Given the following matrix,
$$
A =
\begin{bmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{bmatrix}
$$
assuming eigenvectors exist for $A$, they can be found by first solving for $\lambda$ (i.e. the roots of the equation) in the characteristic equation:
$$
\text{det}(A-\lambda I) = 0
$$
I know that if the determinant of a matrix is equal to zero, the matrix is non-invertible; also, I know that, for a given a matrix $A$, eigenvector $x$ and eigenvalue $\lambda$, $Ax = \lambda x$; hence, with respect to $x$, $A$ is somewhat "equivalent" to $\lambda$, but I'm not entirely sure why solving for the characteristic equation provides the eigenvalues for the matrix $A$.
Given this, my question is: could somebody provide some logic on why the above works?
Also, as an aside, assuming the correct eigenvalues have been found, solving for the system,
$$
\begin{bmatrix}
a_{11}-\lambda & a_{12} \\
a_{21} & a_{22}-\lambda
\end{bmatrix}
\begin{bmatrix}
x_{1} \\
x_{2}
\end{bmatrix}
= \vec{0}
$$
will provide the associated eigenvector for a given $\lambda$.
Is it correct to assume the reasoning behind this is because of the following.
Firstly,
$$
\begin{bmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{bmatrix}
\begin{bmatrix}
x_{1} \\
x_{2}
\end{bmatrix} =
\lambda
\begin{bmatrix}
x_{1} \\
x_{2}
\end{bmatrix}
$$
This implies,
$$
\begin{bmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{bmatrix}
\begin{bmatrix}
x_{1} \\
x_{2}
\end{bmatrix}
–
\begin{bmatrix}
\lambda & 0 \\
0 & \lambda
\end{bmatrix}
\begin{bmatrix}
x_{1} \\
x_{2}
\end{bmatrix}
=
\left(
\begin{bmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{bmatrix}
–
\begin{bmatrix}
\lambda & 0 \\
0 & \lambda
\end{bmatrix}
\right)
\begin{bmatrix}
x_{1} \\
x_{2}
\end{bmatrix}
=
\vec{0}
$$
Thus,
$$
\begin{bmatrix}
a_{11} – \lambda & a_{12} \\
a_{21} & a_{22} – \lambda
\end{bmatrix}
\begin{bmatrix}
x_{1} \\
x_{2}
\end{bmatrix}
=
\vec{0}
$$
Best Answer
You're nearly there.
A square matrix $B$ is non-invertible if and only if there exists a non-zero vector $v$ such that $Bv=0$. For example, necessity follows because $Bv=0$ implies that the function $f(v)=Bv$ is not injective (or "one-to-one") and hence not bijective (which is necessary for $f$ to have an inverse -- see the link). It is easy to see that $B$ is not injective since $B(2v)=2Bv=2\cdot0=0=Bv$, that is, $f$ maps both $v$ and $2v$ onto the same point $0$.
So, if $\lambda$ is an eigenvalue of $A$, and $x$ is its corresponding eigenvector,
$$Ax=\lambda x\Leftrightarrow Ax-\lambda x=0\Leftrightarrow (A-I\lambda)x=0.$$
Hence, $\lambda$ must be such that $B=A-I\lambda$ is non-invertible. Thus $\lambda$ is an eigenvalue of $A$ if and only if it satisfies the characteristic equation $\det(A-I\lambda)=0$.
Aside: If $\lambda$ is real, $x$ is simply a vector that function $f(v)=Av$ maps onto "itself" just stretches it and/or reflects it across the origin. For example, if $\lambda=2$, $f(x)$ simply "stretches" $x$ by two, and if $\lambda=-1$, $f(x)$ reflects $x$ across the origin (rotates it by $180$ degrees).
Edit: You might find these cam-casts of interest.