If in R^{3 }we consider the subset S that contains two arbitrary lines that don't go through the origin, then may I say that Span S is the line joining (x,y,z) to the origin?

Thank you for your consideration of this matter...

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- Sep 24th 2012, 11:25 PMjojo7777777span s
If in R

^{3 }we consider the subset S that contains two arbitrary lines that don't go through the origin, then may I say that Span S is the line joining (x,y,z) to the origin?

Thank you for your consideration of this matter... - Sep 25th 2012, 04:55 AMemakarovRe: span s
- Sep 25th 2012, 07:33 AMjojo7777777Re: span s
If I have the following skew lines:

__x___{1}=(3,-1,0)+t(2,-2,1)__x___{2}=(1,0,0)+s(1,3,0)

, can I tell what is Span S? - Sep 25th 2012, 07:35 AMemakarovRe: span s
And what is S?

- Sep 25th 2012, 07:36 AMjojo7777777Re: span s
S is those two lines

- Sep 25th 2012, 07:44 AMemakarovRe: span s
The following three vectors are in S: (3,-1,0), (1,0,0) and (3,-1,0) + (2,-2,1) = (5,-3,1). Since

, the vectors are linearly independent. Therefore, they span the whole . - Sep 25th 2012, 08:05 AMjojo7777777Re: span s
Thank you very much for your help...!!!

But may I consider this <S> to be now also equal to the plain joining (x,y,z) to the origin?

To be more precisely, in my book I have a following example:

In R^{2}consider the singleton subset S={(x,y)}.Then <S> is the line joining (x,y) to the origin.

So I wonder whether in my example could be a__plain joining (x,y,z) to the origin__? - Sep 25th 2012, 11:26 AMDevenoRe: span s
normally, span(S), where S is a subset of a vector space V is: the smallest subspace U of V containing S.

if you have a single point (x,y,z) in R^{3}, the smallest subspace of R^{3}containing that point is:

U = {(ax,ay,az):a in R}, the line going through that point and the origin.

if you have TWO points, (x,y,z), and (x',y',z'), then they determine a plane in R^{3}.

now, if you have a line L, that does NOT go through the origin in R^{3}, then that is like choosing two points on that line, and determining a plane (that passes through the origin).

for example, let's pick a line L, and pick 4 points sort of randomly on it, and show the two pairs generate the same subspace.

i will use L = (1,1,0) + t(2,0,1).

for my first two points, i'll use t = -1, and t = 1. for my second two points, i'll use t = -3 and t = 4.

now (1,1,0) - (2,0,1) = (-1,1,-1), and (1,1,0) + (2,0,1) = (3,1,1). so U = span({(-1,1,-1),(3,1,1)}). clearly dim(U) = 2, since these 2 vectors are linearly independent.

and (1,1,0) - 3(2,0,1) = (1,1,0) - (6,0,3) = (-5,1,-3), while (1,1,0) + 4(2,0,1) = (1,1,0) + (8,0,4) = (9,1,4). so W = span({(-5,1,-3),(9,1,4)}), and we also have dim(W) = 2.

now, can we find a,b,c,d such that:

a(-1,1,-1) + b(3,1,1) = (-5,1,-3)

c(-1,1,-1) + d(3,1,1) = (9,1,4) ?

the first equation gives us:

-a + 3b = -5

a + b = 1

-a + b = -3

subtracting equation 3 from equation 1, gives us: 2b = -2, so b = -1. thus a = 2 (does this check out? -2 + 3(-1) = -5, 2 -1 = 1, and -2 - 1 = -3, yep).

so 2(-1,1,-1) - (3,1,1) = (-2,2,-2) - (3,1,1) = (-5,1,-3), so (-5,1,-3) is in U.

ok, the second equation gives us:

-c + 3d = 9

c + d = 1

-c + d = 4

from which we rapidly find that c = -3/2, d = 5/2, and as a check: (-3/2)(-1,1,-1) + (5/2)(3,1,1) = (3/2,-3/2,3/2) + (15/2,5/2,5/2)

= (18/2,2/2,8/2) = (9,1,4). so (9,1,4) is in U as well, so W = U.

in fact, it is easy to see that both of these are equal to the subspace generated by {(1,1,0),(2,0,1)} (just use the t-values we used originally).

so when you have a LINE not through the origin (equivalently, two points...which uniquely determine a line), you get a "plane joining the line through the origin" in a unique way (imagine drawing lines going through every point of L through the origin, to get an idea of "what the plane looks like". you'll have to add "one more line": the line through the origin PARALLEL to L. another way to visualize it, is draw a line through the origin parallel to L, and imagine the plane as a sheet of "perpendicular lines" going through both L and the line through the origin parallel to L).

if you have a plane P that does NOT go through the origin (and isn't "degenerate", that is, we have 3 linearly independent vectors (points) in it), in much the same way we find the subpsace generated by P is ALL of R^{3}.

although this may not be said explicitly in your book, you are very close to getting a picture of what "quotient spaces" look like. these are vector spaces where the vectors themselves are LINES or PLANES (or n-PLANES, in higher dimensions) instead of merely points. for example, in R^{3}, the quotient space obtained from a plane, is a lot like a deck of cards (where we have just one dimension: our "height" in the stack), and the quotient space obtained from a line, is like a bundle of straws (where we need 2 dimensions to say "which straw we're at"). - Sep 26th 2012, 06:13 AMjojo7777777Re: span s
Thank you very much...that is an AMAZING explanation...

If you let me I would like to ask one more question:

Just by looking at U = {(ax,ay,az):a in R} how can I tell that it is the line? - Sep 26th 2012, 07:07 AMHallsofIvyRe: span s
Are you saying that x, y, and z are fixed numbers and a is parameter? Then you can tell it is a line because the three components are

**linear**functions of the parameter. (The other way is not true: if U= {(xt^2, yt^2, zt^2): t in R} is linear even though the functions are not linear- it is the same line as {(xs, ys, zs): s in R}.)

More precisely, taking a= 0, p, and q we have A= (0,0,0), B= (px, py, pz), and C= (qx, qy, qz). The vector AB= <px, py, pz> and the vector AC= <qx, qy, qz> are parallel- B is (q/p) times A- for all p and q (p not 0, of course). That is sufficient to show that this is a line. - Sep 28th 2012, 05:58 AMjojo7777777Re: span sQuote:

The other way is not true: if U= {(xt^2, yt^2, zt^2): t in R} is linear

- Sep 28th 2012, 10:04 AMDevenoRe: span s
one has to be careful: there are two uses of the word "linear", which literally means: "like a line".

one use of the word has to do with polynomial expressions, for example: x - a is called a LINEAR factor of x^{2}- a^{2}.

this is because the graph of y = x - a is a straight line.

however, in linear algebra, linear means something slightly "different", the function y = x - a does NOT give "a linear space" (a vector space), because it does not go through the origin (unless a = 0).

subspaces of R^{3}(which we can imagine is the space we live in: it's not, really, but it helps with visualization), have a "linear quality" to them, 1-dimensional subspaces ARE lines (but not all lines are subspaces), 2-dimensional subspaces ARE planes (but not all planes are subspaces), and calculations in these subspaces have a "straight-line-ness" to them (a property we call linearity, meaning adding can be done with linear combinations of "special vectors" (basis elements)).

let's look at a 1-dimensional space in some more detail.

a 1-dimensional subspace has a basis of size 1: {v}. since the only basis vector we have is v (let's say v = (x_{0},y_{0},z_{0})), the only linear combinations are of the form cv = c(x_{0},y_{0},z_{0}) = (cx_{0},cy_{0},cz_{0}).

for example, suppose v = (1,1,2). then all possible "linear combinations" of {v} are: (c,c,2c), where we pick c to be any real number. so what we get for <v> is:

the line L that passes through (0,0,0) and (1,1,2). let's write this line in more "conventional terms", as (0,0,0) + t(1,1,2) = t(1,1,2).

you can imagine this as a vector pointing in the (1,1,2)-direction with it's "tail" at the origin and the "arrowhead" at (1,1,2), and then stretched by a factor of t (if t = -1, the vector points in "the other (opposite) direction" as (1,1,2), and we have the vector (-1,-1,-2)).

all of the vectors in <v> lie on L: we can add points on a line-if we want to add (2,2,4) to (3,3,6), we do this:

(2,2,4) + (3,3,6) = 2(1,1,2) + 3(1,1,2) = 2v + 3v = (2 + 3)v = 5v = (5,5,10).

in fact, "v" is "just along for the ride", all the addition takes place in the "coefficient" of v (to add 2v and 3v above, we added 2+3 and "scalar multiplied" that to v).

BUT:

just because we have a line L, does NOT mean we have a vector space. the math gets a bit more complicated in R^{3}, so let's look at the plane, R^{2}, instead.

say we have a line:

y = mx + b.

this is really the subset of R^{2}:

{(x,mx+b) : x in R} (here, m and b are CONSTANTS).

is this a vector space? well, if it IS, it has to have a 0-vector:

a vector z with v + z = v, for ALL vectors v. let's find out what the 0-vector has to be:

if (x,mx+b) + (z,mz+b) = (x,mx+b), then:

x + z = x (as real numbers)

mx + b = mx + mz + 2b (also as real numbers).

the first equation tells us z = 0. plugging that into the second equation tells us:

mx + b = mx + 2b

b = 2b

b = 0

so, unless b = 0, there ISN'T a 0-vector, and thus no vector space. if, however, b = 0 (we have a line through the origin), then all is well, our 0-vector is (0,m0) = (0,0), as it should be. - Oct 2nd 2012, 12:11 PMjojo7777777Re: span s
I have no words to express my appreciation for your help....

For clarity

Quote:

subspaces of R3 (which we can imagine is the space we live in: it's not, really, but it helps with visualization),

- Oct 2nd 2012, 12:32 PMDevenoRe: span s
yes, the geometry in R

^{3}is euclidean (at least with the standard inner product (dot product) used to define the concept of "angle" and thus "parallel" and "perpendicular").

informally, lines are: "straight", there's no inherent curvature in the space.

in actual point of fact, it appears the space we reside in HAS curvature, but at the scale of perception we normally think of, it is so slight as to be negligible. near large gravitational area (like black holes) this curvature becomes more pronounced, which we perceive as "light-bending".

however, i recommend mastering the "linear approximation" first, before tackling the more complicated mathematics of "locally euclidean" spaces (but trust me, the calculus you learned does get used). - Oct 2nd 2012, 02:51 PMjojo7777777Re: span s
Thank you very much...!!!