anonymous
  • anonymous
help please
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
find the dual basis of {(1,0,0),(0,1,0),(0,0,1) and explain
anonymous
  • anonymous
@zzr0ck3r
anonymous
  • anonymous
@oldrin.bataku

Looking for something else?

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

More answers

anonymous
  • anonymous
its a table
anonymous
  • anonymous
please continue
anonymous
  • anonymous
notice that the 1 shifts each time.
anonymous
  • anonymous
yes
anonymous
  • anonymous
1, 0, 0 0,1,0 0,0,1 thats binary code
anonymous
  • anonymous
yes
anonymous
  • anonymous
the first column is 2^0 second column is 2^1 third column is 2 ^2
anonymous
  • anonymous
how?
anonymous
  • 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
  • anonymous
ok
anonymous
  • anonymous
I'll draw you a picture!
anonymous
  • anonymous
|dw:1439037967941:dw|
anonymous
  • anonymous
i don,'t still get why 2^ of numbers here.
anonymous
  • anonymous
another picture in base 10 (decimal system, which you are more used to) |dw:1439038148182:dw|
anonymous
  • anonymous
There are only 2 numbers 1 and zero that is why it is base 2 or 2^
anonymous
  • anonymous
ok
anonymous
  • anonymous
so how come the numbers they raised to? 2^n
anonymous
  • anonymous
look up positional number systems in wikipedia it should explain it more to you
anonymous
  • anonymous
The value that a digit taks on depends on its position, as I have shown
anonymous
  • anonymous
ok. got it
anonymous
  • anonymous
for example, 110 in decimal system is 1 x 100 + 1 x 10 + 1 x 1
anonymous
  • anonymous
so, you did for base 2 but why base 10?
anonymous
  • 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
  • anonymous
but we are not using base 10 right since you said that only numbers are involved . ?
anonymous
  • anonymous
the value of the number (e.g. 001) depends on itsa place or position) is what I am saying
anonymous
  • anonymous
ok
anonymous
  • 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
  • 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
  • anonymous
so the answer is 7?
anonymous
  • 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
  • anonymous
you get 7 if you add the numbers
anonymous
  • anonymous
ok, so the table with the position is the dual basis?
anonymous
  • 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
  • anonymous
you will get 100% trust me
anonymous
  • 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
  • anonymous
ok thanks . i have a question on inner produte. can i ask?
anonymous
  • anonymous
can i ask?
anonymous
  • 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
  • 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
  • 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
  • anonymous
so what is the duel basis of my question sir?
anonymous
  • 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
  • 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
  • anonymous
is that all?
anonymous
  • 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
  • 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
  • anonymous
can you compute \(1/3+1/8+1/5\)?
anonymous
  • anonymous
(40+15+24)/120=79/120
anonymous
  • anonymous
@Michele_Laino
anonymous
  • anonymous
@Loser66

Looking for something else?

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