Vector Space-Spanning Linear Algebra

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Homework Statement



Bonus] Let E = {“ax+by+cz = d” | a; b; c; d ∈ R} be the set of linear equations
with real coefficients in the variables x, y and z. Equip E with the usual operations
on equations that you learned in high school: addition of equations, denoted here
by “⊕” and multiplication by scalars, denoted here by “~”, as follows:
“ax + by + cz = d” ⊕ “ex + fy + gz = h” = “(a + e)x + (b + f)y + (c + g)z = d + h”
and
∀k ∈ R; k ~ “ax + by + cz = d” = “ka x + kb y + kc z = k d”:
You may assume without proof that E is a vector space.
Find a spanning set for E.
(You must justify your answer.)

Homework Equations





The Attempt at a Solution



I have no idea what this question is asking...isnit the span just span{x,y,z}? o.o
 

Answers and Replies

  • #2
stephenkeiths
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It seems that by {x,y,z} you would mean a vector in [itex]R^{3}[/itex]
However, E seems to be a vector space where the elements (vectors) are equations of the form: ax+by+cz=d (equations of planes). So you want to find a set of planes, [itex]P_{1}[/itex], [itex]P_{2}[/itex], [itex]P_{3}[/itex] such that their linear combinations can represent any element of E (any plane in [itex]R^{3}[/itex]).

If you think the plane x+y+z=1 spans E, try to represent the plane x+2y+z=1 (which is in E) by a linear combination of x+y+z=1. Hopefully you see you'll need more than 1 plane to span E.

Hope this helps!
 
  • #3
1LastTry
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I dont think you can get x+2y+z=1 from x+y+z=1

What I did is: z= 1-y-x so (x,y, 1-y-x)
which is x(1,0,-1) + y(0,1,-1) + (0,0,1)
which means E= span{(1,0,-1), (0,1,-1), (0,0,1)}
And does that mean x+y+z=1 is not in E?
Sorry I am really confused by this, not good with planes
 
  • #4
stephenkeiths
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Linear Algebra can be really hard. Like this problem shows you need to think about 'vectors' abstractly. Every plane in [itex]R^{3}[/itex] is an element of E. And you need to find a set of planes that are a generating set for E. There is a very simple answer to this problem when you get past the abstractness.

Remember every element of E takes the form of ax+by+cz=d for some constants a,b,c, and d.

Now think about what the normal spanning vectors for [itex]R^{3}[/itex] are and try to compare this to what you might need for the planes. Remember you want to find some planes such that taking a linear combination of them, you can get ANY plane.

What does it mean to take a linear combination of 2 planes?
 
  • #5
1LastTry
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Damn...I know what linear combination is, I am just really unsure about how this question is asking the question...it is hard to explain...FOr normal spanning vectors of R^3, isn't it just a(1,0,0) + b(0,1,0) + c(0,0,1)? and I think that in order for a plane to be a vector space doesnt ax+by+cz=d, d has to =0? What do I need for the planes? I am confused about this is because my high school teacher decided not to teach about planes so I only learned lines n 2 and 3 space ..
 
  • #6
stephenkeiths
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You're right about linear combinations in [itex]R^{3}[/itex] (when you say a(1,0,0)+b(0,1,0)+c(0,0,1) you are taking linear combinations of the 3 vectors (1,0,0) (0,1,0) (0,0,1)!). Now look at the problem again and try to find the linear combination of the planes (1x=1) ⊕ 2~(1y=1). What plane does this give you.

The nice thing about linear algebra is you don't have to visualize any planes, or adding/scalar multiplying them. You just have to look at what the equations say. See what you can do.

Notice that when you take the linear combinations a(1,0,0)+b(0,1,0)+c(0,0,1) of those 3 vectors in [itex]R^{3}[/itex] you get a vector (a,b,c) for a,b,c in [itex]R[/itex]. This is the form of ANY vector in [itex]R^{3}[/itex]. What set of planes (equations of the form ax+by+cz=d) do you need to get ANY equation in that form (any plane).

Also remember that the planes themselves don't need to be vector spaces (although if they did you'd be right saying d=0!). This is a vector space of planes. The elements themselves don't necessarily need to be vector spaces.
 
  • #7
1LastTry
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cz=d-ax-by (ax, by, d-ax-by)

ax(1,0,-1)+by(0,1,-1)+d(0,0,1)

does it have anything to do with that^??

For: (1x=1) ⊕ 2~(1y=1) ...I don't quite get what you mean...x=1 +2y=2? which would be x+2y=3? Sorry I don't quite get that...


For any would it be a(1,0,0,0)+b(0,1,0,0) + c(0,0,1,0) + d(0,0,0,-1)? I am just guessing sorry this thing is really confusing the hell out of me.
 
  • #8
stephenkeiths
54
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When they define ⊕ and ~ in the problem they are defining how you add 2 vectors and scalar multiply (remember here vectors are planes). Thats why they say you add like you would add 2 equations in high school.

And the standard equation for a plane is ax+by+cz=d (in vector analysis i think?). That's why I am leaving them in that form, instead of writing them in (x=, y=, z=) form like you are (although I don't think there's anything really wrong with that).

I think you're on the right track. But you need to specify what a(1,0,0,0) means. This appears to be a vector in [itex]R^{4}[/itex] as opposed to an equation for a plane.

If you mean for the entries in the parenthesis to correspond to the coefficients for the plane then the last plane you wrote would be 0x+0y+0z=d ⇔ d=0 which isn't a plane.

You're very close though! This is quite a tricky problem.
 
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  • #9
1LastTry
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a(1,0,0,0) -->I was trying to put everything to one side so ax+by+cz-d=0 so....

I am trying all sorts of things now but I am now stuck

I tried this before (I never mentioned):

you know fromt he question (a+e)x + (b+f)y....? I multiplied (expanded) them out and got them to be ax+by+cz-d + ex+fy+gz-h=0(or d+h if you didn't move it to the other side)

I kept on thinking this has something to do with the solution, but I have no idea so far.

I tried as far as setting the coeficients of the + = to the coefficients of the ~ so (a+e)=ka which obvioiusly didnt work but I just want to let you know I am close to giving up on this. No idea what to do with this one.
 
  • #10
stephenkeiths
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I multiplied (expanded) them out and got them to be ax+by+cz-d + ex+fy+gz-h=0(or d+h if you didn't move it to the other side)

This is how you add equations, just like you surely did in high school! You just add each side together. The ~ is scalar multiplying an equation. For instance 2~x+y+z=1 becomes 2x+2y+2z=2

These 2 things just describe HOW you take linear combinations in this vector space.

if you had the 3 planes x=0, y=0, z=0 then you can take linear combinations to get any plane that goes through the origin. Now you just need a way control the "d" term, on the right side.

Thus I believe [itex]P_{1}[/itex] is x=0, [itex]P_{2}[/itex] is y=0, [itex]P_{3}[/itex] is z=0, and [itex]P_{4}[/itex] is x+y+z=1 is a set of planes that spans E.

If you want to get any plane ax+by+cz=d, take (a-1)~[itex]P_{1}[/itex]+(b-1)~[itex]P_{2}[/itex]+(c-1)~[itex]P_{3}[/itex]+d~[itex]P_{4}[/itex] does it.

So S={[itex]P_{1}[/itex],[itex]P_{2}[/itex],[itex]P_{3}[/itex],[itex]P_{4}[/itex]} spans E

Hopefully this will all make sense with the answer in hand.
 
  • #11
szynkasz
115
2
Should it be:
[tex](a-d)\sim P_1+(b-d)\sim P_2+(c-d)\sim P_3+d\sim P_4[/tex]
 
  • #12
1LastTry
64
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what syznkasz says seems to make sense
 
  • #13
stephenkeiths
54
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I think my statement is correct. Although, the other linear combination works too. Just replace some constants to get it in the form ax+by+cz=d.

*explicitly, let a'=a-d+1, b'=b-d+1, c'=c-d+1, d'=d and it will take the form a'x+b'y+c'z=d' and its shown*

Maybe replace the '+' with '⊕'

But this isn't all that important (IMO). The question wants a set that spans E.
 
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  • #14
1LastTry
64
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so it just p4~d since the others are 0?
 
  • #15
stephenkeiths
54
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Yes.
 
  • #16
1LastTry
64
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nvm i kinda get it but why.the a-1 b-1 ?
 
  • #17
1LastTry
64
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so the span is E=span(0,0,0,x+y+z=1)
 
  • #18
stephenkeiths
54
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nvm i kinda get it but why.the a-1 b-1 ?

Because that gives you (a-1)x+(b-1)y+(c-1)z=0
Then when you add P4 x+y+z=d

you get exactly ax+by+cz=d

Like I mentioned this isn't super important, its just the 'cleanest' way to do it (that I can see atleast).
 
  • #19
1LastTry
64
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so the span is E=span(0,0,0,x+y+z=1)
 
  • #20
stephenkeiths
54
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so the span is E=span(0,0,0,x+y+z=1)

no 0 isn't a plane!

E=span(x=0,y=0,z=0,x+y+z=1)
 
  • #21
1LastTry
64
0
that's what I meant, I was typing on a phone decided to leave the x y z out :P
 
  • #22
1LastTry
64
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I apologize if I am a total dumbass :P before I am just begin to learn this stuff and got confused during the lectures. But You are great help thanks! I get why that is the answer now. I got it with the (a-d)p1 + (b-d)p2.... thing...Thanks a lot!
 
  • #23
1LastTry
64
0
what is the difference between ⊕ and +? My Prof never mentioned this other than that it is for non-standard operations...
 
  • #24
stephenkeiths
54
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Glad I could help!

the ⊕ is just a way to denote 'adding' to 'vectors' in E. Just like ~ was denoting scalar multiplication.

Its used in an attempt to not confuse notation. Usually + and * are used for vector spaces like R. But feel free to use what ever symbol you would like. The point is that they are arbitrary, I could use & to represent add. Its just that it should be explicitly stated what a symbol means. And it should be used only for that purpose.

If you're taking an abstract linear algebra class you will probably see ⊕ again later in the course. It is generally used to denote the 'direct sum' of 2 vector spaces. You will see this when you learn about T-cyclic subspaces.
 
  • #25
1LastTry
64
0
how did you get x+y+z=1? I mean did you get it by first getting x=y=z=0? how did you know to multiple them by (a-1) and etc? I know it is the answer I checked and it made sense.but how did you figure it out?
 

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