[SOLVED] Basis of subspace R

• Mar 28th 2010, 01:02 PM
mybrohshi5
[SOLVED] Basis of subspace R
Find a basis of the subspace of http://webwork2.asu.edu/webwork2_fil...8083707491.png consisiting of all vectors of the form

$\displaystyle \begin{bmatrix} x_1\\6x_1+x_2\\7x_1+3x_2\\-2x_1+6x_2\\ \end{bmatrix}$

i am not sure how to find this since there are variables $\displaystyle x_1$ and $\displaystyle x_2$

Thank you for any help :)
• Mar 28th 2010, 02:10 PM
Bruno J.
Hint :

$\displaystyle \begin{bmatrix} x_1\\6x_1+x_2\\7x_1+3x_2\\-2x_1+6x_2\\ \end{bmatrix} = x_1\begin{bmatrix} 1\\6\\7\\-2\\ \end{bmatrix} +x_2\begin{bmatrix} 0\\1\\3\\6\\ \end{bmatrix}$
• Mar 28th 2010, 02:25 PM
mybrohshi5
The basis are just the two you listed because when

$\displaystyle \begin{bmatrix}1&0&0 \\ 6&1&0 \\ 7&3&0 \\ -2&6&0 \\ \end{bmatrix}$

is in reduced row echelon form i got

$\displaystyle \begin{bmatrix}1&0& 0& 0 \\ 0& 1& 0& 0 \\ 0& 0& 0& 0& \\ 0& 0& 0& 0 \\ \end{bmatrix}$

which showed me that $\displaystyle \begin{bmatrix} 1\\6\\7\\-2\\ \end{bmatrix} and \begin{bmatrix} 0\\1\\3\\6\\ \end{bmatrix}$

are just the basis of the subspace R.

Does that seem correct?

Thanks for the hint :)
• Mar 28th 2010, 02:43 PM
Bruno J.
Well it's clear from the hint I gave you that every vector in your subspace can be written as a linear combination of the two vectors above. So the two vectors form a basis if and only if they are linearly independent. It's very easy to check for linear independence when you have only two vectors; two vectors are linearly dependent if and only if one is a multiple of the other. It's clearly not the case here! Your method is also valid (though I'm not sure why you added a third zero column to the matrix you reduced.)
• Mar 28th 2010, 02:47 PM
mybrohshi5
Thank you that makes sense and actually helps me a lot about understanding these.

One quick (and kind of a stupid question) but could you please quickly explain what you mean by "one is a multiple of the other."

I added the extra columns of zero's cause when i plug the matrix into my calculator to get the rref it needed extra columns of zeros in order to calculate it without my calculator saying error (Rofl)

Thanks again
• Mar 28th 2010, 03:55 PM
Bruno J.
Well, for example, the vectors $\displaystyle (1,1,1), (2,2,2)$ are linearly dependent because $\displaystyle (2,2,2)=2(1,1,1)$. But the vectors $\displaystyle (1,1,1),(2,2,3)$ are linearly independent because $\displaystyle (1,1,1) \neq k(2,2,3)$ for all $\displaystyle k \in \mathbb R$. That's what I mean by one being a multiple of the other.
• Mar 28th 2010, 07:05 PM
mybrohshi5
Perfect thank you so much for the help and explaining that :) i feel silly cause i knew it was easy but i just couldnt remember what a multiple meant haha.