linear algebra

I solved the second problem so my question is just about the first now.

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

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.

e1, e2 and e3 are normally the unity vectors on the three axes of the referential you are in

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.

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-)

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.

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

identity operator on R³ is ((1,0,0),(0,1,0),(0,0,1))

but this doesn't help you, I know :D

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.

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

how would you find I(e1) I(e2) I(e3) with respect to [y1 y2 y3] ?

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

how would you do that expansion?

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?

(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.

>>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.

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?

lj8866,
   Wow are you guys still going!

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

GwynforWeb