# Thread: [SOLVED] Basis of subspace R

1. ## [SOLVED] Basis of subspace R

Find a basis of the subspace of 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

2. Hint :

$\displaystyle \begin{bmatrix} x_1\\6x_1+x_2\\7x_1+3x_2\\-2x_1+6x_2\\ \end{bmatrix} = x_1\begin{bmatrix} 1\\6\\7\\-2\\ \end{bmatrix} +x_2\begin{bmatrix} 0\\1\\3\\6\\ \end{bmatrix}$

3. 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

4. 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.)

5. 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

6. 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.

7. 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.