Diagonalize Matrix: Worked Example & Explanation

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In summary, the conversation is discussing a worked example involving LDU decomposition of a matrix. There are questions about the obtained matrices and how they were found. The solution involves row-reduction and solving equations, and reference is made to a MIT Open CourseWare lecture on the topic for further explanation.
  • #1
roam
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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

[tex]L= \begin{pmatrix}1 & 0 & 0 \\ 3 & 1 & 0 \\ 1 & -1 & 1 \end{pmatrix}[/tex]

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

[tex]L= \begin{pmatrix}1 & 0 & 0 \\ 3 & 8 & 0 \\ 1 & 4 & 5 \end{pmatrix}[/tex]

So where did they get that matrix from? Any explanation is very much appreciated.
 
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  • #2


You are mistaken, L is, in fact, the "L" in LU= A.
 
  • #3


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
[tex]U= \begin{pmatrix}1 & 3 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{pmatrix}[/tex]

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
[tex]L= \begin{pmatrix}1 & 0 & 0 \\ 3 & -1 & 0 \\ 1 & 1 & 5 \end{pmatrix}[/tex]

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.
 
  • #4


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
 
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1. What does it mean to diagonalize a matrix?

Diagonalizing a matrix means finding a new matrix that is similar to the original matrix, but with all non-diagonal elements equal to zero. This new matrix is called a diagonal matrix, and it simplifies calculations involving the original matrix.

2. Why is diagonalization important in linear algebra?

Diagonalization is important in linear algebra because it allows for easier computation and analysis of the original matrix. It also helps in finding eigenvalues and eigenvectors, which are used in many applications such as data analysis, differential equations, and quantum mechanics.

3. How do you diagonalize a matrix?

To diagonalize a matrix, you need to find the eigenvalues and eigenvectors of the matrix. Then, you use these eigenvalues and eigenvectors to create a diagonal matrix by multiplying them in a specific way. The process may involve finding the inverse of a matrix and performing matrix multiplication.

4. What are the benefits of diagonalizing a matrix?

Diagonalizing a matrix has several benefits. It simplifies calculations involving the original matrix, as all non-diagonal elements are equal to zero. It also helps in finding the inverse of the matrix, and in solving systems of linear equations.

5. Can any matrix be diagonalized?

Not all matrices can be diagonalized. A matrix can only be diagonalized if it has n linearly independent eigenvectors, where n is the dimension of the matrix. Additionally, not all matrices have real eigenvalues, which are required for diagonalization. However, a matrix can be diagonalized if it has complex eigenvalues and eigenvectors.

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