help please

- anonymous

help please

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- anonymous

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

- anonymous

@zzr0ck3r

- anonymous

@oldrin.bataku

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## More answers

- anonymous

its a table

- anonymous

please continue

- anonymous

notice that the 1 shifts each time.

- anonymous

yes

- anonymous

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

- anonymous

yes

- anonymous

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

- anonymous

how?

- 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

- anonymous

ok

- anonymous

I'll draw you a picture!

- anonymous

|dw:1439037967941:dw|

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

ok

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

ok. got it

- anonymous

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

- anonymous

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

- 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

- anonymous

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

- anonymous

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

- anonymous

ok

- 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

- 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

- anonymous

so the answer is 7?

- 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.

- anonymous

you get 7 if you add the numbers

- anonymous

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

- 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|

- anonymous

you will get 100% trust me

- 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!

- anonymous

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

- anonymous

can i ask?

- 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

- 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 $$b^i(v)=\langle b_i,v\rangle$$

- anonymous

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

- anonymous

so what is the duel basis of my question sir?

- 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

- 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: $$\delta_{ij}=\left\{\begin{matrix}1&\text{if }i=j\\0&\text{otherwise}\end{matrix}\right.$$

- anonymous

is that all?

- 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 \(x,y\)

- anonymous

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

- anonymous

can you compute \(1/3+1/8+1/5\)?

- anonymous

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

- anonymous

@Michele_Laino

- anonymous

@Loser66

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