2

enter image description here enter image description here

For this matrix, I got $U_{31}=1$, but the answer says $U_{31}=1/2$. Since the first two elements of the third row are 0, I think this two answers actually are equivalent. I am just wondering if both of them are correct? Since it is not in the Row Reduced echelon form, I suppose the answer is not unique. Thanks!

J.doe
  • 1,417
  • 1
  • 17
  • 31
  • 1
    It should just be $1/2$. Since it requires $l_{ii}=1$, you cannot multiply a row by a constant. By this requirement, the final $LU$ should be unique. – KittyL May 07 '16 at 19:51

1 Answers1

1

We have:

$$A = \begin{bmatrix}1&-1&0\\2 & 2 & 3 \\ -1 & 3 & 2\end{bmatrix} = LU=\begin{bmatrix}1 & 0 & 0\\l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1\end{bmatrix} \begin{bmatrix}u_{11} & u_{12} & u_{13}\\0 & u_{22} & u_{23} \\ 0 & 0 & u_{33}\end{bmatrix}$$

The individual calculations in order they are performed are:

  • $u_{11} = 1, u_{12} = -1, u_{13} = 0$
  • $l_{21} u_{11} = 2 \implies l_{21} = 2, ~l_{21}u_{12}+u_{22} = 2 \implies u_{22} = 4, ~l_{21}u_{13} + u_{23} = 3 \implies u_{23} = 3$
  • $l_{31}u_{11} = -1 \implies l_{31} = -1, ~ l_{31} u_{12} + l_{32}u_{22} = 3 \implies l_{32} = \dfrac{1}{2}$
  • $l_{31}u_{13} + l_{32}u_{23} + u_{33} = 0 +\dfrac{3}{2} + u_{33} = 2 \implies u_{33} = \dfrac{1}{2}$

This results in:

$$A = \begin{bmatrix}1&-1&0\\2 & 2 & 3 \\ -1 & 3 & 2\end{bmatrix} = LU = \begin{bmatrix}1 & 0 & 0\\2 & 1 & 0 \\ -1 & \dfrac{1}{2} & 1\end{bmatrix} \begin{bmatrix}1 & -1 & 0\\0 & 4 & 3 \\ 0 & 0 & \dfrac{1}{2}\end{bmatrix}$$

Moo
  • 11,311
  • So what about the permutation matrix? If we don't interchange rows, the permutation matrix for a $nxn$ matrix, would always be $I_n$? – J.doe May 07 '16 at 22:09
  • You only use a permutation matrix when it is necessary, but here we do not need one. If needed, it would be part of the process and you'd have to keep track of row interchanges. – Moo May 07 '16 at 22:46
  • Does that mean a permutation matrix is not unique? – J.doe May 07 '16 at 22:57
  • Well, it is unique for any particular problem, see examples here: http://math.stackexchange.com/questions/485513/what-are-pivot-numbers-in-lu-decomposition-please-explain-me-in-an-example – Moo May 07 '16 at 23:00
  • P is used to keep track of row interchanges. Suppose we have two rows (not the 1st row), with pivot 1, and we can interchange either one of them with the first row. Different choice leads to different P, isn't it? – J.doe May 07 '16 at 23:55
  • IIRC, you pick the one with largest magnitude for numerical stability. Maybe you can post a problem of a particular example. – Moo May 07 '16 at 23:56
  • Hi, I've attached a pic of the problem in the question. Thanks! – J.doe May 08 '16 at 02:14
  • $P = \left( \begin{array}{ccc} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \ \end{array} \right)$ – Moo May 08 '16 at 02:40
  • Why do we choose to interchange the first two rows? We can also interchange the third row with the first row, right? – J.doe May 08 '16 at 03:32