[SOLVED] Basis of subspace R

**mybrohshi5** · March 28th 2010, 01:02 PM

Find a basis of the subspace of

consisiting of all vectors of the form

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

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

Thank you for any help

**Bruno J.** · March 28th 2010, 02:10 PM

Hint :

$<br /> \begin{bmatrix} x_1\\6x_1+x_2\\7x_1+3x_2\\-2x_1+6x_2\\ \end{bmatrix} = x_1\begin{bmatrix} 1\\6\\7\\-2\\ \end{bmatrix} +x_2\begin{bmatrix} 0\\1\\3\\6\\ \end{bmatrix} <br />$

**mybrohshi5** · March 28th 2010, 02:25 PM

The basis are just the two you listed because when

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

is in reduced row echelon form i got

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

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

are just the basis of the subspace R.

Does that seem correct?

Thanks for the hint

**Bruno J.** · March 28th 2010, 02:43 PM

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.)

**mybrohshi5** · March 28th 2010, 02:47 PM

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

Thanks again

**Bruno J.** · March 28th 2010, 03:55 PM

Well, for example, the vectors $(1,1,1), (2,2,2)$ are linearly dependent because $(2,2,2)=2(1,1,1)$ . But the vectors $(1,1,1),(2,2,3)$ are linearly independent because $(1,1,1) \neq k(2,2,3)$ for all $k \in \mathbb R$ . That's what I mean by one being a multiple of the other.

**mybrohshi5** · March 28th 2010, 07:05 PM

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.

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