Sufficient condition for eigenvalues

Question

Suppose that I have a matrix $A$ and I want to find its eigenvalues. I then set up the equation $$Av=\lambda v$$ which becomes $$(A-\lambda I)v =0$$ Now I now that this is true if the determinant of $A-I$ is 0, and this is a necessary condition, but how do I know that this is sufficient for finding all eigenvalues?

Answer 1

You messed up on the details. So, starting at $$Av=\lambda v$$, we subtract and insert the identity operator: $$Av-\lambda Iv=0$$. Now, using linearity of matrix addition, we can "factor out" the vector, to be left with $$(A-\lambda I)v=0$$.

This says that $v$ is in the null space of $A-\lambda I$. If we want a nontrivial solution (which we need a non 0 vector for eigenvector), we need the null space to be nontrivial, this is the case if and only if the operator is not invertible. A matrix/operator is not invertible if and only if the determinant is zero, so it is necessary and sufficient that $\lambda$ is an eigenvalue if $$\det (A-\lambda I)=0$$