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 $x_{1} + x_{2} + x_{3} = 0$


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?

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.