## anonymous one year ago help please

1. anonymous

find the dual basis of {(1,0,0),(0,1,0),(0,0,1) and explain

2. anonymous

@zzr0ck3r

3. anonymous

@oldrin.bataku

4. anonymous

its a table

5. anonymous

6. anonymous

notice that the 1 shifts each time.

7. anonymous

yes

8. anonymous

1, 0, 0 0,1,0 0,0,1 thats binary code

9. anonymous

yes

10. anonymous

the first column is 2^0 second column is 2^1 third column is 2 ^2

11. anonymous

how?

12. anonymous

you know how to count in tens. you know the first column is 0-9 the second column is 10 - 19 etc., remember the abacus? it is similar for the binary system

13. anonymous

ok

14. anonymous

I'll draw you a picture!

15. anonymous

|dw:1439037967941:dw|

16. anonymous

i don,'t still get why 2^ of numbers here.

17. anonymous

another picture in base 10 (decimal system, which you are more used to) |dw:1439038148182:dw|

18. anonymous

There are only 2 numbers 1 and zero that is why it is base 2 or 2^

19. anonymous

ok

20. anonymous

so how come the numbers they raised to? 2^n

21. anonymous

look up positional number systems in wikipedia it should explain it more to you

22. anonymous

The value that a digit taks on depends on its position, as I have shown

23. anonymous

ok. got it

24. anonymous

for example, 110 in decimal system is 1 x 100 + 1 x 10 + 1 x 1

25. anonymous

so, you did for base 2 but why base 10?

26. anonymous

I just showed you the same principle applies in base 2 as it applies in base 10 (decimal) which you were more familiar with

27. anonymous

but we are not using base 10 right since you said that only numbers are involved . ?

28. anonymous

the value of the number (e.g. 001) depends on itsa place or position) is what I am saying

29. anonymous

ok

30. anonymous

So of 1 is placed in the units column in the decimal system then it can only take the value between 0 and 9

31. anonymous

your question was about the dual basis of this, I have offered an answer in relation to that. i.e that the digits given are base 2

32. anonymous

33. anonymous

No. It said find the dual basis of this, and explain. you have found the dual basis. reproduce my table i n relation to 2^0, "^1 etc, and then state that the value of oo1,010 depends on their position.

34. anonymous

you get 7 if you add the numbers

35. anonymous

ok, so the table with the position is the dual basis?

36. anonymous

for something to have a "dual basis" it must have two possible outcomes.i.e. binary. you show that as follows: |dw:1439039911382:dw|

37. anonymous

you will get 100% trust me

38. anonymous

now you are asked to explain: so here goes:( the value of the number zero or 1 depends on the position) that's it!

39. anonymous

ok thanks . i have a question on inner produte. can i ask?

40. anonymous

41. anonymous

compute the I2 inner product of x and y where x=1,1/2,1,3,1/4 and y=1/3,1/4,1/5

42. anonymous

that's not what the dual basis is. if you have a real inner product space $$V,\langle\cdot,\cdot\rangle$$ then an ordered basis $$\{b_i\}$$ induces a basis for the dual space $$V^*$$ of linear maps $$V\to R$$, such that $$b^i(v)=\langle b_i,v\rangle$$

43. anonymous

this is equivalent to the idea of the transpose, where we identify column vectors with $$v\in V$$ and so row vectors with $$u\in V^*$$

44. anonymous

so what is the duel basis of my question sir?

45. anonymous

in other words, for an orthonormal basis $$\{e_i\}$$, like you have, our dual basis looks like $$e^i$$, so that $$e^i(e_i)=1$$ and $$e^i(e_j)=0$$ for $$i\ne j$$. in other words, our dual basis lets us pick out things parallel to the corresponding original basis vectors

46. anonymous

so a dual basis for $$\{(1,0,0),(0,1,0),(0,0,1)\}$$ looks like $$\{\delta_{1j},\delta_{2j},\delta_{3j}\}$$ where $$\delta_{ij}$$ denotes the Kronecker delta: $$\delta_{ij}=\left\{\begin{matrix}1&\text{if }i=j\\0&\text{otherwise}\end{matrix}\right.$$

47. anonymous

is that all?

48. anonymous

now in $$\ell^2(\mathbb N)$$ we're talking about a sequence space and we have an inner product like: $$\langle x,y\rangle=\sum_{k=0}^\infty x_ky_k$$ for real sequences $$x,y$$

49. anonymous

$$\langle\left(1,\frac12,1,3,\frac14,0,0,0,\dots\right),\left(\frac13,\frac14,\frac15,0,0,0,\dots\right)\rangle=1\cdot\frac13+\frac12\cdot\frac14+1\cdot\frac15=\dots$$

50. anonymous

can you compute $$1/3+1/8+1/5$$?

51. anonymous

(40+15+24)/120=79/120

52. anonymous

@Michele_Laino

53. anonymous

@Loser66