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I want to ask if adding the rows would somehow affect my eigenvalues , I just added 1st row to 3rd row and after finding the eigenvalues for both $A$ and $A'$ I didn't get the same result.

Adding the rows doesn't not affect the determinant nor the rank, so why should eigenvalues be different ?

$A=$$\begin{bmatrix}1&2&1\\6&-1&0\\-1&-2&-1 \end{bmatrix}$

$A'=$$\begin{bmatrix}1&2&1\\6&-1&0\\0&0&0 \end{bmatrix}$

Eigenvalues of A' : $0,\sqrt{13},-\sqrt{13}$

Eigenvalues of A : $0,3,-4$

Widawensen
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Oleg
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1 Answers1

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The reason is that determinant and rank alone don't determine the eigenvalues. Maybe you know that the trace of a matrix is the sum of its eigenvalues. Now, adding rows in a matrix usually changes its trace. You can convince yourself of this with your example.

jflipp
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  • oh , yes . for me it is a strange property , namely because I don't understand it , why should sum of trace = sum of eigenvalues ? – Oleg Dec 05 '14 at 08:41
  • @Oleg Maybe this helps: http://math.stackexchange.com/questions/546155/proof-that-the-trace-of-a-matrix-is-the-sum-of-its-eigenvalues And yeah, there is no such invariant-under-addition-of-rows property for eigenvalues – GDumphart Dec 05 '14 at 08:42