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Given the subspace $W= \{(x_{1}, x_{2}, x_{3}): x_{1} + x_{2} + x_{3} = 0\}$

Is the Set $S= \{(-1, -1, 2), (-3, 2, 1)\}$ a basis for W.

What I did was, I first checked if it was linearly independent and found that it was. Next I know I need to check if it spans the subspace but I'm having trouble with it. I don't exactly know how to span a subspace.

The second question is Explain why a basis of R3 cannot consist entirely of vectors $(x_{1}, x_{2}, x_{3}$) where $x_{1} + x_{2} + x_{3} = 0$

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Verifying linear independence is the correct start. To see why $S$ spans $W$, let $\vec{x}\in W$ be arbitrary. We know that $\vec{x}$ is of the form $(x_1,x_2,x_3)$ where $x_1+x_2+x_3=0$. So, we have that $$\vec{x} = x_1 (1,0,-1) + x_2(0,1,-1) $$

EDIT: Why do we know that the above equation holds? This is a matter of describing $W$ as the span of a set of linearly independent vectors. Notice that since $x_1+x_2+x_3=0$, we can completely determine $x_3$ by ranging values of $x_1$ and $x_2$. For example, consider $x_1=1$ and $x_2=0$. In this case, $x_3=-1-0=-1$, so this yields $\vec{x}=(1,0,-1)$. Another case is $x_1=0$ and $x_2=1$. Here, $x_3=0-1=-1$ so this yields $\vec{x}=(0,1,-1)$. In fact, this combination is linearly independent, which you can verify.

By definition of span, we need to show that there are $c_1,c_2\in \mathbb{R}$ such that $$\vec{x} = c_1 (-1, -1, 2) + c_2 (-3, 2, 1)$$ so, we obtain the system of equations \begin{align*} x_1 &= -c_1 -3 c_2 \\ x_2 &= -c_1 + 2c_2 \\ -x_1-x_2 &= 2c_1+c_2 \end{align*} The solution to this system, Mathematica tells me, is \begin{align*} c_1 &= \frac{1}{5}(-2x_1 - 3x_2) \\ c_2 &= \frac{1}{5}(-x_1+x_2) \end{align*} Now, $\vec{x}=(x_1,x_2,x_3)$ was arbitrary, so this holds for all such $\vec{x}\in W$. Therefore, $S$ is a basis of $W$.

There are many ways to explain why vectors in $W$ cannot form a basis for $\mathbb{R}^3$, so I'll give you some intuition: notice that $x_3$ is totally determined by the values of $x_1$ and $x_2$. In other words, every vector $\vec{x}\in W$ can be described by two variables. Can all vectors in $\mathbb{R}^3$ be described by two variables?

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  • $\begingroup$ Yes makes sense. Thanks. Also if you don't mind, could you explain how you got the vector for the subspace? $\endgroup$ – samir91 Oct 30 '14 at 22:42

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