Anyone good with inner products and their axioms?

- KonradZuse

Anyone good with inner products and their axioms?

- katieb

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- tyteen4a03

Just ask your question.

- KonradZuse

5. Please refer to Denition 1 on page 335 of the textbook. Given u = (u1; u2; u3) and v =
(v1; v2v3). Determine whether or not the following are inner products on R3. For those that
are not, list the axioms that do not hold.

- KonradZuse

(a) (u,v) =
u[1] v[1] + u[3] v[3]
(b) (u,v) =
2 2 2 2 2 2
u[1] v[1] + u[2] v[2] + u[3] v[3]
(c) (u,v) =
2 u[1] v[1] + u[2] v[2] + 4 u[3] v[3]
(d) (u,v) =
u[1] v[1] - u[2] v[2] + u[3] v[3]

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## More answers

- KonradZuse

I'm not 100% sure about Axiom 4, but it seems like so far 1 and 2 satify them all... thoughts? @tyteen4a03

- KonradZuse

@Hero save me.

- Hero

Bro, I took linear algebra already, but I'd have to re-familiarize myself with this.

- Hero

Hang on for a minute

- Hero

@KonradZuse, where in the book is that question?

- KonradZuse

not in the book.

- KonradZuse

sorry it's kind of screwed up in formatting.

- KonradZuse

1 second.

- KonradZuse

##### 1 Attachment

- KonradZuse

I posted part of my answer of part b, as well as part c and d. I can post everything if oyu wihs.

- KonradZuse

okay hold on.

- KonradZuse

here happy?

##### 1 Attachment

- KonradZuse

I just want to know if I'm doing this correctly/ what exactly is axiom 4 asking?

- KonradZuse

saying*

- KonradZuse

@UnkleRhaukus

- Zarkon

I guess I'll reply...a lot of your work is not correct..I'll let you think about that for a while

- UnkleRhaukus

where are the axioms you are referencing ?

- KonradZuse

one second.

- KonradZuse

##### 1 Attachment

- UnkleRhaukus

(a)
\[(\textbf u,\textbf v)=u_1 v_1+u_3v_3 \]

- KonradZuse

I was thinking that might fail because of the new u2v2, but not too sure?

- UnkleRhaukus

\[(\textbf u,\textbf v)=u_1 v_1+u_3v_3=u_3v_3+u_1 v_1=(\textbf v,\textbf u)\] the symmetry axiom holds

- KonradZuse

ok.

- KonradZuse

I actually posted my work up a bit. Should I repost it?

- KonradZuse

I did a and B I just was confused about axiom 4, and so far it doesn't seem like anything is going to fail. Thoughts?

- UnkleRhaukus

\[(\textbf u+\textbf v,\textbf w)=(\textbf u+\textbf v)_1\textbf w_1+(\textbf u+\textbf v)_3\textbf w_3\]\[\qquad\qquad\quad=u_1w_1+v_1w_1+u_3w_3+v_3w_3\]\[\qquad\qquad\quad=u_1w_1+u_3w_3+v_1w_1+v_3w_3\]\[\qquad\qquad\quad=(\textbf u,\textbf w)+(\textbf v,\text w)\]
the additivity axion holds

- UnkleRhaukus

\[ (k\textbf u,\textbf v)=ku_1 v_1+ku_3v_3\]\[\qquad\quad=k(u_1 v_1+u_3v_3)\]\[\qquad\quad=k(\textbf u,\textbf v)\]
the homogeneity axiom holds

- KonradZuse

##### 1 Attachment

- KonradZuse

not sure what was wrong with my work. I did exactly what the book/my other homework did for it.

- UnkleRhaukus

\[(\textbf v,\textbf v)=v_1v_1+v_3v_3\]\[\qquad\quad=v_1^2+v_3^2â‰¥0\]
\[(\textbf v,\textbf v)=0\iff \textbf v=\textbf 0\]
the positivity axiom holds

- Zarkon

that is the problem..axiom 4

- Zarkon

your work done on axiom 4 is not correct

- Zarkon

remember \(u\) and \(v\) are 3-dimensional vectors

- Zarkon

let \(u=v=<0,1,0>\)
what is \((u,v)\)

- Zarkon

or same thing ...what is \((v,v)\)

- KonradZuse

I will check it out.

- KonradZuse

I guess I'm still confused how to prove axiom 4....

- Zarkon

you don't prove axiom 4...it is an axiom

- KonradZuse

errr "satisfy axiom 4"

- Zarkon

(a) is not an inner product...I gave you a counter example

- Zarkon

((0,1,0),(0,1,0))=0*0+0*0=0
but (0,1,0) is not the zero vector

- KonradZuse

ic so axiom 4 isn't = 0 thus fails?

- Zarkon

it fails...
\((v,v)=0\Leftrightarrow v=\vec{0}\)

- UnkleRhaukus

i see what you are saying now @Zarkon \[(\textbf v,\textbf v)=v_1v_1+v_3v_3\]\[\qquad\quad=v_1^2+v_3^2â‰¥0\]\[(\textbf v,\textbf v)=0\iff \textbf v=(0,v_2,0)\not\equiv\textbf 0\]the positivity axiom does not hold in the general case

- Zarkon

correct...but one only needs so show a single counter example to show that it is not an inner product (i'm lazy that way :) )

- KonradZuse

I'm so confused though :)

- KonradZuse

I guess I can see how (0,1,0)(0,1,0) = 1 not 0. So all we have to do is multiply and see if = 0?

- UnkleRhaukus

how did you get = 1 ?

- KonradZuse

I thought that's what he said above, I guess I'm just too konfused hahaha :(

- UnkleRhaukus

the rule was; to add the product of the first components to the product of the third components ,
the rule satisfied the 4th axiom iff the rule only generates zero when the vector is the zero vector.
Zarkon has shown that that the second component of the vector could be anything , therefore there are many vectors that generate zero , not just the zero vector)

- KonradZuse

hmm ic.

- KonradZuse

So does Axiom 4 only fail for the first one?

- Zarkon

another one of your problems fails axiom 4

- KonradZuse

hmm time to find it, and see if any other's fail anything....

- KonradZuse

it seems like the others pass the other axioms, now to try and figure out which one fails #4 :P

- KonradZuse

Does D by any chance fail Axiom 2?

- Zarkon

no...but it fails another axiom

- Zarkon

I'll tell you that only one of the four is an inner product

- KonradZuse

geh..

- KonradZuse

C also seems to fail one.

- KonradZuse

my program just crashed so h/o.

- KonradZuse

all of my work is gone, I'm flipping out, hold on please.

- KonradZuse

SO d fails some axiom, but what about C @Zarkon ?

- Zarkon

(c) is an inner product...the only one

- KonradZuse

Damn now I got to look into b... Maybe that also fails axiom 4. D must fail axiom 3, maybe 4, IDk. Need to rework it, but flipping out trying to finish these other problems. Thanks for all the help I really appreciate it!

- Zarkon

\[(u,v)=u_1^2v_1^2+u_2^2v_2^2+u_3^2v_3^2\]
\[(ku,v)=(ku_1)^2v_1^2+(ku_2)^2v_2^2+(ku_3^2)v_3^2\]
\[=k^2(u_1^2v_1^2+u_2^2v_2^2+u_3^2v_3^2)=k^2(u,v)\]

- KonradZuse

axiom 3

- KonradZuse

do you think you could help me with my other problems? I would greatly appreciate it.

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