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 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.