Use Lagrange Interpolating Method for Spectral Decomposition.
Ok, the professor through this example on the board with really no explanation. Asking questions didn't help, so I'm hoping someone here can be more helpful.
Given:
matrix A:
|-6 -5 -5 |
|15 14 15 |
|-5 -5 -6 |
Spec A = {-1, -1, 4}
m(x) = (x+1)(x-4) //prof made a note there were no multiple roots, and A is diagonalizable. So?
Two distinct e-values:
x = -1 and x = 4:
x = -1 case
want: p1(-1) = 1, p1(4) = 0 //why do we want this?
p1(x) = (x-4)/(-1-4) = (-1/5)(x-4)
E1 = p1(x) = (-1/5)(A-4I) =
| 2 1 1 |
|-3 -2 -3 |
|1 1 2 |
Repeat for x =4 case.
-----
Ok, I'm so lost it hurts.
Why do I care what the m(x) value is?
Why do we want p1(-1) = 1 and p1(4) = 0 ?
Where does the fraction in this come from :
p1(x) = (x-4)/(-1-4)
Where does A-4I come from:
E1 = p1(x) = (-1/5)(A-4I) =
I'm not exaggerating, the prof literally just wrote this example on the board and explained nothing.
Please help me understand this.
Ok, the professor through this example on the board with really no explanation. Asking questions didn't help, so I'm hoping someone here can be more helpful.
Given:
matrix A:
|-6 -5 -5 |
|15 14 15 |
|-5 -5 -6 |
Spec A = {-1, -1, 4}
m(x) = (x+1)(x-4) //prof made a note there were no multiple roots, and A is diagonalizable. So?
Two distinct e-values:
x = -1 and x = 4:
x = -1 case
want: p1(-1) = 1, p1(4) = 0 //why do we want this?
p1(x) = (x-4)/(-1-4) = (-1/5)(x-4)
E1 = p1(x) = (-1/5)(A-4I) =
| 2 1 1 |
|-3 -2 -3 |
|1 1 2 |
Repeat for x =4 case.
-----
Ok, I'm so lost it hurts.
Why do I care what the m(x) value is?
Why do we want p1(-1) = 1 and p1(4) = 0 ?
Where does the fraction in this come from :
p1(x) = (x-4)/(-1-4)
Where does A-4I come from:
E1 = p1(x) = (-1/5)(A-4I) =
I'm not exaggerating, the prof literally just wrote this example on the board and explained nothing.
Please help me understand this.