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 GramSchmidt)
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 ndimensional vector, and d is a constant, $ is the ndimensional zero vector and I is the nxn identity matrix.
Ax=dx
Axdx=$
(AdI)x=$
det(AdI)=0 solve for all values of d
(this is called the characteristic equation which gives the ndegree 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.
