KonradZuse
  • KonradZuse
Anyone good with inner products and their axioms?
Linear Algebra
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

tyteen4a03
  • tyteen4a03
Just ask your question.
KonradZuse
  • KonradZuse
5. Please refer to De nition 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
  • 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]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.