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 $A\mathbf X = \lambda\mathbf x$.
A: Assume that $A\mathbf{k}=\lambda\mathbf{k}$ for a non-zero vector $\mathbf{k}$ and a scalar $\lambda\in \mathbb{F}$. 
This is equivalent to
$$(A-\lambda I)\mathbf{k}=\mathbf{0}.$$
Now if $A-\lambda I$ is invertible then we can left multiply by $(A-\lambda I)^{-1}$:
$$\begin{align}
(A-\lambda I)^{-1}(A-\lambda I)\mathbf{k} &=(A-\lambda I^{-1})\mathbf{0}
\\ \Rightarrow I\mathbf{k}=\mathbf{k}&=\mathbf{0}.
\end{align}$$
Now this is a contradiction therefore we must have that $A-\lambda I$ is not invertible. This is the logic I work with anyway and now the other answers discuss where the determinant comes in.
Now the $\det(A-\lambda I)$ is a polynomial in $\lambda$ and so by the Fundamental Theorem of Algebra, there exists a number $\mu$ such that $A-\mu I$ is not invertible.
Therefore, using properties of linear maps, $A-\mu I$ sends something non-zero, say $\mathbf{v}$, to zero:
$$
\begin{align}
(A-\mu I)\mathbf{v}&=0
\\ \Rightarrow A\mathbf{v}&=\mu \mathbf{v}.
\end{align}$$
This $\mathbf{v}$ is an eigenvector for the eigenvalue $\mu$.
A: Let $\lambda$ choose such that $det(A-\lambda I)=0$. Now, consider the system: $(A-\lambda I)X=0$. Because of $det(A-\lambda I)=0$, the matrix $K=A-\lambda I$ isn't invertible. So, the system $KX=0$ has a non-zero solution.
Hint: the following are equivalent:
$(1)$ The matrix $S$ is invertible
$(2)$ $det(S)\neq 0$
$(3)$ The system $SX=0$ has a solution $X\neq 0$
A: A singular matrix has a zero determinant and isn't invertible, so that non-trivial solutions can exist. Recall the formula for the matrix inverse.
