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Noxide
#5
Apr18-10, 02:20 AM
P: 118
Given:
A is symmetric

Conclusion:
if A is symmetric then that means A equals its transpose and is of size nxn (THIS IS ALWAYS TRUE)
if A is symmetric then it has n independent eigenvectors (THIS IS ALWAYS TRUE)

The matrix factorization of A is SDS^(-1)
The n columns of S are the n independent eigenvectors and for a symmetric matrix those eigenvectors are orthogonal when A is symmetric and can be made orthonormal (which makes finding the factorization of A a lot easier, via Gram-Schmidt)
The diagonal entries of the Diagonal matrix D are the eigenvalues associated with the eigenvectors

So, to find those diagonal entries and those independent eigenvectors the general form is as follows where A is an nxn matrix, x is an n-dimensional vector, and d is a constant, $ is the n-dimensional zero vector and I is the nxn identity matrix.

Ax=dx
Ax-dx=$
(A-dI)x=$
det(A-dI)=0 solve for all values of d
(this is called the characteristic equation which gives the n-degree polynomial which is used to determine the values of d (your eigenvalues that satisfy the equation))





Edit: Hope this helps, I'm taking intro linear algebra this semester so if anything is wrong please let me know.