KonradZuse
Anyone good with inner products and their axioms?
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tyteen4a03
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Just ask your question.
KonradZuse
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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
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(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
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I'm not 100% sure about Axiom 4, but it seems like so far 1 and 2 satify them all... thoughts? @tyteen4a03
KonradZuse
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@Hero save me.
Hero
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Bro, I took linear algebra already, but I'd have to re-familiarize myself with this.
Hero
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Hang on for a minute
Hero
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@KonradZuse, where in the book is that question?
KonradZuse
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not in the book.
KonradZuse
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sorry it's kind of screwed up in formatting.
KonradZuse
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1 second.
KonradZuse
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KonradZuse
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I posted part of my answer of part b, as well as part c and d. I can post everything if oyu wihs.
KonradZuse
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okay hold on.
KonradZuse
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here happy?
KonradZuse
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I just want to know if I'm doing this correctly/ what exactly is axiom 4 asking?
KonradZuse
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saying*
KonradZuse
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@UnkleRhaukus
Zarkon
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I guess I'll reply...a lot of your work is not correct..I'll let you think about that for a while
UnkleRhaukus
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where are the axioms you are referencing ?
KonradZuse
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one second.
KonradZuse
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UnkleRhaukus
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(a)
\[(\textbf u,\textbf v)=u_1 v_1+u_3v_3 \]
KonradZuse
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I was thinking that might fail because of the new u2v2, but not too sure?
UnkleRhaukus
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\[(\textbf u,\textbf v)=u_1 v_1+u_3v_3=u_3v_3+u_1 v_1=(\textbf v,\textbf u)\] the symmetry axiom holds
KonradZuse
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ok.
KonradZuse
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I actually posted my work up a bit. Should I repost it?
KonradZuse
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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
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\[(\textbf u+\textbf v,\textbf w)=(\textbf u+\textbf v)_1\textbf w_1+(\textbf u+\textbf v)_3\textbf w_3\]\[\qquad\qquad\quad=u_1w_1+v_1w_1+u_3w_3+v_3w_3\]\[\qquad\qquad\quad=u_1w_1+u_3w_3+v_1w_1+v_3w_3\]\[\qquad\qquad\quad=(\textbf u,\textbf w)+(\textbf v,\text w)\]
the additivity axion holds
UnkleRhaukus
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\[ (k\textbf u,\textbf v)=ku_1 v_1+ku_3v_3\]\[\qquad\quad=k(u_1 v_1+u_3v_3)\]\[\qquad\quad=k(\textbf u,\textbf v)\]
the homogeneity axiom holds
KonradZuse
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KonradZuse
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not sure what was wrong with my work. I did exactly what the book/my other homework did for it.
UnkleRhaukus
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\[(\textbf v,\textbf v)=v_1v_1+v_3v_3\]\[\qquad\quad=v_1^2+v_3^2≥0\]
\[(\textbf v,\textbf v)=0\iff \textbf v=\textbf 0\]
the positivity axiom holds
Zarkon
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that is the problem..axiom 4
Zarkon
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your work done on axiom 4 is not correct
Zarkon
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remember \(u\) and \(v\) are 3-dimensional vectors
Zarkon
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let \(u=v=<0,1,0>\)
what is \((u,v)\)
Zarkon
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or same thing ...what is \((v,v)\)
KonradZuse
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I will check it out.
KonradZuse
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I guess I'm still confused how to prove axiom 4....
Zarkon
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you don't prove axiom 4...it is an axiom
KonradZuse
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errr "satisfy axiom 4"
Zarkon
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(a) is not an inner product...I gave you a counter example
Zarkon
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((0,1,0),(0,1,0))=0*0+0*0=0
but (0,1,0) is not the zero vector
KonradZuse
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ic so axiom 4 isn't = 0 thus fails?
Zarkon
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it fails...
\((v,v)=0\Leftrightarrow v=\vec{0}\)
UnkleRhaukus
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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
Zarkon
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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
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I'm so confused though :)
KonradZuse
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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
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how did you get = 1 ?
KonradZuse
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I thought that's what he said above, I guess I'm just too konfused hahaha :(
UnkleRhaukus
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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
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hmm ic.
KonradZuse
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So does Axiom 4 only fail for the first one?
Zarkon
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another one of your problems fails axiom 4
KonradZuse
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hmm time to find it, and see if any other's fail anything....
KonradZuse
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it seems like the others pass the other axioms, now to try and figure out which one fails #4 :P
KonradZuse
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Does D by any chance fail Axiom 2?
Zarkon
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no...but it fails another axiom
Zarkon
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I'll tell you that only one of the four is an inner product
KonradZuse
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geh..
KonradZuse
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C also seems to fail one.
KonradZuse
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my program just crashed so h/o.
KonradZuse
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all of my work is gone, I'm flipping out, hold on please.
KonradZuse
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SO d fails some axiom, but what about C @Zarkon ?
Zarkon
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(c) is an inner product...the only one
KonradZuse
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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
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\[(u,v)=u_1^2v_1^2+u_2^2v_2^2+u_3^2v_3^2\]
\[(ku,v)=(ku_1)^2v_1^2+(ku_2)^2v_2^2+(ku_3^2)v_3^2\]
\[=k^2(u_1^2v_1^2+u_2^2v_2^2+u_3^2v_3^2)=k^2(u,v)\]
KonradZuse
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axiom 3
KonradZuse
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do you think you could help me with my other problems? I would greatly appreciate it.