we must have $x_2= -x_1- x_3$ and $x_4= (1/3)x_3- (1/3)x_1$. That is, $<x_1, x_2, x_3, x_4>= <x_1, -x_1- x_3, x_3, (1/3)x_3- (1/3)x_4>= x_1<1, -1, 0, -1/3>+ x_3<0, -1, 1, 1/3>$ which makes it clear what a basis is.

Mark44 is saying that $x_1= x_3- 3x_4$ and $x_2= -4x_3+ 3x_4$ so that $<x_1, x_2, x_3, x_4>= <x_3- 3x_4, -4x_3+ 3x_4, x_3, x_4>= x_3< 1, -4, 1, 0>+ x_4<-3, 3, 0, 1>$. That gives another basis for the same subspace.