Question

Determine is the following is an inner product space on R^2

Answer 1

\[A=\left(\begin{matrix}a_{1} \\ a_{2}\end{matrix}\right)\] \[B=\left(\begin{matrix}b_{1} \\ b_2\end{matrix}\right)\]
A dot B= \[a_1b_1-a_2b_1-a_1b_2+3_2b_2\]

Answer 2

yes? and

Answer 3

I'm having trouble verifying clause 1= A dot B=B dot A

Answer 4

ohh ok h/o

Answer 5

I...don't know. The equation I wrote is the defined dot product of A dot B

Answer 6

ok let's see if we can reverse engineer it

Answer 7

what is the det(AB) (the matrix composed of the two?

Answer 8

Can there be a determinant of a non-square matrix?

Answer 9

\[ \left [AB \right] = \left [a_1,b_1;a_2,b_2 \right]^T\]

Answer 10

???

Answer 11

that's the matrix I was talking about. The one composed of the two

Answer 12

\[\left[\begin{matrix}a_1 & b_1 \\ a_2 & b_2\end{matrix}\right]\]

Answer 13

Ahh. Determinant would be... a1b2-a2b1 yes?

Answer 14

yea kind of reminds you of their dot product right

Answer 15

also that 3 is that an a up there?

Answer 16

yes, but it's missing some terms from the defined dot product. Also yes, last term is 3a2b2

Answer 17

ok the 3... I'm not sure how to reverse that this method won't help give me a min to think about the new problem

Answer 18

ok

Answer 19

I should stop you here and say that I've now verified clauses 1,2, and 3. I'm having trouble proving clause 4 to be greater than or equal to 0

Answer 20

so lets see we can write \(A \dot\ B\) as:
\[1(a_1b_1)-1(a_2b_1)-1(a_1b_2)+3(a_2b_2)\] right?

Answer 21

Sorry, been working.

Answer 22

lol it's ok

Answer 23

so try doing \(A \dot\ A\) see what you get

Answer 24

Clause 4 states that V dot V is greater than or equal to zero. How do I prove that V dot V is greater than or equal to zero for all cases? I'm about to check if it's for all real V, or V greater than zero or whatnot.

Answer 25

A dot A gives me \[a_1^{2}-2a_2a_1+3a_2^{2}\]

Answer 26

and can you factor that?

Answer 27

Umm.

Answer 28

(a1+a2)(a1-3a2)?

Answer 29

also clause 4 is that d(A,B)≥0 for me is that yours too? with it =0 iff A=B

Answer 30

I have a dot A is greater than or equal to zero. and A dot A is equal to zero if, and only if v=0

Answer 31

Also, my above doesn't factor. The last term becomes negative in my factoring

Answer 32

hmm interesting. on the property and yea if it doesn't factor what could that imply?

Answer 33

That...it doesn't equal zero?

Answer 34

ok so then by mine it is not an IPS but hmm let me see if I can factor it and look at your def.

Answer 35

I'm going to bring in some backup just in case. @e.mccormick can you take a look at this?

Answer 36

Hey mccormick, thanks for coming.

Answer 37

/wave Looks like you are having fun.

Answer 38

Of course, Linear Algebra is just AAMAzing *rolls eyes*

Answer 39

Of course. Now, I can state that clause four is true for specific cases, but how do I say that's it's true for all v?

Answer 40

Well, for it to be an inner product, the result must be nonnegative. That is one thing.

Answer 41

Well yes. But how do I express that? The equation I got for A dot A is above

Answer 42

yes because as per my Clause 4 \(\sqrt{A\dot\ A}=0\)

Answer 43

My clause four states: A dot A is greater than or equal to zero, AND A dot A is equal to zero if, and only if, v=0

Answer 44

hmm... could we use both?

Answer 45

oh wait your 4 is my 1 sorry found it

Answer 46

no wonder i was confused

Answer 47

So, you are basically proving the four axioms of inner products, which are essentially the same as dot products, for this particualr case, right?

Answer 48

yep but with an odd definition of dot product. It's a weighted Euclidean inner product like problem

Answer 49

Right. With that weird definition on the dot product, I can't say that A dot A= A^2 and therefore greater than or equal to zero

Answer 50

basically you need to confirm all four, or break one. If one is a problem to test, move on and save it for last. See if any of the other three break.

Answer 51

b=1/3 (a-I Sqrt[2] a) but because it has an imaginary root that means it is negative

Answer 52

right?

Answer 53

he already did the other 3

Answer 54

Ah.... well... I did skip to the end here. LOL

Answer 55

lol cheater ;P

Answer 56

I don't understand your answer Fibonacci. There is no B in this problem, just A and A.

Answer 57

oops change the b to \(a_2\)

Answer 58

Right, but does it matter if a2 is negative? It's the dot product that cannot be negative, not one specific term.

Answer 59

Though it gives me a thought

Answer 60

Scratch my thought. Anyone else got one?

Answer 61

\[a_2=\frac{1}{3}(a-ia\sqrt{2})~and~a_2=\frac{1}{3}(a-ia\sqrt{2})\]

Answer 62

I don't quite understand the reasoning. Wait, if both a1 and a2 do not exist, then a1^2 is negative, and so is a2^2. And a1a2 is also negative. Therefore...

Answer 63

nah there are 2 \(a_2\)

Answer 64

So how would I express that?

Answer 65

well I think I would go for the \(A \dot\ A =0\) clause if you have an imaginary number that isn't in \( R^2\)

Answer 66

???

Answer 67

Both parts of clause 4 have to be true

Answer 68

If \(A \dot\ A=0\) at the points I mentioned above then 1) A is not contained in \(R^2\) and 2) \(A \not=0\)

Answer 69

I still don't understand. How would you write out clause 4 from start to finish?

Answer 70

Well to start, No, it is not an inner product space for \(R^2\) because...

Answer 71

do you agree?

Answer 72

So FibonacciChick666, you are saying it is not an inner product because non-zero vectors can have a norm of zero?

Answer 73

yes?... I'm confusing myself now

Answer 74

I think so, because that would invatidate it.

Answer 75

it's by contradiction surely

Answer 76

I'm lost

Answer 77

ok so have you had a proof writing class?

Answer 78

No, unfortunately. I did proofs back in geometry about 4 years ago, and we've done some proofs in linear algebra. No formal proof course though

Answer 79

okk that makes this especially hard(i just had the same problem this semester) so what I'm sure e.mccormick is typing is a proof

Answer 80

we will guide you through said proof ok?

Answer 81

If there is a way to make \(a_2>a_1\) and negative when it comes out of the root, which I think the \(a_2=\frac{1}{3}(a-ia\sqrt{2})~and~a_2=\frac{1}{3}(a-ia\sqrt{2})\) is showing, then you get 3 times a negative, and larger than the rest of the \(a_1^{2}-2a_2a_1+3a_2^{2}\). Wouldn't that break a few things?

Answer 82

Alright. After this question I've got one more for the board tonight. I can type it here for you two if you'd like to help (it's on cross product proofs) or I can make a new question.

Answer 83

new question would be preferable due to the long list we currently have going :)

Answer 84

ok, but after we figure this out. Now, Mccormick. I have to prove a2 larger than a1 (or smaller, if negative) for this to fail?

Answer 85

If we just assume A>0 for the start. WTS \(A \dot\ A\not=0\)

Answer 86

plug and chug

Answer 87

If I could just assume for a proof like this i wouldn't need to work on the problem

Answer 88

then get some values where it = 0

Answer 89

How?

Answer 90

we can assume because we will show the other in another step

Answer 91

That doesn't make sense to me

Answer 92

This would be easiest if I could come up with a number for a1 and a number for a2 such that (a1^2)+2(a1)(a2)+)a2^2)=0 where a1=/=0 or a2=/=0

Answer 93

well actually we wouldn't need another step because it violated our initial condition

Answer 94

Inductive proofs make an assumption and either confirm it or prove it wrong. If they confirm it, it is a proof. it they deny it it is a proof by contradiction that eliminates the posibility.

Answer 95

meaning?

Answer 96

You are allowed to assume things. You just need to catch up to what was assumed or blow it away.

Answer 97

So if I let A be greater than or equal to zero, how would I express that as a step and have it make sense?

Answer 98

Assume A>0