Question

Linear Algebra. "*"---> means multiply
Let {v1, v2,.... Vn}be an orthonormal basis for Rn,
For any x and y in Rn prove that
x*y = (x*v1)(v1*y) + (x*v2)(v2*y) +...
.+ (x*Vn)(Vn* y):

Answer 1

x = x1v1 + x2v2 + x3v3+...+xnvn
y = y1v1 + y2v2 + y3v3 +...+ynvn
------------------------------
might be something like this

Answer 2

not quite sure what you mean by multiply tho, im reading it as a dot product tho

Answer 3

yea its suppose to be a dot

Answer 4

im not sure what to make of your notation for it tho; my xns and yns would correspond to the component parts that add up the basis vectors to create x and y with.

what would (x*vn)(vn*y) mean?

Answer 5

https://b36cb25a-a-62cb3a1a-s-sites.googlegroups.com/site/andrewfdouglas/Last_assignment_2580.pdf?attachauth=ANoY7covCE5s6dk8lJaJOVPYfG7YXgwMubJj7Ov1VDqjvdJ0RzAnVqVY-udUPx2R1spIg8jl2ilmSjlGgY6rOHTAvXlVcNR8D8gIpa1UMhdce6806iZA1ZHU_Ua7qoMP6ylvJ_t3hw0SMrp8XP71VxJSDEiKGehQDL13KsMjKfkBiL2Bxgq6FZjmhRdgk7U4x4vsH6kDyMdk4eAfCl6SrsWuDgLqtMlavk-rpWGi_8DK-OUtTC9U7wg%3D&attredirects=0

Answer 6

its question 8

Answer 7

its okay if you dont get it

Answer 8

id have to take it to paper to run a few ideas

Answer 9

start with the orthoN , v1 = 1,0 v2 = 0,1 and try to work an example

Answer 10

get a feel for it

Answer 11

lets pick some random vectors for x and y

x = 2,3

y = -3,8

lets use the vectors v1=1,0 and v2 = 0,1

x = <2,3> + <2,3>
<1,0> <0,1>
----------------
2 + 3

y = <-3,8> + <-3,8>
<1,0> <0,1>
----------------
-3 + 8

2(-3) + 8(3) = -6 + 24 = 18


x = 2,3
y = -3,8
-----
-6+24 = 18

well, the concept works out

Answer 12

yea but i think it means to prove that both sides equal each other just not sure how

Answer 13

yeah, its prolly something hidden in a definitional thrm or something thats eluding us

Answer 14

not sure what it gets us, but this is where i get to lol

x = a
y = b
u = any unit vector

\[|a|\cos\alpha =a\cdot u \]\[|b|\cos\beta =b\cdot u \]------------
\[|a||b|\cos\alpha \cos\beta =(a\cdot u)(b\cdot u) \]\[|a||b|\cos\gamma =a\cdot b \]

the idea is that \[|a||b|\cos\alpha \cos\beta = |a||b|\cos\gamma\]


\[\cos\alpha \cos\beta = \cos\gamma\]\[\cos\alpha \cos\beta = \cos(\alpha-\beta )\]
\[\cos\alpha \cos\beta = \cos\alpha\cos\beta +\sin\alpha\sin\beta\]
\[0 = \sin\alpha\sin\beta;\ \alpha=0,\pi\ or\ \beta=0,\pi\]