# Mathematica problem, nontrivial solution for matrix equation Ax=0

by Uncle_John
Tags: equation, mathematica, matrix, nontrivial, solution
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,552 The equation Ax= 0 has a non-trivial solution if and only if A is not one-to-one. That is the same as saying that its determant is 0 and that it has 0 as an eigenvalue. The standard way to find an eigenvalue, $\lambda$ for matrix A is to solve the equation $det(A- \lambda I)= 0$. If A is an n by n matrix, that will be a polynomial equation of degree n and so has n solutions (not necessarily all distinct, not necessarily real). IF $\lambda$ really is an eigenvalue, then $Ax= \lambda x$ or $Ax- \lambda x= (A- \lambda I)x= 0$ has, by definition of "eigenvalue", a non-trivial solution. That is, some of the equations you get by looking at individual components will be dependent. Note that x= 0 always will be a solution, just not the only one. Perhaps if you posted a specific example, we could point out errors. The most obvious one, if you "keep getting x=0", is that what you think is an eigenvalue really isn't!