Algebra, the basis of a solution space

[FunkReverend]

Homework Statement

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

Homework Equations

The basis must span W and be independent.

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?

 

[Mark44]

Homework Statement

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

Homework Equations

The basis must span W and be independent.

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?

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.

 

[HallsofIvy]

To add just a little bit, you are saying that for any vector in the solution space, &lt;x_1, x_2, x_3, x_4&gt;, we must have x_2= -x_1- x_3 and x_4= (1/3)x_3- (1/3)x_1. That is, &lt;x_1, x_2, x_3, x_4&gt;= &lt;x_1, -x_1- x_3, x_3, (1/3)x_3- (1/3)x_4&gt;= x_1&lt;1, -1, 0, -1/3&gt;+ x_3&lt;0, -1, 1, 1/3&gt; 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 &lt;x_1, x_2, x_3, x_4&gt;= &lt;x_3- 3x_4, -4x_3+ 3x_4, x_3, x_4&gt;= x_3&lt; 1, -4, 1, 0&gt;+ x_4&lt;-3, 3, 0, 1&gt;. That gives another basis for the same subspace.