Finding a basis of R^3

Printable View

May 1st 2011, 10:07 PM
tangibleLime

Finding a basis of R^3

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

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

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...

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

Then I solved this like I solve for a kernel.

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

Which gave me a result of,

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

However, the answer is:

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

Where did I go wrong? Any help is extremely appreciated. Thanks!
May 1st 2011, 10:59 PM
FernandoRevilla

Quote:

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.
May 1st 2011, 11:00 PM
tangibleLime

Ah, thank you for the clarification!
May 2nd 2011, 04:29 PM
HallsofIvy

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

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.

All times are GMT -8. The time now is 12:13 AM.

Copyright © 2005-2013 Math Help Forum. All rights reserved.

Copyright © 2005-2013 Math Help Forum. All rights reserved.