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.
Best Answer
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.