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tyteen4a03 Group TitleBest ResponseYou've already chosen the best response.0
Just ask your question.
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
(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]
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
I'm not 100% sure about Axiom 4, but it seems like so far 1 and 2 satify them all... thoughts? @tyteen4a03
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
@Hero save me.
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
Bro, I took linear algebra already, but I'd have to refamiliarize myself with this.
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
Hang on for a minute
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
@KonradZuse, where in the book is that question?
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
not in the book.
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
sorry it's kind of screwed up in formatting.
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
1 second.
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
I posted part of my answer of part b, as well as part c and d. I can post everything if oyu wihs.
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
okay hold on.
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
here happy?
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
I just want to know if I'm doing this correctly/ what exactly is axiom 4 asking?
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
saying*
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
@UnkleRhaukus
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
I guess I'll reply...a lot of your work is not correct..I'll let you think about that for a while
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
where are the axioms you are referencing ?
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
one second.
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
(a) \[(\textbf u,\textbf v)=u_1 v_1+u_3v_3 \]
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
I was thinking that might fail because of the new u2v2, but not too sure?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[(\textbf u,\textbf v)=u_1 v_1+u_3v_3=u_3v_3+u_1 v_1=(\textbf v,\textbf u)\] the symmetry axiom holds
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
I actually posted my work up a bit. Should I repost it?
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[(\textbf u+\textbf v,\textbf w)=(\textbf u+\textbf v)_1\textbf w_1+(\textbf u+\textbf v)_3\textbf w_3\]\[\qquad\qquad\quad=u_1w_1+v_1w_1+u_3w_3+v_3w_3\]\[\qquad\qquad\quad=u_1w_1+u_3w_3+v_1w_1+v_3w_3\]\[\qquad\qquad\quad=(\textbf u,\textbf w)+(\textbf v,\text w)\] the additivity axion holds
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[ (k\textbf u,\textbf v)=ku_1 v_1+ku_3v_3\]\[\qquad\quad=k(u_1 v_1+u_3v_3)\]\[\qquad\quad=k(\textbf u,\textbf v)\] the homogeneity axiom holds
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
not sure what was wrong with my work. I did exactly what the book/my other homework did for it.
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[(\textbf v,\textbf v)=v_1v_1+v_3v_3\]\[\qquad\quad=v_1^2+v_3^2≥0\] \[(\textbf v,\textbf v)=0\iff \textbf v=\textbf 0\] the positivity axiom holds
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
that is the problem..axiom 4
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
your work done on axiom 4 is not correct
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
remember \(u\) and \(v\) are 3dimensional vectors
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
let \(u=v=<0,1,0>\) what is \((u,v)\)
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
or same thing ...what is \((v,v)\)
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
I will check it out.
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
I guess I'm still confused how to prove axiom 4....
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
you don't prove axiom 4...it is an axiom
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
errr "satisfy axiom 4"
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
(a) is not an inner product...I gave you a counter example
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
((0,1,0),(0,1,0))=0*0+0*0=0 but (0,1,0) is not the zero vector
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
ic so axiom 4 isn't = 0 thus fails?
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
it fails... \((v,v)=0\Leftrightarrow v=\vec{0}\)
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
i see what you are saying now @Zarkon \[(\textbf v,\textbf v)=v_1v_1+v_3v_3\]\[\qquad\quad=v_1^2+v_3^2≥0\]\[(\textbf v,\textbf v)=0\iff \textbf v=(0,v_2,0)\not\equiv\textbf 0\]the positivity axiom does not hold in the general case
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
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 :) )
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
I'm so confused though :)
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
how did you get = 1 ?
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
I thought that's what he said above, I guess I'm just too konfused hahaha :(
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
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)
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
hmm ic.
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
So does Axiom 4 only fail for the first one?
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
another one of your problems fails axiom 4
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
hmm time to find it, and see if any other's fail anything....
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
it seems like the others pass the other axioms, now to try and figure out which one fails #4 :P
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
Does D by any chance fail Axiom 2?
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
no...but it fails another axiom
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
I'll tell you that only one of the four is an inner product
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
C also seems to fail one.
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
my program just crashed so h/o.
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
all of my work is gone, I'm flipping out, hold on please.
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
SO d fails some axiom, but what about C @Zarkon ?
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
(c) is an inner product...the only one
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
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!
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.2
\[(u,v)=u_1^2v_1^2+u_2^2v_2^2+u_3^2v_3^2\] \[(ku,v)=(ku_1)^2v_1^2+(ku_2)^2v_2^2+(ku_3^2)v_3^2\] \[=k^2(u_1^2v_1^2+u_2^2v_2^2+u_3^2v_3^2)=k^2(u,v)\]
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
axiom 3
 one year ago

KonradZuse Group TitleBest ResponseYou've already chosen the best response.0
do you think you could help me with my other problems? I would greatly appreciate it.
 one year ago
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