*Slightly* Altered Eigenvalue Problem Consider the 2 Matricies:
$$
A=\left( \begin{array}{ccc}
3 & 1 \\
6 & -2 \end{array} \right)
$$
$$
B=\left( \begin{array}{ccc}
-1 & 1 \\
0 & 0 \end{array} \right)
$$
Find $x\in \mathbb{R}^2$ s.t. $\mathbf{A}x=\lambda\mathbf{B} x$. The problem also says "note: lambda may depend on $x$".
So my first instinct was set the determinant of $\mathbf{A}-\lambda\mathbf{B}$ to zero and solve $\lambda$. And I do get an answer that works ($\lambda = 3$).
Can anyone make sense of this note though? I'm worried I might be missing solutions.
 A: Since $Ax=\lambda Bx$, we have $Ax-\lambda Bx=0$ or $(A-\lambda B)x=0$
When we solve for eigenvalues, we have $Ax=\lambda x$ so $Ax-\lambda x=0 \rightarrow (A-\lambda I)x=0$. Thus, $\lambda$ is an eigenvalue iff $A-\lambda I$ is singular or $det(A-\lambda I)=0$.
Thus, in this case, it is correct that $\lambda$ will be an "eigenvalue" if $A-\lambda B$ is singular or $det(A-\lambda B)=0$
A: Your '' instinct'' is correct only if we want $Ax=\lambda Bx $  for all $x$. 
But, if we search some $x$ for which the equation is true, than we have to solve
$$
\begin{bmatrix}
3&1\\
6&-2
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}
=
\begin{bmatrix}
-1&1\\
0&0
\end{bmatrix}
\begin{bmatrix}
\lambda x\\
\lambda y
\end{bmatrix}
$$
in this case we have
$$
\begin{cases}
3x+y=-\lambda x+\lambda y\\
6x-2y=0
\end{cases}
$$
and this gives $y=3x$ and $\lambda=3$ as you found. But in general $\lambda$ can be dependent on $x$ (or $y$) as you can easily see if the second row of $B$ is not all null. 
A: When $\mathbf{B}$ is invertible (which isn't the case here), $$\mathbf{A}x=\lambda\mathbf{B} x$$ can be rewritten
$$\mathbf{B}^{-1}\mathbf{A}x=\lambda x,$$ which is an ordinary Eigenproblem.
That should be enough for you to understand the note.
