# Linear Algebra subspaces and spans

• testme

## Homework Statement

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 below
by “⊕” and multiplication by scalars, denoted below by “$\odot$”.

You may assume without proof that E is a vector space.
Find a spanning set for E.

## Homework Equations

“ax+by +cz = d”⊕“ex+fy +gz = h” = “(a+e)x+(b+f)y +(c+g)z = d+h”
∀k ∈ R; k $\odot$ “ax + by + cz = d” = “ka x + kb y + kc z = k d”

## The Attempt at a Solution

to be completely honest I'm not sure. My first thoughts were to use something of the form
E = span{(1,0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)}

Though I know that's wrong because that would be the span of any vector in R4

Sorry that my attempt isn't all that great and I'm probably way off, I'm just fairly confused with what I'm supposed to find as the span.

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## Homework Statement

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 below
by “⊕” and multiplication by scalars, denoted below by “$\odot$”.

You may assume without proof that E is a vector space.
Find a spanning set for E.

## Homework Equations

“ax+by +cz = d”⊕“ex+fy +gz = h” = “(a+e)x+(b+f)y +(c+g)z = d+h”
∀k ∈ R; k $\odot$ “ax + by + cz = d” = “ka x + kb y + kc z = k d”

## The Attempt at a Solution

to be completely honest I'm not sure. My first thoughts were to use something of the form
E = span{(1,0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)}

Though I know that's wrong because that would be the span of any vector in R4

Sorry that my attempt isn't all that great and I'm probably way off, I'm just fairly confused with what I'm supposed to find as the span.

I think you have a good idea. While what you describe spans ##R^4##, it is certainly much like the given problem. Instead of (1,0,0,0) try writing it as 1x+0y+0z=0 to get the corresponding equation in your problem, and similarly for the others. What you know about how to do things in ##R^4## should be very helpful.

So then we'd have

1x + 0y + 0z = 0
0x + 1y + 0z = 0
0x + 0y + 1z = 0
0x + 0y + 0z = 1? <--- confused as to how that works

So then we'd have

1x + 0y + 0z = 0
0x + 1y + 0z = 0
0x + 0y + 1z = 0
0x + 0y + 0z = 1? <--- confused as to how that works

Your vector space is the set of all linear equations in x,y,z. That last one qualifies as a linear equation. It happens to be inconsistent but is still a linear equation.

So then would the span be

E = span{x = 0, y = 0, z = 0, 0 = 1}

Also if that's right then how can I justify that?

So then would the span be

E = span{x = 0, y = 0, z = 0, 0 = 1}

Also if that's right then how can I justify that?

No, that isn't right. We aren't talking about solutions of equations but the equations themselves. The vectors in your vector space are equations. Just like in ##R^4## where you might label your vectors ##e_1=(1,0,0,0),\, e_2 = (0,1,0,0),\, e_3 =(0,0,1,0),\, e_4=(0,0,0,1)## and prove ##\{e_1,e_2,e_3,e_4\}## is a basis, in your problem you might call $$v_1=1x + 0y + 0z = 0,v_2=0x + 1y + 0z = 0,v_3=0x + 0y + 1z = 0,v_4=0x + 0y + 0z = 1$$You need to prove that the equations represented by ##\{v_1,v_2,v_3,v_4\}## form a spanning set for all the equations, using the given operations.

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“ax+by +cz = d”⊕“ex+fy +gz = h” = “(a+e)x+(b+f)y +(c+g)z = d+h”
∀k ∈ R; k ⊙ “ax + by + cz = d” = “ka x + kb y + kc z = k d”

so using those two something like... (sorry if this is completely off, I'm not really sure how to prove that they satisfy all equations)

v1 ⊕ v2 ⊕ v3 ⊕ v4 = (a + e + i + l)x + (b + f + j + m)y + etc etc...

and show that it makes sense or is that completely off?

Your problem states you may assume without proof that ##E## is a vector space. You are only asked to find a spanning set. You have the "obvious" candidate ##S=\{v_1,v_2,v_3,v_4\}##. What do you have to do to show ##S## is a spanning set for ##E##?

We'd need to show that a linear combination of v1 v2 v3 and v4 will give us E?

We'd need to show that a linear combination of v1 v2 v3 and v4 will give us E?

That's the general idea but you need to phrase it more mathematically. Something like "If ...then ..." where you fill in the dots to give a statement you can prove. If might help to look at the definition in your text and maybe some examples there.

Something like..

Let h, j, k, and l be scalars, h, j, k, l ∈ R.
If
hv1 + jv2 + kv3 + lv4 = E
hv1 + jv2 + kv3 + lv4 = "ax + by + cz = d"
then S is a spanning set of E?

(in essense are we saying that an addition of planes that have been multiplied by some scalar give us E?) It's a weird concept to grasp

We'd need to show that a linear combination of v1 v2 v3 and v4 will give us E?

Something like..

Let h, j, k, and l be scalars, h, j, k, l ∈ R.
If
hv1 + jv2 + kv3 + lv4 = E
hv1 + jv2 + kv3 + lv4 = "ax + by + cz = d"
then S is a spanning set of E?

(in essense are we saying that an addition of planes that have been multiplied by some scalar give us E?) It's a weird concept to grasp

This is all to vague and sloppy to work with. You need to use proper terms and definitions. When you said "a linear combination of v1 v2 v3 and v4 will give us E" that actually makes no sense. A linear combination of 4 vectors (elements in a vector space) will give another element in the vector space. It can't be the same as E because E is a vector space. A vector isn't equal to a vector space. So I will say it again: Look up in your text the definition of what it means for a set of vectors ##S## to be a spanning set for a vector space ##E##. Then state it for your set of vectors ##S=\{v_1,v_2,v_3,v_4\}## and the vector space ##E##. You must do that to understand what you have to prove.

Well we know every span is a subspace.. so then can we check that S is a subspace of E?

Sorry if I'm not getting it, we don't have many notes on spans

Edit: Well.. I've been thinking about that and I don't think it's right, I'm going to try to find other definitions online and see if I can piece it together.

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