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I am supposed to find basis of the subspace of vector space $ \mathbb{R}^{3} $ of solutions of this linear system of equations: $y = \left\{ \begin{array}{ll} x_{1}+2x_{2}-x_{3}=0 \\ 2x_{1}+7x_{2}-2x_{3}=0 \\ -x_{1}+3x_{2}+x_{3}=0 \end{array} \right.$
I solve this system and I got: $x_{1}=x_{3}$ and $x_{2}=0$
$\vec x=\begin{bmatrix} x_1 \\0\\x_1\end{bmatrix} = x_{1}\begin{bmatrix} 1\\0\\1 \end{bmatrix} + 0 \begin{bmatrix} 0\\0\\0 \end{bmatrix}$

Is the basis : $ \begin{bmatrix} 1\\0\\1 \end{bmatrix}$ ?

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  • $\begingroup$ Is the first equation $x_1 + 2x_2 - x_3 = 0$? $\endgroup$ – Christopher Liu Dec 11 '13 at 19:43
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    $\begingroup$ If the computations are correct, a basis is $\langle (1,0,1)\rangle$. A vector (here at least), can't be a basis. $\endgroup$ – Git Gud Dec 11 '13 at 19:44
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    $\begingroup$ Checked the math, that is the correct basis. $\endgroup$ – Christopher Liu Dec 11 '13 at 19:47
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    $\begingroup$ No, a vector isn't a basis. A basis is a set. Assuming the calculations are correct, the answer would be $\left\langle \begin{bmatrix} \\1\\0\\1 \end{bmatrix}\right\rangle$or whatever notation you're using. $\endgroup$ – Git Gud Dec 11 '13 at 19:47
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    $\begingroup$ you certainly can, @MarcinMajewski, yet it is customary to use rounded parentheses, and it never minds whether the vector is a row or column one unless you have defined something definite about this. $\endgroup$ – DonAntonio Dec 11 '13 at 19:47
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The question was basically answered in comments, I will add an answer to remove it from the list of unanswered question.

By adding the first and the third equation we get $5x_2=0$ which implies $x_2=0$. If we now plug $x_2=0$ into remaining equations, we see, that they are all multiples of $x_1-x_2=0$.

Hence all solutions are of the form $(x_1,0,x_1)$. I.e., the solutions form the subspace $\{(x_1,0,x_1); x_1\in\mathbb R\}$.

Every vector in this subspace is a multiple of $(1,0,1)$, which means that vector $(1,0,1)$ generates the subspace.

Hence this subspace is one-dimensional and basis consists of a single vector $(1,0,1)$.

Of course, we could take any non-zero multiple of $(1,0,1)$ instead. For example, vector $(-2,0,-2)$ generates the same subspace.

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