Finding a solution to $C\vec{x} = (A - \lambda I_A)\vec{x} = 0$ is the equivalent of considering the determinant of $C$ when it is zero. This means the matrix is linearly dependent and has infinite sol'ns.
My question is, the determinant is $zero$ when the matrix has no solutions as well. So doesn't this return eigenvectors that don't even cross each other?
Maybe I am considering that $\vec{0}$ is included under the spectrum so this wouldn't be true...is my reasoning correct?
On
A square matrix $C$, say over $\mathbb{R}$, is invertible if and only if its determinant is non-zero; if and only if $C\mathbf{x}=\mathbf{0}$ has a unique solution. The unique solution is $\mathbf{x}=\mathbf{0}$ clearly- if and only if columns/ rows are linearly independent. Otherwise there is an infinite number of solutions - in this case, as you say, the columns/rows of $C$ are linearly dependent- any non-zero solution $\mathbf{x}$ has as components the coefficients of linear dependence of column vectors.
To see this observe that, if columns of $C$ are $\mathbf{c}_1,\mathbf{c}_2,\dots \mathbf{c}_k$, and $\mathbf{x}=[x_1;x_2;\dots ,x_k]$, then
$C\mathbf{x}=x_1\mathbf{c}_1+x_2\mathbf{c}_2+\dots +x_k\mathbf{c}_k$.
You seem to be thinking about $A\mathbf{x} = \mathbf{b}$ where $\mathbf{b} \neq \mathbf{0}$. In this case then yes, it's possible that there are no solutions $\mathbf{x}$ which solve the equation when $\det A = 0$. However, if $\mathbf{b} = \mathbf{0}$ and $\det A = 0$ then there is always a non-zero solution to $A\mathbf{x} = \mathbf{0}$. In other words, singular matrices always have a non-trivial nullspace.