# 查看完整版本 : Basis + Coordinate Matrices

## Basis + Coordinate Matrices

Question
[IMG]http://i1382.photobucket.com/albums/ah257/jeffreytam01/Capture_zpsvkrivlgw.png[/IMG]
[IMG]http://i1382.photobucket.com/albums/ah257/jeffreytam01/Capture1_zpsprdbxdxf.png[/IMG]

My Attempt

I am not sure the expression in Question 1a, what function is that?
Also, Question 2aii, how do you go further or get the answer?
Thank you so much:loveliness:

14c14c 2015-9-14 11:14 PM

2aii. Do you know factor theorem?
p(1)=0 => p contains (x-1) as a factor
Let Q(x) be s.t. Q(x)(x-1)=p(x)
by compareing degree, since deg(p)=2 and deg(x-1)=1, deg(Q(x))=2-1=1
=> Q(x)=ax+b for some a, b
therefore p must be of the form (x-1)(ax+b)

14c14c 2015-9-14 11:27 PM

I don't know what is coordinate matrix and I think it is a bit strange if it is 2x2 instead of 1x3 or 3x1

[quote]原帖由 [i]14c14c[/i] 於 2015-9-14 11:27 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=425758387&ptid=25094339][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
I don't know what is coordinate matrix and I think it is a bit strange if it is 2x2 instead of 1x3 or 3x1 [/quote]
I think I have figured it out already.
I get: (-5/2)
(-3)
(3/2)

Coordinate matrix is same as coordinate vector, but in matrix form.

Btw, I don't get 1a. Do is W? The capital F and f represent?

14c14c 2015-9-15 08:05 PM

[quote]原帖由 [i]好男仔[/i] 於 2015-9-15 09:06 AM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=425775813&ptid=25094339][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

I think I have figured it out already.
I get: (-5/2)
(-3)
(3/2)

Coordinate matrix is same as coordinate vector, but in matrix form.

Btw, I don't get 1a. Do is W? The capi ... [/quote]
Yes I can see your answer but I think the coordinate should be ((-5/2),(-3),(3/2))
either a column vector(3x1 matrix) or row vector(1x3 matrix) instead of a 2x2 matrix

And your 1a... I don't know what you are talking about
the question already gives you the definition of F[a,b]
it consist of all functions from [a,b] to R
F[a,b] is a vector space since if f, g:[a,b]->R, then a1f+a2g is also a function from [a,b] to R
W is the set of all function from [a,b] to R satsifying f(6)=10
then obviously it is not a subspace since one can take the constant function g(x)=10
then g is in W, but the function 2g is not in W since (2g)(6)=2(g(6))=20
in fact any function f in W is already ok by considering 2f

14c14c 2015-9-15 08:16 PM

I mean I cannot see the reason that the "coordinate matrix" has to place this way

and 1a果個我係指我睇唔明你既答案

[quote]原帖由 [i]14c14c[/i] 於 2015-9-15 08:16 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=425814514&ptid=25094339][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
I mean I cannot see the reason that the "coordinate matrix" has to place this way

and 1a果個我係指我睇唔明你既答案 [/quote]
1a, I get it thanks. I didn't realise that F([a,b],R) means a function of F has the domain [a,b] and range of R before. So I got confused haha.

For 3a, I changed it into a 3X1 matrix already. It is in this order because it has to be in the same order as the matrice given in the set B. So that it gives the desired matrix when they multiply and added. Its similar to 3b, I guess. Because 3b is given the coordinate matrix.

[[i] 本帖最後由 好男仔 於 2015-9-15 09:17 PM 編輯 [/i]]