Degree of the characteristic polynomial of a matrix pencil On "Applied Numerical Linear Algebra" by J.W. Demmel I found the following proof.
Definition. Let $A$, $B$ be $n \times n$ matrices, if $P(\lambda) = det(A − \lambda B)$ is not identically zero, the pencil $A − \lambda B$ is called regular.
Theorem. Let $A − \lambda B$ be regular. If B is singular, then $ deg (P(\lambda)) = rank(B)$.
Proof. If $B$ is singular, then take $P(\lambda) = det(A − \lambda B)$, write the SVD of $B$ as $B = U\Sigma V^T$, and substitute to get $P(\lambda) = det(A − \lambda U\Sigma V^T) = det(U(U^T AV − \lambda \Sigma)V^T) = ± det(U^T AV − \lambda\Sigma)$.
Since $rank(B) = rank(\Sigma)$, only $rank(B)$ $\lambda$′s appear in $U^T AV − \lambda\Sigma$, so the degree of the polynomial $det(U^T AV − \lambda\Sigma)$ is rank(B).
I cannot find errors in this proof, but if you consider:
$$
    A = \begin{bmatrix}
    1 & 2\\
    0 & 2\\
    \end{bmatrix}
$$
$$
    B = \begin{bmatrix}
    1 & 0\\
    1 & 0\\
    \end{bmatrix}
$$
then $rank(B) = 1$ while $P(\lambda) = 2(1-\lambda)+2\lambda = 2$ is constant. Why does this happen?
 A: Hope this answer is still relevant. The presented proof is a constructive proof, which means that we can use the result directly to calculate the characteristic polynomial of your example.
The singular value decomposition of the $B$ matrix is equal to
$$
B = U \Sigma V^T = 
\left(
\frac{\sqrt{2}}{2}
\begin{bmatrix}
 -1 & -1 \\ -1 & 1
\end{bmatrix}
\right)
\begin{bmatrix}
 \sqrt{2} & 0 \\ 0 & 0
\end{bmatrix}
\begin{bmatrix}
-1&0\\0&1
\end{bmatrix}.
$$
We can now multiply the matrix pencil from the left and right by $U^T$ and $V^T$.
The transformed pencil is equal to
$$
U^T (A-\lambda B) V = 
\frac{\sqrt{2}}{2} \begin{bmatrix}
1 & -4 \\ 1 & 0
\end{bmatrix} - \lambda
\begin{bmatrix}
\sqrt{2} & 0 \\ 0 & 0
\end{bmatrix}.
$$
The determinant of this matrix pencil is than equal to
$$
\det\left(
\frac{\sqrt{2}}{2}
\begin{bmatrix}
1 - 2\lambda & -4 \\ 1 & 0
\end{bmatrix} 
\right) = 2.
$$
Which is exactly what you found. This implies that the presented proof is wrong.
The correct theorem should be:
Theorem:
Let $A-\lambda{B}$ be regular. If $B$ is singular, then $\text{deg}(a-\lambda B)$ is at most equal to $\text{rank}(B)$.
