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i would know if there is some specific relationship between eigenvalues of matrices and eigenvalues of matrix given by some linear combination from these matrices.for example let us consider following case :

A =

 2     1     3
 0     1     4
 5    12    14

[E D V]=svd(A)

E =

   -0.1715   -0.4127   -0.8946
   -0.1829   -0.8789    0.4405
   -0.9681    0.2391    0.0753


D =

   19.7277         0         0
         0    2.1456         0
         0         0    1.4884


V =

   -0.2627    0.1726   -0.9493
   -0.6068    0.7354    0.3016
   -0.7502   -0.6553    0.0885

for second matrix

B =

 0    12    13
10     9    11
20     5     1

[E1,D1,V1]=svd(B)

E1 =

   -0.4725   -0.7037   -0.5305
   -0.6285   -0.1529    0.7626
   -0.6178    0.6938   -0.3701


D1 =

   27.2453         0         0
         0   17.1917         0
         0         0    1.7720


V1 =

   -0.6842    0.7182    0.1265
   -0.5291   -0.3694   -0.7639
   -0.5019   -0.5896    0.6328

and finally

 C=A+B;
>> [E2,D2,V2]=svd(C)

E2 =

   -0.4193   -0.7946   -0.4392
   -0.4765   -0.2191    0.8514
   -0.7727    0.5663   -0.2868


D2 =

   42.5915         0         0
         0   13.1294         0
         0         0    2.5662


V2 =

   -0.5851    0.7903    0.1817
   -0.5483   -0.2204   -0.8067
   -0.5975   -0.5717    0.5623

clearly if i add first two i will get greater values on diagonal,then it is on the third matrix diagonals,so my question is will linear combination of matrices also affect on their eigenvalues?let say

we have some matrices say $A$ and $B$ so that

$k*A+m*B=c$

does it also creates some sort of linear combinations between their eigenvalues?or does there exist some special matrices,so that Eigenvalue of matrix given by sum of two matrix is just sum of their eigenvalues?thanks in advance

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For diagonal matrices as well as matrices that share the same eigenvectors what you claim is true. I am not sure of the necessary condition.

The eigenvalue is the scalar by which the eigenvector is expanded or contracted by the original matrix $A$. If you have two matrices $A$ and $B$, let $\lambda_{1A}$ and $\lambda_{1B}$ be (one of) their eigenvalues and the corresponding eigenvectors be $v_{A}$ and $v_{B}$. The eigenvector of $A+B$ may not be the linear combination of $v_{A}$ and $v_{B}$ as $v_{B}$ will be twisted (not just multiplied) by A perhaps having a component along all other eigenvectors; and $v_{A}$ will be similarly twisted by B.

For perhaps a deeper answer beyond my understanding, I found this reference. http://arxiv.org/abs/math/9908012

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  • $\begingroup$ thanks in advance ,i will review tomorrow ,now it is too much late,thanks for help $\endgroup$ – dato datuashvili Sep 18 '13 at 19:12

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