Why is the determinant useful when finding eigenvectors? I understand that the eigenvectors of a linear transformation are the vectors that, when transformed, are simply scaled by a factor of the corresponding eigenvalue. Therefore
$$
A\vec{x}=\lambda\vec{x}.
$$
We can replace the scaling of the vector $\vec{x}$ by $\lambda$ with its equivalent linear transformation, $\lambda I$. Then we can rearrange this equation as follows:
$$
A\vec{x}-(\lambda I)\vec{x}=\begin{bmatrix}0\\0\end{bmatrix}\\ 
(A-\lambda I)\vec{x}=\begin{bmatrix}0\\0\end{bmatrix}\\ 
$$
For this to be true either $\vec{x}$ must be a zero vector or the transformation $A-\lambda I$ must have $\vec{x}$ in its null-space. The course I'm following says that we can solve for $\lambda$ by finding when the determinant of $A-\lambda I$ is zero, but why does that guarantee that $\vec{x}$ will become the zero vector when transformed by $A-\lambda I$? I understand that for $\vec{x}$ to be an eigenvector this transformation must have a determinant of zero, but I don't see how the inverse can be true; having a determinant of zero doesn't gaurantee that $\vec{x}$ will be transformed into the zero vector. For example, let
$$
(A-\lambda I) = \begin{bmatrix}1&2\\0&0\end{bmatrix}, \vec{x}=\begin{bmatrix}1\\1\end{bmatrix}
$$
Then, although
$\det\begin{bmatrix}1&2\\0&0\end{bmatrix} = 0$, it does not gaurantee $(A-\lambda I)\vec{x}$ to be the zero vector.
$$
\begin{bmatrix}1&2\\0&0\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}3\\0\end{bmatrix} 
$$
We can see that the transformation gives us $\begin{bmatrix}3\\0\end{bmatrix}$ and not the zero vector. Why does finding when the determinant of this transformation is equal to zero help us find the eigenvectors? Am I missing something?
 A: The condition $\det(A-\lambda I)=0$ guarantees that there exists a nonzero $\vec{x}$ such that $(A-\lambda I)\vec{x}=0$. It does not mean that any $\vec{x}$ will work, and your $\vec{x}$, with your $A-\lambda I$, does not work. There is an $\vec{x}$ that does, however, namely
$$
\vec{x}=\begin{bmatrix}-2 \\1\end{bmatrix}.
$$
Any scalar multiple of this vector will also work.
The usual situation is a bit different from yours. Usually you have $A$, but $\lambda$ is not known. So $\det(A-\lambda I)$ will be an expression involving $\lambda$. In fact, it will be a polynomial in $\lambda$ whose degree is the dimension of $A$. The roots of this polynomial are the eigenvalues, and each one determines a space of eigenvectors. If I'm understanding correctly, you must already know $\lambda$ since your matrix $A-\lambda I$ is a constant matrix.
There seems to be some confusion at the end of your post where you were expecting $(A-\lambda I)\vec{x}$ to lie in the span of $\vec{x}$. In fact, you should be expecting $(A-\lambda I)\vec{x}$ to be the zero vector.
Perhaps you meant that $A\vec{x}$ should be in the span of $\vec{x}$? If so, that would be correct, but you haven't told us what $A$ is—you've only told us what $A-\lambda I$ is.
Added: Here's a complete example. We will compute the eigenvalues and eigenvectors of the matrix
$$
A=\begin{bmatrix}1 & 2\\ 3 & 0\end{bmatrix}.
$$
We want to find pairs $(\lambda,\vec{x})$ such that $A\vec{x}=\lambda\vec{x}$, or $(A-\lambda I)\vec{x}=\vec{0}$.
Now
$$
\det\begin{bmatrix}1-\lambda & 2\\ 3 & -\lambda\end{bmatrix}=\lambda^2-\lambda-6,
$$
which has roots $\lambda=-2,\ 3$. If $\lambda=-2$, then
$$
(A-\lambda I)=\begin{bmatrix} 3 & 2\\ 3 & 2\end{bmatrix}.
$$
Now
$$
\begin{bmatrix} 3 & 2\\ 3 & 2\end{bmatrix}\begin{bmatrix}2\\ -3\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix},
$$
and so we have found an eigenvector. Indeed
$$
\begin{bmatrix}1 & 2\\ 3 & 0\end{bmatrix}\begin{bmatrix}2\\ -3\end{bmatrix}=\begin{bmatrix}-4\\ 6\end{bmatrix}=-2\begin{bmatrix}2\\ -3\end{bmatrix}.
$$
If you try out the other eigenvalue, $\lambda=3$, you should find that $\vec{x}=\begin{bmatrix}1\\ 1\end{bmatrix}$ is an eigenvector.
Second addition: There seems to be a fundamental misconception here. Except in rare cases, not every vector is an eigenvector. So we can't (and don't) require that $A-\lambda I$ map every $\vec{x}$ to $\vec{0}$.
On the other hand $A-\lambda I$ always maps $\vec{0}$ to $\vec{0}$, no matter what $\lambda$ is. That's not what we're interested in either. We don't consider the zero vector to be an eigenvector.
To seek an eigenvector means to seek a particular non-zero $\vec{x}$ such that $A-\lambda I$ maps $\vec{x}$ to $\vec{0}$. If $\det(A-\lambda I)\ne0$ then there is no such non-zero $\vec{x}$. But if $\det(A-\lambda I)=0$, then there does exist such a non-zero $\vec{x}$, although we still have to figure out what $\vec{x}$ is by solving the linear system $(A-\lambda I)\vec{x}=\vec{0}$.
So in your example, $\det\begin{bmatrix}1 & 2\\ 0 & 0\end{bmatrix}=0$ guarantees that there is a non-zero $\vec{x}$ such that $\begin{bmatrix}1 & 2\\ 0 & 0\end{bmatrix}\vec{x}=\vec{0}$, but $\begin{bmatrix}1\\ 1\end{bmatrix}$ is not such an $\vec{x}$.
Let's find such an $\vec{x}$ by writing $\vec{x}=\begin{bmatrix}x\\ y\end{bmatrix}$ and solving for $x$ and $y$:
$$
\begin{bmatrix}1 & 2\\ 0 & 0\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}0\\ 0\end{bmatrix}.
$$
The second row of this matrix equation holds true no matter what $x$ and $y$ are, so the only condition is that provided by the first row, $1x+2y=0$. This gives $x=-2y$, so any vector of the form
$$
\vec{x}=\begin{bmatrix}-2y\\ y\end{bmatrix}
$$
is an eigenvector, as long as $y$ is not zero. (If $y$ is zero, we get the zero vector, which we already knew about and aren't interested in.)
What goes wrong if $\det(A-\lambda I)\ne0$? In that case $A-\lambda I$ is an invertible matrix. So
$$
(A-\lambda I)\vec{x}=\vec{0}
$$
has the same solution set as
$$
(A-\lambda I)^{-1}(A-\lambda I)\vec{x}=(A-\lambda I)^{-1}\vec{0},
$$
which reduces to $\vec{x}=\vec{0}$,
the solution we aren't interested in.
