Is there a way to diagonalize a symmetric matrix without using a calculator?

Click For Summary
SUMMARY

The discussion focuses on diagonalizing the symmetric matrix A, defined as A = [[1, 2, 3], [2, 5, 7], [3, 7, 11]]. The characteristic polynomial derived from the matrix is λ³ - 17λ² + 9λ - 1. Participants confirm that the roots of this polynomial are complex and challenging to compute without a calculator. An alternative approach suggested involves demonstrating the positive definiteness of the matrix by analyzing the eigenvalue equation using calculus to find critical points.

PREREQUISITES
  • Understanding of symmetric matrices
  • Familiarity with characteristic polynomials
  • Knowledge of eigenvalues and eigenvectors
  • Basic calculus for analyzing functions
NEXT STEPS
  • Study methods for finding eigenvalues of symmetric matrices
  • Learn about positive definiteness and its implications
  • Explore calculus techniques for finding critical points of polynomials
  • Investigate numerical methods for approximating roots of polynomials
USEFUL FOR

Students in linear algebra, mathematicians working with matrix theory, and anyone interested in understanding the properties of symmetric matrices and their diagonalization.

Locoism
Messages
77
Reaction score
0

Homework Statement


I need to diagonalize the matrix A=
1 2 3
2 5 7
3 7 11

The Attempt at a Solution



Subtracting λI and taking the determinant, the characteristic polynomial is
λ3 - 17λ2 + 9λ - 1 (I have checked this over and over)

The problem now is it has some ugly roots, none that I would never be able to find without a calculator (which is sort of the objective here). Anyways, is there some other way to find P such that
P-1AP = PTAP is a diagonal matrix?

*edit* I would just add that the matrix A is the matrix associated to the quadratic form q(v) = x2 + 5y2 + 11z2 + 4xy + 6xz + 14yz
 
Last edited:
Physics news on Phys.org
Locoism said:

Homework Statement


I need to diagonalize the matrix A=
1 2 3
2 5 7
3 7 11

The Attempt at a Solution



Subtracting λI and taking the determinant, the characteristic polynomial is
λ3 - 17λ2 + 9λ - 1 (I have checked this over and over)

The problem now is it has some ugly roots, none that I would never be able to find without a calculator (which is sort of the objective here). Anyways, is there some other way to find P such that
P-1AP = PTAP is a diagonal matrix?

*edit* I would just add that the matrix A is the matrix associated to the quadratic form q(v) = x2 + 5y2 + 11z2 + 4xy + 6xz + 14yz

It is true the roots are ugly. And there's no way to diagonalize that without some calculator assistance. On the other hand is that really the question? Do you want to show it's positive definite? Then you just need to show the eigenvalue equation has three positive roots. You can do that with calculus. Find the critical points of the eigenvalue equation. Basically, sketch a graph of it.
 

Similar threads

Replies
14
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
4
Views
7K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
15K
Replies
17
Views
8K
  • · Replies 5 ·
Replies
5
Views
5K
Replies
25
Views
5K
  • · Replies 7 ·
Replies
7
Views
4K