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If I start with matrix A given by

$A = \begin{bmatrix}a & b \\ c & d \end{bmatrix}$

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 eig(A) = eig(B) + 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.

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  • $\begingroup$ If $B$ and $C$ commute, at best one can say 'eig($A$)$\subseteq$eig($B$)+eig($C$)'. $\endgroup$ – Arundhathi Jun 24 '16 at 6:26
  • $\begingroup$ @M.Vinay You are correct. I will edit that to prevent any confusion. $\endgroup$ – ThatsRightJack Jun 24 '16 at 8:13
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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.

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  • $\begingroup$ Thanks for the answer. I'm able to follow your example, but if the matrix A is given by block elements a,b,c,d, where none of them are equal, does that make a difference in your example? $\endgroup$ – ThatsRightJack Jun 24 '16 at 5:46
  • $\begingroup$ Also, if I'm not mistaken, your example above seems to show that eig(A)=eig(B)+eig(C) holds? I obviously don't want to draw any generic conclusions, but if it is true under certain conditions, then I can at least check if it holds for my specific application. $\endgroup$ – ThatsRightJack Jun 24 '16 at 5:53
  • $\begingroup$ OP, sorry for not being clear. It seems that the example shows that eig(A)$\neq$eig(B)+eig(C) as you can see from the eigenvectors. Thus, since we have a counterexample, we are done. $\endgroup$ – Claire Jun 24 '16 at 5:59
  • $\begingroup$ I'm obviously not an expert in eigen problems, but is seems that all vectors are the same and the eigenvalues of B and C add up to A? Is there no significance to that? $\endgroup$ – ThatsRightJack Jun 24 '16 at 6:06
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    $\begingroup$ @Erica I thought that might be the cause of confusion. I will change that wording to prevent confusion. It sounds like M.Vinay is deferring to you for presenting the correct answer? About 5 comments up, M.Vinay gives a good mathematical explanation. Would you mind integrating that into your answer and making the necessary edits so that I can accept your answer. Also, do you have any comments about the idea of a "relative" matrix? Is that something that is well known or of any significance? (These last two questions, I'm just curious). Thanks for your help and time Erica. $\endgroup$ – ThatsRightJack Jun 24 '16 at 21:23
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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.

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