Solutions

Vector Space Problems p. 250 1-4

Each problem simply requires you to decide whether or not the set is a vector space (as well as being a subspace of Euclidean space, Rⁿ) so technically, all you have to provide is a yes or no answer. We will also provide some discussion.

Mathematically we have two directions to go in:

* if we believe it is a vector space then we need to prove that the two closure properties hold (not required by this particular homework assignment). This has to be done symbolically. Examples do not prove anything, although they help you to get a better understanding.

or

* provide a numerical counterexample showing it is not a vector space.

Step 1: read the description of the set until you understand what it really is. You may need to jot down some examples. If you don't understand what it means for something to be in the set, you will not get anywhere.

Step 2: do you feel it is or isn't a vector space?? If you think it isn't then ask:

Is the 0 vector in your set?? If not then you are done because the 0 vector (whatever it is) has to be in a set for it to be a vector space.

Can you come up with a counterexample?? Pick two members , add them together and see if they are still in the set. Scalar multiply an example and see if it is still in the set. If in either case the answer is no then you are done.

Step 3: you suspect it is a vector space. Then working with symbols (variables), take two generic elements (x₁ and x₂) of the set, add them together and see if what you get satisfies the definition of the set. Make sure and use the fact that each one of them was assumed to be in the set. If you don't, you won't get anywhere.

Next, scalar multiply either of your generic vectors by k and see if the result is still in the set.

Problem 1A { (a , b , 2) } (a plane, 2 units above the x-y plane )

Following the discussion above, we check to see if the 0 vector is in our set. Here, the 0 vector is (0,0,0) But our set requires that the third component be 2 so the 0 vector is not in our set. We are done. It's not a vector space.

OR we could have tried for a counterexample. Two members of this set are (1,1,2) and

(0,0,2). Their sum is (1,1,4). This is not in our set because the 3^rd component is not 2. So we have a counterexample and a second proof that it is not a vector space.

Problem 1B. { (a,b,c) | c = a + b } also a plane

Is the 0 vector in it? This is (0,0,0) and its 3^rd component does equal the sum of the first 2 so yes. This doesn't mean we have a vector space, just that we are not done. Having the 0 vector in your set is a necessary but not sufficient condition (look at the parabola from class Tuesday)

Let's try a couple of examples. (1,2,3) and (2,7,9) are both in the set (why?). Their sum is (3,9,12) This is also in the set because the third component is the sum of the second + first (12 = 3 +9) So we begin to suspect it is a vector space. In the context of this homework assignment, we answer YES based on knowing we have a plane through the origin.

Let's see what a formal proof looks like:

1) closure under addition. Let x1 = (a1, b1, a1+b1) and x2 = (a2, b2, a2+b2) be two members of the set. Is x1+ x2 in the set??

We compute x1 + x2 = (a1 + a2, b1 + b2, a1+a2+b1+b2). Is this in the set??? To be in the set means that the third component is the sum of the first two components. Is that true here?? YES, so we do have closure under addition.

2). closure under scalar multiplication. Let k be any scalar. Is kx1 in the set if x1 is??

kx1 = k(a1 , b1, a1+b1) = (ka1, kb1, k(a1+b1)). Is this in the set?? To be in the set, again, means that the third component is the sum of the first two. Clearly we do here so we have closure under scalar multiplication

Thus we have closure under addition and scalar multiplication so we have a vector space and we are done.

Problem 1c { (a,b,c) | c > 0 } This is the upper half of R³

Is it a vector space??? is the 0 vector in it?? is (0,0,0) in it? NO since to be in the set requires that the third component be positive.

If you did not go that route than let's think about the closure properties. Closure under scalar multiplication requires that k x1 be in the set for any value of k (positive or negative). Let's try an example:

take x1 = (1,2,3) which satisfies the definition since 3 >0 Now take k=-1. Then

k x1 = (-1,-2, -3) which is not in the set since -3 is not positive.

So we have a counterexample and thus we don’t have a vector space.

Problem 2A { (a,b,c) | a = c = 0 } This is really the y axis, a line through the origin. We thus suspect it is a vector space and answer YES.

The set is all vectors with 1^st and 3^rd components 0.

If we wanted to prove this, we would prove closure under addition and scalar multiplication.

x1 = (0, b1, 0) and x2 = (0, b2 , 0) as arbitrary members of the set.

closure under addition: x1 + x2 = (0, b1 + b2 , 0 ) which is in the set since the first and third component are 0

closure under scalar multiplication: k x1 = (0, kb1, 0 ) which also has 1^st and 3^rd components of 0 so it is also in the set.

Thus we have satisfied both properties of a vector space and we are done with the proof.

Problem 2B { (a,b,c) | a = -c }

What is this? a = -c is the same as a + c = 0 which is the same as x + z = 0. This is a plane through the origin.

( you could put the solutions in vector form y(0,1,0) + x (-1,0,1) to get more detail but it's not relevant )

Since it’s a plane through the origin we suspect it’s a vector space and answer YES.

To prove this, as earlier , requires proving closure under addition and scalar multiplication.

closure under addition: x1 + x2 = (a1 + a2, b1 + b2, c1 + c2). Is this in the set? To be in the set requires that

the first and third components are negatives of one another. Is this true?

Is a1 + a2 = - (c1 + c2)?? Yes, because we are assuming that x1 and x2 are in the set so we know from that

a1 = -c1 and a2 = -c2 so adding these equations gives us what we want.

closure under scalar multiplication: k x1 = (ka1, kb1, kc1). Is this in the set? Is the first component the negative of the third? This is the same as asking

is ka1 = -kc1 ??

x1 was assumed to be in the set so a1 = -c1 Multiplying both sides by k gives us what we need and we are done. We have shown both closure properties to be satisfied. We have a vector space.

Problem 2C { (a,b,c) | b = 2a + 1 }

What is this? In the x-y plane, we have the line y=2x + 1. Adding a third component gives us a plane.

This plane does not go through the origin. (0,0,0) does not satisfy 0 = 2(0) + 1. So we do not have a vector

space. We are done.

On problems 3 and 4 we will more briefly summarize the results.

{ (a,b,c,d) } in these problems

Problem 3A a – b = 2 not a vector space. The origin is not in it

Problem 3B c = a+ 2b d = a – 3b Is a vector space. The equations are linear and homogeneous. Prove this for Friday!! Use symbols as above

Problem 3C a= 0 and b=-d Is a vector space. The equations are linear and homogeneous.

Prove for friday

Problem 4A a=b=0 Is a vector space.

Problem 4B a = 1, b=0, a+d = 1 Is not a vector space. The origin is not in the set.

Problem 4C Is not a vector space. Not closed under scalar multiplication. Pick k to be -1. Then kx1 will not have a positive first component if x1 did. a>= 0 and b=<0

For Friday:

rework all problems 1-4:

clearly find and use counterexamples where appropriate

do actual proofs in the cases where the set is a vector space.

work neatly, include words and mathematics. You are making logical arguments.

Clarity and precision are key elements. Hand it in