Show Span of Vector Sum of Lengths 2 & 4 is 2-6

[ehrenfest]

Homework Statement

Let A be the set of all vectors with length 2 and let B be the set of all vectors of length 4.

How do you show that the span of the sum of a vector in A and a vector in B is all vectors with lengths between 2 and 4?

EDIT: change 4 o 6

I tried drawing triangles but that got me nowhere. Do I actually need to write out the components or something?

Let me know if I did not explain the problem okay.

Homework Equations

The Attempt at a Solution

 

[CompuChip]

Suppose $a \in A, b \in B$. Can you estimate the length ||a + b|| in terms of ||a|| and ||b||? Or you could find a limiting case, and show that it is a limiting case (e.g. write down two vectors for which the norm of the sum is maximal and show that it is smaller for any two others).

By the way, are you sure it's between 2 and 4? E.g. if a and b are parallel it will have length 2 + 4 = 6.

 

[ehrenfest]

Ah. You're right. It should be

"How do you show that the span of the sum of a vector in A and a vector in B is all vectors with lengths between 2 and 6?"

You can get the upper and lower bounds with the triangle inequality.

I am just not sure how to prove that it spans the annular region between the upper and lower bounds.

 

[JasonRox]

All you need to show that the two vectors, one of length 2 and another of length 4, can create all vectors of length between 2 and 6.

What is span? Do you know what that is?

 

[CompuChip]

Another hint: choose the angle between the vectors.

 

[ehrenfest]

Sorry. I am not sure what you mean CompuChip? There are three vectors in question.

Anyway, let $$\vec{v}$$ be an arbitrary vector with length between 2 and 6. All I need to show is that $$\vec{v}$$ is two units of distance from a circle of radius 4 around the origin. Drawing the picture this is clearly true.

The closest distance from $$\vec{v}$$ to the circle is less than 2 since v must be perpendicular to the circle somewhere and it is just intuitive (can someone explain that any better?)

There are then distances from v to the circle greater than 4 for the same reason (can someone explain why that is true succinctly?).

And since the distance of v from the circle vary continuously, we are done.

 

[HallsofIvy]

Sorry. I am not sure what you mean CompuChip? There are three vectors in question.

Anyway, let $$\vec{v}$$ be an arbitrary vector with length between 2 and 6. All I need to show is that $$\vec{v}$$ is two units of distance from a circle of radius 4 around the origin. Drawing the picture this is clearly true.

On the contrary, it doesn't even make sense! What do you mean by a vector being two units of distance from a circle of radius 4 around the origin? What do you mean by the distance from a vector to a circle?

The closest distance from $$\vec{v}$$ to the circle is less than 2 since v must be perpendicular to the circle somewhere and it is just intuitive (can someone explain that any better?)

There are then distances from v to the circle greater than 4 for the same reason (can someone explain why that is true succinctly?).

And since the distance of v from the circle vary continuously, we are done.

 

[ehrenfest]

The length of our vector minus the circle vector is the distance from our vector to a point on the circle!