2
$\begingroup$

So I have this:

$$ F = \{(x_1, x_2, x_3) : x_1 - x_2 + x_3 = 0\} $$

I'm having trouble finding out the vectors whose linear combination generate the given subspace...

Basically I need something like: $x_1(a_1,b_1,c_1) + x_2(a_2,b_2,c_2) + x_3(a_3,b_3,c_3)$ so that given any value for $x_1, x_2$ and $x_3$ the equation gives me a vector in $F$ and so that each of the vectors above is in $F$

I can't find any ...

$\endgroup$
1
$\begingroup$

Note that if you know $x_1$ and $x_2$, then $x_3$ is uniquely determined. (So, this is a two-dimensional space.)

You can write $x_3=x_2-x_1$. Therefore, any such vector can be written as $$ \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}x_1\\x_2\\x_2-x_1\end{bmatrix}=\begin{bmatrix}x_1\\0\\-x_1\end{bmatrix}+\begin{bmatrix}0\\x_2\\x_2\end{bmatrix}=x_1\begin{bmatrix}1\\0\\-1\end{bmatrix}+x_2\begin{bmatrix}0\\1\\1\end{bmatrix} $$ So, a reasonable choice of basis is the two vectors $\langle 1,0,-1\rangle^T$ and $\langle0,1,1\rangle^T$.

$\endgroup$
  • $\begingroup$ It's quite funny as I get 3 different answers and they all look and seem correct to me, I still need to find a subspace that is complementary to F. Any thoughts on that? $\endgroup$ – Carlos Sá Mar 27 '17 at 17:59
  • $\begingroup$ A couple of ways to think about it. Your two-dimensional subspace is a plane; to be orthogonal to that means to be parallel to the plane's normal vector. Since your plane is $x_1-x_2+x_3=0$, the normal vector is $\langle1,-1,1\rangle^T$. You can pretty easily check that this vector is orthogonal to the two basis vectors above. And, it has to span the whole of the orthogonal complement, since $\mathbb{R}^3$ is 3-dimensional and your space is 2-dimensional. $\endgroup$ – Nick Peterson Mar 27 '17 at 18:03
  • $\begingroup$ That was it! After a night of thinking about it I saw what you were saying. Thanks alot :) $\endgroup$ – Carlos Sá Mar 28 '17 at 8:15
2
$\begingroup$

$F$ is the solution space of the set of linear equations $$x_1-x_2+x_3=0.$$

We always solve sets of linear equations by Gauss elimination, because it is guaranteed to find the complete set of solutions. In this case there is no work to be done, the equations -- the equation ! -- is already in row reduced form. So we can read off the solutions, they are $$x_3=\lambda,\ x_2=\mu,\ x_1=-\lambda+\mu\ \text{ for arbitrary} \lambda,\mu$$ and hence $$(x_1,x_2,x_3)=\lambda(-1,0,1)+\mu(1,1,0).$$

Gauss elimination guarantees not only do these two vectors $$(-1,0,1),\ \text{and}\ (1,1,0)$$span $F$, they actually form a basis of $F$.

$\endgroup$
2
$\begingroup$

we have $F=\{(x_1, x_2, x_3) : x_1 - x_2 + x_3 = 0\}=\{(x_1, x_2, x_3) : x_1 + x_3 =x_2 \}$ it means: $$\begin{align} x \in F&\leftrightarrow x=(x_1,x_1+x_3,x_3)\\ &\leftrightarrow x =(x_1,x_1,0)+(0,x_3,x_3) \\ &\leftrightarrow x=x_1(1,1,0) + x_3(0,1,1) \\ &\leftrightarrow x \in \langle(1,1,0),(0,1,1) \rangle\end{align}$$ therefore the vectors $(1,1,0)$ and $(0,1,1)$ generate $F$

$\endgroup$
  • $\begingroup$ So if I was to find an arbitrary H(complementary subspace to F) how could I do it? I can't seem to see the canonical vectors of $R^3$ in the generators $\endgroup$ – Carlos Sá Mar 27 '17 at 17:49

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.