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 [quote="mole"]I found a "Cholesky's method" in my maths textbook, this looks like what you want:
code:
We have a matrix A  A  = a11 a12 a13      a21 a22 a23      a31 a32 a33  And we want two matrices B and BT B  = b11  0   0      b21 b22  0      b31 b32 b33   BT = b11 b21 b31      0  b22 b32      0   0  b33

[b]First row/column:[/b] (b11 to bj1) b11 = sqrt(a11) bj1 = aj1/b11 [b]Diagonal:[/b] (b11 to bjj) bjj = sqrt(ajj - sum [s = 1 to j - 1] (bjs2) [b]Everything else:[/b] bpj = 1/bjj(apj - sum [s = 1 to j - 1] (bps bsj) I assume afterwards that you'd find B-1B-1T to get A-1, like you do with the other decompositions. [This message has been edited by mole (edited 11-30-2003 09:09 AM).][/quote]
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math geek
Posted: Fri Dec 19, 2003 4:06 pm    Post subject: 1

mole had it perfectly, with the sole comment that this algorithm applies only to symmetric and positively defined real matrices. And maybe to Hermitic matrices if working over the Complex plane.

[having read the thread again, i realised i mistook Cholesky for LU-decomposition - the will-reply-tomorow still applies though.]

actualy Cholesky decomposition works best with one of the L(ower) and U(pper) having 1's on their main diagonals. The algorhythm was taught to me at Numerical Computing(possibly incorrectly translated) and i have it at home. I think i can post more details tomorow.

And since my brain isn't in top shape right now, is inverting a triangular matrix so much easier that decomposing is feasible? EDIT: YES it is!

[This message has been edited by math geek (edited 12-20-2003 01:04 PM).]
jesternl
Posted: Wed Dec 03, 2003 5:42 pm    Post subject: 0

I thought decomposing was something Mozart did after he died
mole
Posted: Sun Nov 30, 2003 3:48 am    Post subject: -1

I found a "Cholesky's method" in my maths textbook, this looks like what you want:

code:

We have a matrix A

A = a11 a12 a13
a21 a22 a23
a31 a32 a33

And we want two matrices B and BT
B = b11 0 0
b21 b22 0
b31 b32 b33

BT = b11 b21 b31
0 b22 b32
0 0 b33

First row/column: (b11 to bj1)
b11 = sqrt(a11)
bj1 = aj1/b11

Diagonal: (b11 to bjj)
bjj = sqrt(ajj - sum [s = 1 to j - 1] (bjs2)

Everything else:
bpj = 1/bjj(apj - sum [s = 1 to j - 1] (bps bsj)

I assume afterwards that you'd find B-1B-1T to get A-1, like you do with the other decompositions.

[This message has been edited by mole (edited 11-30-2003 09:09 AM).]
Chuck
Posted: Fri Nov 28, 2003 4:31 pm    Post subject: -2

What's so hard? Do a screen print of the matrix, paste it into Paint, do a flip horizontal, and do a flip vertical.
Beartalon
Posted: Fri Nov 28, 2003 4:17 pm    Post subject: -3

That's OK. I'm terribly rusty with this level of math.
mith
Posted: Thu Nov 27, 2003 6:28 pm    Post subject: -4

Does anyone know anything about Cholesky decomposition? I'm wanting to write a program that involves inverting matricies, and I read somewhere that it is more stable that Gauss-Jordan elimination, but I haven't found anywhere detailing the method.

May be back with more random questions later, I'm terribly rusty with programming.