Starting Strang's Introduction to Linear Algebra, 4th ed. 2009.

Already confused at the first "worked example" 1.1A on page 6.

"The linear combinations of v=(1,1,0) and w=(0,1,1) fill a plane. Describe that plane. Find a vector that is not a combination of v and w.
...
The vector (1,2,3) is not in the plane because 2 =/= 1+3. "

Huh?

Question

Starting Strang's Introduction to Linear Algebra, 4th ed. 2009.

Already confused at the first "worked example" 1.1A on page 6.

"The linear combinations of v=(1,1,0) and w=(0,1,1) fill a plane. Describe that plane. Find a vector that is not a combination of v and w.
...
The vector (1,2,3) is not in the plane because 2 =/= 1+3. "

Huh?

zepdrix · Answer

It's not in the plane because there is no combination of v and w which gives you that vector.

$\large\rm a\vec v+b\vec w=\left(\begin{matrix}1 \ 2\ 3\end{matrix}\right)$

There are no values for a and b, no multiples of v and w which give us (1,2,3).
But I also am confused by this 2 =/= 1+3 business, hmm. :(

zepdrix · Answer

I was never very good with this matrix business though :P
Lemme see if @kainui is around.

zepdrix · Answer

Bahhh he's offline >.<

holsteremission · Answer

"The linear combinations of $\mathbf v=(1,1,0)$ and $\mathbf w=(0,1,1)$ fill a plane..."

This refers to the fact that $\mathbf v$ and $\mathbf w$ are linearly independent, which, if you haven't covered the definition of independence yet, basically means you can't scale and add these two vectors in any way to get the zero vector without having to flat-out multiply $\mathbf v$ and $\mathbf w$ by $0$. Mathematically, there are no not-all-zero constants $c_1,c_2$ such that $c_1\mathbf v+c_2\mathbf w=\mathbf0$.

"... Describe that plane ..."

Any plane that is spanned (i.e. generated) by two vectors is the set of all linear combinations of those vectors. Geometrically, a plane is defined by three points, two of which can be the two generating vectors $\mathbf v$ and $\mathbf w$. Any third vector that lies in the plane could then be any combination of $\mathbf v$ and $\mathbf w$, say $\mathbf v+\mathbf w$.

"... Find a vector that is not a combination of v and w."

Any vector that is not a combination of $\mathbf v$ and $\mathbf w$ will not lie in the plane (since the plane consists of all combinations of these vectors), so you're looking for any vector $x$ that is impossible to achieve with any choice of constants $c_1,c_2$, i.e. you can choose any $x$ such that
$$c_1\begin{pmatrix}1\1\0\end{pmatrix}+c_2\begin{pmatrix}0\1\1\end{pmatrix}=\begin{pmatrix}x_1\x_2\x_3\end{pmatrix}$$is not satisfied.

Say we simplify the left hand side into a single vector:
$$c_1\begin{pmatrix}1\1\0\end{pmatrix}+c_2\begin{pmatrix}0\1\1\end{pmatrix}=\begin{pmatrix}c_1\c_1+c_2\c_2\end{pmatrix}$$Then $(1,2,3)$ can't be in the plane because that would require $c_1=1$ and $c_2=3$, but $c_1+c_2=1+3\neq2$.

zepdrix · Answer

Oooo neato :O

kainui · Answer

Nothing I can really add, but if you're familiar with it, in 3D space the cross product of two vectors is a new vector that's orthogonal to those, which is something you could use.

In general though you need to use the determinant (or equivalently Levi-Civita tensor) to find a vector that's perpendicular to the others.

mdhunstiger · Answer

I don't understand why $$x _{2}$$ is$$c _{1}+c _{2}$$
I've attached a screen cap of Strang's answer.  Are those "four particular vectors" just drawn from a hat?

zepdrix · Answer

$$\large\rm c \left(\begin{matrix}1 \ 1\ 0\end{matrix}\right)+d\left(\begin{matrix}0 \ 1\ 1\end{matrix}\right)= \left(\begin{matrix}c \ c\ 0\end{matrix}\right)+\left(\begin{matrix}0 \ d\ d\end{matrix}\right)=\left(\begin{matrix}c+0\c+d\0+d\end{matrix}\right)$$Does that help for your first question?

In regards to your second question,
Yes, he chose different values for $\large\rm c$ and $\large\rm d$ to give some examples of vectors in the plane.

When $\large\rm c=\pi$ and $\large\rm d=\pi$ we have,$$\large\rm \left(\begin{matrix}c+0\c+d\0+d\end{matrix}\right)=\left(\begin{matrix}\pi+0\ \pi+\pi\0+\pi\end{matrix}\right)$$

mdhunstiger · Answer

Yes, thank you.  I was confused because the way it's written makes it sound like the second component is always a combination of the first and third.  So, is that true or not?  That's just an artifact, in this case, of the first vector's first component being a 0 and the second vector's third component being a 0.  Is that correct?

zepdrix · Answer

Yes, correct. :)
The zeros gave us that interesting connection.