Eigenvalues and eigenvectors of the addition of matrices If I start with matrix 
$$A = \begin{bmatrix}a & b \\ c & d \end{bmatrix}$$
and I express it as a sum
$$A = \begin{bmatrix} w & x \\ y & z \end{bmatrix} + \begin{bmatrix}(a/w-1)w & (b/x-1)x \\ (c/y-1)y & (d/z-1)z \end{bmatrix}$$
I'll now call the first and second matrix on the RHS $B$ and $C$, respectively. For clarity,
$$B = \begin{bmatrix} w & x \\ y & z \end{bmatrix}$$
$$C = \begin{bmatrix}(a/w-1)w & (b/x-1)x \\ (c/y-1)y & (d/z-1)z \end{bmatrix}$$
My question is on the resulting eigenvectors and eigenvalues. Would the original eigenvalues and eigenvectors of $A$ be preserved in the sum of $B$ and $C$? In other words, would 
$$\mbox{eig} (A) = \mbox{eig} (B) + \mbox{eig} (C)$$
since $A=B+C$ in the example above?
Also, does matrix $C$ have a special name? It looks like a "relative" matrix?
I should also note, there is nothing special about the matrix other than it is square.
 A: Suppose we have a rather dull matrix
$$A=\begin{bmatrix}
1&2\\
2&1
\end{bmatrix}\quad B=\begin{bmatrix}
1&1\\
1&1 
\end{bmatrix}
\quad C=\begin{bmatrix}
0&1\\
1&0
\end{bmatrix}$$
The eigenvalues of $A$ are $\lambda_1=3, \lambda_2=-1$ with eigenvectors $v_1=\begin{bmatrix}
1\\
1
\end{bmatrix}, v_2=\begin{bmatrix}
-1\\
1
\end{bmatrix}$
The eigenvalues of $B$ are $\lambda_1=2,\lambda_2=0$ with eigenvectors $v_1=\begin{bmatrix}
1\\
1
\end{bmatrix}, v_2=\begin{bmatrix}
-1\\
1
\end{bmatrix}$
The eigenvalues of $C$ are $\lambda_1=-1,\lambda_2=1$ with eigenvectors $v_1=\begin{bmatrix}
-1\\
1
\end{bmatrix}, v_2=\begin{bmatrix}
1\\
1
\end{bmatrix}$
While it seems that $v_1+v_2$ are equal for all three matrices, there is no similar conclusion about the relationships among the eigenvectors as you have described.
A: Based on the comments following the answer given by @Erica, I decided to post a solution which I feel is more correct. I asked @Erica to make the changes, but she never took the opportunity to. I copied the following statement from @M. Vinay in the discussion comments following the answer of @Erica. I feel this is the correct answer and the example from @Erica supports it if you change the eigenvalue and eigenvector labels for $C$.
@M. Vinay states:

If B and C have a common eigenvector, then A=B+C also has the same
  eigenvector and a corresponding eigenvalue that is the sum of the
  corresponding eigenvalues of B and C. For example, let v be an
  eigenvector of B corresponding to eigenvalue λ, and also of C
  corresponding to eigenvalue μ. Then Av=(B+C)v=Bv+Cv=λv+μv=(λ+μ)v. That
  is, v is an eigenvector of A, with eigenvalue λ+μ.

