Question

Anyone good with inner products and their axioms?

Answer 1

Just ask your question.

Answer 2

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.

Answer 3

(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]

Answer 4

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

Answer 5

@Hero save me.

Answer 6

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

Answer 7

Hang on for a minute

Answer 8

@KonradZuse, where in the book is that question?

Answer 9

not in the book.

Answer 10

sorry it's kind of screwed up in formatting.

Answer 11

1 second.

Answer 12

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

Answer 13

okay hold on.

Answer 14

here happy?

Answer 15

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

Answer 16

saying*

Answer 17

@UnkleRhaukus

Answer 18

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

Answer 19

where are the axioms you are referencing ?

Answer 20

one second.

Answer 21

(a)

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

Answer 22

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

Answer 23

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

Answer 24

ok.

Answer 25

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

Answer 26

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?

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

that is the problem..axiom 4

Answer 32

your work done on axiom 4 is not correct

Answer 33

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

Answer 34

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

what is \((u,v)\)

Answer 35

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

Answer 36

I will check it out.

Answer 37

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

Answer 38

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

Answer 39

errr "satisfy axiom 4"

Answer 40

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

Answer 41

((0,1,0),(0,1,0))=0*0+0*0=0

but (0,1,0) is not the zero vector

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

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 :) )

Answer 46

I'm so confused though :)

Answer 47

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?

Answer 48

how did you get = 1 ?

Answer 49

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

Answer 50

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)

Answer 51

hmm ic.

Answer 52

So does Axiom 4 only fail for the first one?

Answer 53

another one of your problems fails axiom 4

Answer 54

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

Answer 55

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

Answer 56

Does D by any chance fail Axiom 2?

Answer 57

no...but it fails another axiom

Answer 58

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

Answer 59

geh..

Answer 60

C also seems to fail one.

Answer 61

my program just crashed so h/o.

Answer 62

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

Answer 63

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

Answer 64

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

Answer 65

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!

Answer 66

\[(u,v)=u_1^2v_1^2+u_2^2v_2^2+u_3^2v_3^2\]

\[(ku,v)=(ku_1)^2v_1^2+(ku_2)^2v_2^2+(ku_3^2)v_3^2\]

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

Answer 67

axiom 3

Answer 68

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