Q:

Eigenvalue problem intuition

Ak = $\lambda$k
(A - $\lambda$I)k = 0
det(A - $\lambda$I) = 0
I understand that we are looking for non trivial solutions where k 
$\neq$ 0. So then we choose $\lambda$ such that (A - $\lambda$I) = 0. Why then the determinant? Why must the eigenvalues be chosen such that
(A - $\lambda$I) is singular? What is actually happening?    

A:

Saying that $A$ has an eigenvector $\mathbf k$ with eigenvalue $\lambda$ means exactly the same as saying that the matrix $(A - \lambda I)$ has $\mathbf k$ as an eigenvector with eigenvalue $0$. This is what the manipulations $A\mathbf{k} = \lambda \mathbf k \Longleftrightarrow (A - \lambda I)\mathbf k = 0$ tells you.
Now, if the matrix $A - \lambda I$ has a non-trivial eigenvector of eigenvalue $0$, that means precisely that it is singular (it's there somewhere in the big list called "the invertible matrix theorem", which I assume is in your book). That means that we can check for it by calculating the determinant.
Once we calculate the determinant as a function of $\lambda$ (this function is called the characteristic polynomial of $A$), we can go hunt for those specific values of $\lambda$ that makes $(A-\lambda I)$ into a singular matrix. Once we have such a $\lambda$, we know that it must be an eigenvalue of some eigenvector of $A$. Once we know that, we can try to calculate the eigenvector by solving $A\mathbf X = \lambda\mathbf x$.

