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tyteen4a03
 2 years ago
Best ResponseYou've already chosen the best response.0Just ask your question.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.05. 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
 2 years ago
Best 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]

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

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0Bro, I took linear algebra already, but I'd have to refamiliarize myself with this.

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0@KonradZuse, where in the book is that question?

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0sorry it's kind of screwed up in formatting.

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

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

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

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0where are the axioms you are referencing ?

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0(a) \[(\textbf u,\textbf v)=u_1 v_1+u_3v_3 \]

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

UnkleRhaukus
 2 years ago
Best 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

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I actually posted my work up a bit. Should I repost it?

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I 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
 2 years ago
Best 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

UnkleRhaukus
 2 years ago
Best 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

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

UnkleRhaukus
 2 years ago
Best 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

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.2that is the problem..axiom 4

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.2your work done on axiom 4 is not correct

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.2remember \(u\) and \(v\) are 3dimensional vectors

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

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.2or same thing ...what is \((v,v)\)

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I will check it out.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I guess I'm still confused how to prove axiom 4....

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.2you don't prove axiom 4...it is an axiom

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0errr "satisfy axiom 4"

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

Zarkon
 2 years ago
Best 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

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0ic so axiom 4 isn't = 0 thus fails?

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

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0i 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

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.2correct...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
 2 years ago
Best ResponseYou've already chosen the best response.0I'm so confused though :)

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0I 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
 2 years ago
Best ResponseYou've already chosen the best response.0how did you get = 1 ?

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

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0the 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
 2 years ago
Best ResponseYou've already chosen the best response.0So does Axiom 4 only fail for the first one?

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

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0hmm time to find it, and see if any other's fail anything....

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

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Does D by any chance fail Axiom 2?

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.2no...but it fails another axiom

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

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0C also seems to fail one.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0my program just crashed so h/o.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0all of my work is gone, I'm flipping out, hold on please.

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0SO d fails some axiom, but what about C @Zarkon ?

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

KonradZuse
 2 years ago
Best ResponseYou've already chosen the best response.0Damn 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
 2 years ago
Best ResponseYou've already chosen the best response.2\[(u,v)=u_1^2v_1^2+u_2^2v_2^2+u_3^2v_3^2\] \[(ku,v)=(ku_1)^2v_1^2+(ku_2)^2v_2^2+(ku_3^2)v_3^2\] \[=k^2(u_1^2v_1^2+u_2^2v_2^2+u_3^2v_3^2)=k^2(u,v)\]

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