Diagonalize Matrix: Worked Example & Explanation

[roam]

Homework Statement

This is a worked example:

[PLAIN]http://img24.imageshack.us/img24/7783/44818829.gif

The Attempt at a Solution

So, in the answer I don't understand how they obtained

L= \begin{pmatrix}1 & 0 & 0 \\ 3 & 1 & 0 \\ 1 & -1 & 1 \end{pmatrix}

I don't think this "L" here is the lower triangular matrix used in the LU factorization of A. Because I followed the LU decomposition algorithm and ended up with

L= \begin{pmatrix}1 & 0 & 0 \\ 3 & 8 & 0 \\ 1 & 4 & 5 \end{pmatrix}

So where did they get that matrix from? Any explanation is very much appreciated.

 

[HallsofIvy]

You are mistaken, L is, in fact, the "L" in LU= A.

 

[Mark44]

roam,
I don't follow this example, either. When a problem asks to diagonalize a matrix, I reflexively think of eigenvalues and eigenvectors. In this problem the entries on the main diagonal of D aren't the eigenvalues of A (which happen to be about 11.6, 2.5, and -.17).

The only thing I understand about this problem is that they have row-reduced A to an upper triangular matrix U, where
U= \begin{pmatrix}1 & 3 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{pmatrix}

Assuming that A = LU, and noticing that U is invertible, you can solve this equation for L, with L = AU^-1.
Doing this, I get
L= \begin{pmatrix}1 & 0 & 0 \\ 3 & -1 & 0 \\ 1 & 1 & 5 \end{pmatrix}

I don't know how they got the diagonal matrix unless there's some technique for factoring a lower triangular matrix (L here) into another lower triangular matrix and a diagonal matrix that I don't know about. It is probably significant that their lower triangular matrix has 1's on the main diagonal.

 

[hgfalling]

It seems that this is an example of LDU decomposition. There's some stuff about it in the following lecture from MIT's Open CourseWare, including an example and an algorithm that might help you:

http://ocw.mit.edu/NR/rdonlyres/Mathematics/18-335JFall-2004/BA2C6B59-4639-4FC4-ACEF-B0362FB16CC3/0/lecture11.pdf