Proving Eigenvalues of a Matrix

Question

I just want to confirm this, to prove that something is an eigenvalue, we just need to find a vector satisfying the equation $A x = \lambda x$, right?

Answer 1

Yes*, but this is often a challenging enterprise, since apriori, it is not known what $\lambda$ is. A different way to go about this:

let $x$ be a vector. Then

$$Ax-\lambda x =0 \implies (A-\lambda I)x=0$$

with $x$ nontrivial, has a solution if and only if the determinant $\mathrm{det}(A-\lambda I)=0$. In this case, you can fetch the eigenvalues, and solve for the eigenvector.

For a nondeterminant (and slightly more abstract) approach, see Down With Determinants.

*as in the comments, make sure that the solution is nontrivial.