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i actually need help with it though...
any ideas Hero?
I have no idea what this is, but it looks unpleasant. Good luck with that.
haha thanks. surely someone can do this :(
have you solved this kind of problems in class?
mostly, orthonormal basis, projections etc everything orthogonals..but never came across something like this.
oh okay, don't worry we can work together to solve this. Ready? :)
so, the first question what is the dimension of this base?
well if we have two elements that are different in V we get 0 however if we have two of the same elements and dot them we get ij i.e v1 . v1=1...etc |dw:1377679746655:dw|
thats what i think its saying...
and a basis is when the vectors are lin independent of each other...
then im stuck
Ok, don't worry we will solve it together just hold on a sec...
Alright sorry for taking long, was helping someone else. So what you said previously is correct, so the span of a vector space are set of linearly independent vectors which form the basis of the vector space, right?
correct, so would that mea we just have one linearly independent set?
since the computed values of the elements are simply numerical
we need the dim span V i.e. the number of vectors that are linearly independent in the vector space of V the key to answering the question is how the dot product relates to linearly independency, any thought on that?
well if two vectors are linearly independent then they are orthogonal (dot product is 0)
and what the question stated is that all the vectors in V are orthogonal except they are the same two vectors ( i = j ) that means the dim span V is the number of vectors in V because they are all orthogonal and the dim span V = r let me know if that is clear
ohh yea right. thats clear. so since each vector is linearly independent in V since vi . vj=0 then this means that the dimspanV=r ahhh yea now i got it
so when i=j was irrelevant?
Glad you did!
yea it would be otherwise they wouldn't be lin independent
the i = j only stated that the only time the dot product is not zero is when the two vectors involved are the same it is very relevant
oh true true
can we go to the (b) part?
yea lets do it i'm keen!
so we need to know what inner product is? then we can be able to get the needed expression, any thought before we go ahead?
isn't it very similar to the dot product?
just in a vector space?
yes, that is right good! ;)
as you stated it will be the dot product with the vectors defined on the white board do you have any question on that?
is the first symbol vector u?
and isnt' k a scalar here?
no it is a scalar, sorry dot product result in scalar, right? My drawing in not that not as I'm not an artist.
i think i had misinterpreted the question
i can see the reason for your confusion, lol.
can you write it again. i'm getting more confused haha
each of v should be a vector and u should also be a vector let's backtrack
ok, in this case where v is a vector, we cannot really use the dot product form of inner product rather, we can go the way of matrix definition
so don't bother with ur previous drawing?
this is so because when a 1 X r matrix multiplies a r X 1 matrix the result should be a 1 x 1 matrix which is k1 v1 + ...... + krvr dont bother with the other one
your question helped to know that it was in error because i assumed u to be a scalar, lol. :D
haha yea ;) k i follow
so we need to find the k's we could have easily found that using the inverse function but we do not have a square matrix to work with
how can we get it into a square matrix form?
not a clue
you know that v1 itself will have its own dimension as it is a vector
let's assume that the dimension is n so that v1 = (v1 1, v1 2, v1 3, ..........., v1 n)
the dimension of each vector in V in n?
in v1, we have scalars though so you mean|dw:1377682078766:dw|
each of the column vectors are the v's written in column vector form
let me know if you have a question on that
it is just a matrix way of writing u = k1v1 + k2v2 + k3v3 + k4v4 + ...... + krvr
so we start off like this.|dw:1377682157043:dw|
where a general vector is...
brb i gotta have dinner. you can keep going on and i'l catch up if you want. won't be too long.
thanks for your help so far
what you are writing is right however, v1, v2 must each be independently have a basis of r for the dimension of the vector space V in which they are contained to be r so we need to be able to write the equation in a form that we can easily find the k's
ok, i will just go ahead and hopefully you can join me after your dinner or you will see the solution on the whiteboard
k i'm back
yep i followed that
hold on, I'm going to attach a file where I've done all the working outs one moment...
there you go!
it seems to be a blank page
The multiplication on the right generates matrix with k1, k2, ....., kr so that is a way of using inner product to find ki as asked in the question :)
its a document you need to save it in order to open it
Do you see it now? :)
nup still getting a blank page...
Are you sure? When you click on the attachment it leads you to a new tab, right? Then a small box appears on the left side which says for you to save the document that I sent you.
You click on save to whatever place you want to save the document into, and then the microsoft word document will automatically open.
yep still getting a blank page. maybe save it as a pdf?
If this is not working then, I may need your email address so I can send it through the email and it will work-hopefully. If you don't mind!
how bout .pdf?
yea got it!
Ok, awesome! :)
i see only 5 pages of actual writing out of the 10 pages? is that what is meant to happen?
Go through the document, and let me know if you have any questions. Also yes it is supposed to be 5 pages, my bad I added the extra 5. lol
the first whiteboard doesn't matter? cause u is a vector right?
Also I'm going to leave and help some other students, if you have any questions ask. And yes it doesn't matter I just put it there to summarise.
thanks so much!
No worries, I'm glad I could help. This stuff is really easy once you get the hang of it. Just learn the rule, practise, practise, practise, because practise makes it perfect! For now, good luck with your studies! :)
thankyou! i hope this becomes easy but the only reason i'm sticking at it because i find it so interesting
you are right, it is indeed very interesting!
:) thanks again
No worries, budd.