for those who boss linear algebra.

- anonymous

for those who boss linear algebra.

- Stacey Warren - Expert brainly.com

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- katieb

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- anonymous

##### 1 Attachment

- anonymous

i actually need help with it though...

- anonymous

@.Sam.

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## More answers

- anonymous

any ideas Hero?

- Mendicant_Bias

I have no idea what this is, but it looks unpleasant. Good luck with that.

- anonymous

haha thanks. surely someone can do this :(

- anonymous

have you solved this kind of problems in class?

- anonymous

mostly, orthonormal basis, projections etc everything orthogonals..but never came across something like this.

- anonymous

oh okay, don't worry we can work together to solve this. Ready? :)

- anonymous

YES

- anonymous

Ok|dw:1377679515993:dw|

- anonymous

so, the first question
what is the dimension of this base?

- anonymous

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|

- anonymous

thats what i think its saying...

- anonymous

and a basis is when the vectors are lin independent of each other...

- anonymous

then im stuck

- anonymous

Ok, don't worry we will solve it together just hold on a sec...

- anonymous

kayy :)

- anonymous

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?

- anonymous

correct, so would that mea we just have one linearly independent set?

- anonymous

since the computed values of the elements are simply numerical

- anonymous

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?

- anonymous

well if two vectors are linearly independent then they are orthogonal (dot product is 0)

- anonymous

good thought

- anonymous

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

- anonymous

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

- anonymous

so when i=j was irrelevant?

- anonymous

Glad you did!

- anonymous

yea it would be otherwise they wouldn't be lin independent

- anonymous

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

- anonymous

oh true true

- anonymous

nice one

- anonymous

:)

- anonymous

can we go to the (b) part?

- anonymous

yea lets do it i'm keen!

- anonymous

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?

- anonymous

isn't it very similar to the dot product?

- anonymous

just in a vector space?

- anonymous

yes, that is right good! ;)

- anonymous

|dw:1377681129822:dw|

- anonymous

as you stated it will be the dot product with the vectors defined on the white board
do you have any question on that?

- anonymous

is the first symbol vector u?

- anonymous

and isnt' k a scalar here?

- anonymous

no it is a scalar, sorry dot product result in scalar, right? My drawing in not that not as I'm not an artist.

- anonymous

i think i had misinterpreted the question

- anonymous

i can see the reason for your confusion, lol.

- anonymous

can you write it again. i'm getting more confused haha

- anonymous

each of v should be a vector and u should also be a vector let's backtrack

- anonymous

yes, sure.

- anonymous

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

- anonymous

so don't bother with ur previous drawing?

- anonymous

|dw:1377681518747:dw|

- anonymous

alrighty

- anonymous

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

- anonymous

your question helped to know that it was in error because i assumed u to be a scalar, lol. :D

- anonymous

haha yea ;) k i follow

- anonymous

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

- anonymous

how can we get it into a square matrix form?

- anonymous

not a clue

- anonymous

you know that v1 itself will have its own dimension as it is a vector

- anonymous

correct

- anonymous

let's assume that the dimension is n so that v1 = (v1 1, v1 2, v1 3, ..........., v1 n)

- anonymous

the dimension of each vector in V in n?

- anonymous

is n*

- anonymous

in v1, we have scalars though so you mean|dw:1377682078766:dw|

- anonymous

|dw:1377681989326:dw|

- anonymous

each of the column vectors are the v's written in column vector form

- anonymous

let me know if you have a question on that

- anonymous

it is just a matrix way of writing u = k1v1 + k2v2 + k3v3 + k4v4 + ...... + krvr

- anonymous

so we start off like this.|dw:1377682157043:dw|

- anonymous

|dw:1377682236614:dw|

- anonymous

|dw:1377682354786:dw|

- anonymous

where a general vector is...

- anonymous

brb i gotta have dinner. you can keep going on and i'l catch up if you want. won't be too long.

- anonymous

thanks for your help so far

- anonymous

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

- anonymous

ok, i will just go ahead and hopefully you can join me after your dinner or you will see the solution on the whiteboard

- anonymous

k i'm back

- anonymous

yep i followed that

- anonymous

hold on, I'm going to attach a file where I've done all the working outs one moment...

- anonymous

alrighty

- anonymous

##### 1 Attachment

- anonymous

there you go!

- anonymous

it seems to be a blank page

- anonymous

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 :)

- anonymous

its a document you need to save it in order to open it

- anonymous

Do you see it now? :)

- anonymous

nup still getting a blank page...

- anonymous

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.

- anonymous

You click on save to whatever place you want to save the document into, and then the microsoft word document will automatically open.

- anonymous

yep still getting a blank page. maybe save it as a pdf?

- anonymous

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!

- anonymous

how bout .pdf?

- anonymous

##### 1 Attachment

- anonymous

yea got it!

- anonymous

Ok, awesome! :)

- anonymous

i see only 5 pages of actual writing out of the 10 pages? is that what is meant to happen?

- anonymous

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

- anonymous

the first whiteboard doesn't matter? cause u is a vector right?

- anonymous

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.

- anonymous

thanks so much!

- anonymous

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! :)

- anonymous

thankyou! i hope this becomes easy but the only reason i'm sticking at it because i find it so interesting

- anonymous

you are right, it is indeed very interesting!

- anonymous

:) thanks again

- anonymous

No worries, budd.

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