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linear algebra

Posted on 2003-03-24
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I have two questions, but what would help more is if someone explained the concepts behind the problems.

y1 = [1,1,1]T
y2 = [1,1,0]T
y3 = [1,0,0]T

and I is the identity operator on R^3

Find coordinates of I(e1), I(e2), I(e3) with respect to [y1,y2,y3]

and find a matrix A such that Ax is the coordinate vector of x with respect to [y1,y2,y3].

If L(c1y1 + c2y2 + c3y3) = (c1 + c2 + c3)y1 + (2c1 + c3)y2 - (2c2 + c3)y3 find a matrix representing L with respect to basis [y1,y2,y3]

and write x as a linear combination of y1, y2, y3 and use the matrix determined above to determine L(x).

x = [7,5,2]T



The second problem is L is the linear operator mapping P2 into R^2 defined by
L(p(x)) = [integral from 0 to 1 of p(x) dx, p(0) ] T

Find a matrix A such that L(alpha + beta * x) = A[alpha, beta]T.


I already have the solutions to these, but I do not understand how they were attained. Could someone explain the process behind how to solve these problems? That would be most helpful. Thank you.

-lj8866
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Question by:lj8866
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23 Comments
 

Author Comment

by:lj8866
ID: 8200320
I solved the second problem so my question is just about the first now.
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Author Comment

by:lj8866
ID: 8200336
I solved the second problem so my question is just about the first now.
0
 
LVL 15

Expert Comment

by:VGR
ID: 8205757
If L(c1y1 + c2y2 + c3y3) = (c1 + c2 + c3)y1 + (2c1 + c3)y2 - (2c2 + c3)y3 find a matrix representing L with respect to basis [y1,y2,y3]

unless I understood wrongly your english Maths jargon (I'm French), this means :

L=((c1,2c1,0),(c2,0,-2c2),(c3,c3,-c3))

I must admit I've problems apprehending your way of presenting linear algebra
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Author Comment

by:lj8866
ID: 8205773
i'm not sure what it means, I am simply copying out of my textbook. This is why I am quite confused.

I thought the method to finding the matrix was to evalute
L(y1) L(y2) and L(y3) rather than L(e1) L(e2) L(e3)
but I don't see that here.
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LVL 15

Expert Comment

by:VGR
ID: 8205903
e1, e2 and e3 are normally the unity vectors on the three axes of the referential you are in
0
 

Author Comment

by:lj8866
ID: 8205936
yes, but if you're changing the basis and doing the transformation then you take L(y1) L(y2) L(y3) rather than of e1 e2 e3. But I do not see how you can do that here.

0
 
LVL 15

Expert Comment

by:VGR
ID: 8205944
I find your problem very badly stated as we don't have the type of "constants" c1,c2,c3 (vectors?matrices?integers?reals?)

also I can't translate-understand the meaning of "coordinates of I(e1), I(e2), I(e3) " and "the coordinate vector of x with respect to [y1,y2,y3]."

Sorry, you'll have to wait for someone passing by which is more accustomed to that way of writing Maths as simple as linear algebra and matrix operations 8-)

0
 

Author Comment

by:lj8866
ID: 8205951
yes, but if you're changing the basis and doing the transformation then you take L(y1) L(y2) L(y3) rather than of e1 e2 e3. But I do not see how you can do that here.

0
 

Author Comment

by:lj8866
ID: 8205959
I means the identity operator on R^3. I don't know what they mean either. They should be some sort of standard notation.

Sorry for the doubles
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LVL 15

Expert Comment

by:VGR
ID: 8205997
identity operator on R³ is ((1,0,0),(0,1,0),(0,0,1))
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LVL 15

Expert Comment

by:VGR
ID: 8205999
but this doesn't help you, I know :D
0
 

Author Comment

by:lj8866
ID: 8206009
lol. yea i knew that already :)

I was wondering what the I(x) operator was also, because that's the first step of the problem.
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LVL 15

Expert Comment

by:VGR
ID: 8206044
theoretically, I(x)=x
so I is an operator represented by a matrix "identity" being the one I posted (big stuff! :D )
0
 

Author Comment

by:lj8866
ID: 8206064
how would you find I(e1) I(e2) I(e3) with respect to [y1 y2 y3] ?
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LVL 4

Expert Comment

by:n_fortynine
ID: 8213108
How about expanding e1 e2 e3 as linear combinations of y1 y2 y3 e.g.

e1 = (1,0,0) = y3
e2 = (0,1,0) = y2 - y3
e3 = (0,0,1) = y1 - y2
0
 

Author Comment

by:lj8866
ID: 8230381
how would you do that expansion?
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LVL 4

Expert Comment

by:n_fortynine
ID: 8231634
So assume you are given the standard basic vectors:

e1 = 1,0,0..0
e2 = 0,1,0..0
..
en = 0,0,0..1

and their corresponding y(i) values, e.g.

T(1,0,0..0) = T(e1) = y1 = (2,4,...4) (m values)
T(0,1,0..0) = T(e2) = y2 = (1,-2,..2) (m values)
...
T(0,0,0..1) = T(e3) = yn = (3,0,..-2) (m values)

So every vector in the form (x1,x2,...xn) in R(n) can be expanded as a linear combination of the e1,e2,..en right?

(x1,x2,..xn) = x1(1,0,0..0) + x2(0,1,0..0) + ... + xn(0,0,0..1)
      = x1e1 + x2e2 + ... + xnen

Now you want to find a matrix a to be the reduced function for T, i.e. use T = Ax (A times x) to transform a vector x from R(n) to R(m).

Apply the transformation to (x1,x2,...xn) we have:
T(x1,x2,..xn) = T(x1e1 + x2e2 + ...+ xnen)
   = x1*T(e1) + x2*T(e2) + ... + xn*T(en)

(Notice that T(x) by definition always follow the addition and scalar multiplication properties, i.e.
  T(u + v) = T(u) + T(v)
  T(ru) = rT(u)
)

So now you can replace T(e(i)) with y(i), you'll end up with
T(x) = T(x1,x2,..xn) = x1y1 + x2y2 + ... + xnyn

Since Ax = T(x), T(x) is in the column space (CS) of A, thus y(i)'s have to be the columns of A.
(If you haven't learned about CS then here's an *cheating* explanation:
  Let A be an m x n matrix with y(i) as columns then.
 (Please copy and paste into Notepad or something to be able to read this)
                 | x1 |
                 | x2 |
   [y1 y2 ... yn]| .. | = x1y1 + x2y2 + ... + xnyn (an mx1 matrix)
                 | .. |
                 | xn |
   This works for any matrix.
 )
 So you've realized that A is only formed by making y(i) the columns of it, or in another words, transpose the matrix on the right hand side of the T[1,0,0..0] etc.

Now the problem you're having is you're not given the standard basis, in your case you are given
[1,1,1]
[1,1,0]
[1,0,0]
Don't worry the path to go is: since these vectors are linearly independant and they span R3, you can effectively express ANY vector in R3 with them right?

So now the only thing you need to do is to go and try to express the standard basis vectors as a Linear combination of these. In some case you have to go through and find
[1,1,1][a]   [1]
[1,1,0][b] = [0]
[1,0,0][c]   [0]
----------------
[1,1,1][e]   [0]
[1,1,0][f] = [1]
[1,0,0][g]   [0]
----------------
[1,1,1][h]   [0]
[1,1,0][i] = [0]
[1,0,0][j]   [1]

but here you can cheat and see that

e1 = (1,0,0) = y3
e2 = (0,1,0) = y2 - y3 = [1,1,0] - [1,0,0]
e3 = (0,0,1) = y1 - y2 = [1,1,1] - [1,1,0]

just like I said. Any more question about this part?
0
 
LVL 4

Expert Comment

by:n_fortynine
ID: 8231653
(continue)
so the matrix A would have y3 as the first column, y2-y3 as the second column and y1-y2 as the third etc.
0
 
LVL 4

Expert Comment

by:n_fortynine
ID: 8231674
>>e1 = (1,0,0) = y3
>>e2 = (0,1,0) = y2 - y3 = [1,1,0] - [1,0,0]
>>e3 = (0,0,1) = y1 - y2 = [1,1,1] - [1,1,0]

Aye, sorry for my just getting up. No, whatever I said until that point is correct but this is bogus haha.

e1 = [1,0,0] so T(e1) = T(1,0,0) = y3
e2 = [0,1,0] = [1,1,0] - [1,0,0] so T(e2) = y2 - y3
e3 = [0,0,1] = [1,1,1] - [1,1,0] so T(e3) = y1 - y2
(It's the T of e(i) not e(i) itself, sorry)

AND THEN A would have y3 as the first column, y2-y3 as the second column and y1-y2 as the third.
0
 
LVL 4

Expert Comment

by:n_fortynine
ID: 8231746
For the second part you have x = [7,5,2]T
then you can express x as:

[7]     [1]     [1]     [1]
[5]T = 2[1]T + 3[1]T + 2[0]T = 2y1 + 3y2 + 2y3
[2]     [1]     [0]     [0]

Then L(x) = L(2y1 + 3y2 + 2y3) = blah blah blah (expand them out you have the formula.

The only thing left now is to find a matrix to represent L with respect to [y1,y2,y3]. This is simple:
L(c1y1 + c2y2 + c3y3) = (c1 + c2 + c3)y1 + (2c1 + c3)y2 - (2c2 + c3)y3

So you have
[1  1  1][c1]
[2  0  1][c2][y1 y2 y3] = L(c1y1 + c2y2 + c3y3) right?
[0 -2 -1][c3]
-------------
 (this part)

so haven't you got the matrix you want?
0
 
LVL 4

Accepted Solution

by:
n_fortynine earned 320 total points
ID: 8231749
btw it took me some time to look it up so please as well post the answers from your book to see if I got this right. (I *really* want to know) =)
0
 
LVL 31

Expert Comment

by:GwynforWeb
ID: 8301221
lj8866,
   Wow are you guys still going!

   I think you owe n_fortynine both a response and a lot more than 30pts.

GwynforWeb
0
 

Author Comment

by:lj8866
ID: 8319100
I looked at your work; it looks impressive but I don't really understand it. I really don't care about it anymore since I realized it was way more complicated than what I needed to know. I had fun doing the problem though :)

I increased the points a bit and I'll give you an A for the problem.

Oh yea, part I looks correct I think. I got two answers:

[0  0  1]   [0  0  1]
[0  0 -1]   [0  1 -1]
[1 -1  0]   [1 -1  0]

I know one of those is correct.

Part II I don't know if it's correct but I got
[0  0  1]
[0  1 -1]
[1 -1  0]
0

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