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I have two matrices

$$
A=beginpmatrix
a & 0 & 0 \
0 & b & 0 \
0 & 0 & c
endpmatrix
quad
text and
quad
B=beginpmatrix
d & e & f \
d & e & f \
d & e & f
endpmatrix
$$

In reality mine are more like 1000 x 1000 matrices but the only thing that is important for now is that the left matrix is diagonal and the right one has one row that repeats itself.

Obviously the eigenvalues of the left matrix are its diagonal components. I want to create a new matrix C

$$C = A+B=beginpmatrix
a & 0 & 0 \0 & b & 0 \0 & 0 & c endpmatrix+beginpmatrix d & e & f \d & e & f \d & e & f endpmatrix=beginpmatrix a+d & e & f \d & b+e & f \d & e & c+f endpmatrix$$

I am now wondering how the eigenvalues of this new matrix C are related to the eigenvalues of the diagonal matrix A. Can I use an argument that uses row reduction in order to relate the eigenvalues of both matrices?

The reason why I am asking is that my 1000 x 1000 matrix (implemented in mathematica) that is described as above gives me almost the same eigenvalues as the corresponding diagonal matrix (only a few eigenvalues differ) and I really cannot think of any reason why that should be the case.

EDIT:

I implemented a simple code in mathematica to illustrate what I mean. One can see that every eigenvalue of the diagonal matrix A appears in C:

 dim = 50;

 A = DiagonalMatrix[Flatten[RandomInteger[0, 10, 1, dim]]];

 mat = RandomReal[0, 100, 1, dim];
 B = ArrayFlatten[ConstantArray[mat, dim]];

 c = A + B;

 Abs[Eigenvalues[A]]
 Round[Abs[Eigenvalues[c]], 0.01]

 (*10, 10, 10, 10, 10, 10, 9, 9, 9, 9, 9, 9, 8, 8, 8, 8, 7, 7, 7, 7, 7, 
 6, 6, 6, 6, 5, 5, 5, 5, 5, 4, 4, 4, 4, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 
 1, 1, 1, 0, 0, 0*)

 (*2084.89, 10., 10., 10., 10., 10., 9.71, 9., 9., 9., 9., 9., 8.54, 
 8., 8., 8., 7.72, 7., 7., 7., 7., 6.61, 6., 6., 6., 5.44, 5., 5., 5., 
 5., 4.29, 4., 4., 4., 3.51, 3., 3., 3., 3., 2.28, 2., 2., 2., 2., 
 1.21, 1., 1., 0.33, 0., 0.*)

The reason is that your second matrix is a rank-one update of your first matrix:
$$
Bequiv uv^t
$$
where $u=(1,1,1)$ and $v=(d,e,f)$. Therefore, the new eigenvalues are typically a small perturbation of the old ones, and there are some known formulas for special cases. See e.g. these lectures or the references in this math.OF post.

It doesn't happen here:

SeedRandom[0];
aa = RandomReal[-10, 10, 1000, 1000];
bb = ConstantArray[RandomReal[-10, 10, 1000], 1000];

eva = Eigenvalues@aa;
evc = Eigenvalues[aa + bb];

ListPlot[ReIm@eva, ReIm@evc, ImageSize -> Large, MaxPlotPoints -> 1000]

OTOH, it does happen here:

bb = ConstantArray[RandomReal[-1, 1 1*^-8, 1000], 1000];

eva - Eigenvalues[aa + bb] // Abs // Max
(* 5.4818*10^-7 *_)

Of course, the explanations should be obvious.