# Thread: Finding a basis of R^3

1. ## Finding a basis of R^3

Find a basis of R^3 defined by the equation,

$\displaystyle 2x_{1}+3x_{2}+x_{3} = 0$

I took the above equation and solved for x3, assuming that was saying that x3 could be created using -2 copies of x1 minus 3 copies of x2...

$\displaystyle 2x_{1}+3x_{2} = -x_{3}$
$\displaystyle -2x_{1}-3x_{2} = x_{3}$

Then I solved this like I solve for a kernel.

$\displaystyle \begin{bmatrix}x_{1}\\ x_{2}\\ x_{3}\end{bmatrix}=\begin{bmatrix}t\\ s\\ -2t-3s\end{bmatrix}= t\begin{bmatrix}1\\ 0\\ -2\end{bmatrix}+s\begin{bmatrix}0\\ 1\\ -3\end{bmatrix}$

Which gave me a result of,

$\displaystyle \begin{bmatrix}1\\ 0\\ -2\end{bmatrix},\begin{bmatrix}0\\ 1\\ -3\end{bmatrix}$

However, the answer is:

$\displaystyle \begin{bmatrix}-3\\ 2\\ 0\end{bmatrix},\begin{bmatrix}-1\\ 0\\ 2\end{bmatrix}$

Where did I go wrong? Any help is extremely appreciated. Thanks!

2. Originally Posted by tangibleLime
Where did I go wrong? Any help is extremely appreciated. Thanks!

You are completely right. Take into account that the solution is not unique.

3. Ah, thank you for the clarification!

4. Also note that the problem should be "Find a basis of the subspace of R^3 defined by the equation,
$\displaystyle 2x_1+ 3x_2+ x_3= 0$"

Since that is one equation constraining the three variables, the subspace has dimension 3- 1= 2 and so your basis should contain 2 vectors, just as you say.