If $v_1,v_2...v_n$ are eigenvectors of an $n\times n$ matrix, is it possible to find $\sum_{i=0}^{n} v_i$ without explicitly finding the eigen vectors?

All the vectors are normalized to 1 and $+1$ or $-1$ in front of it doesn't matter.

It is most probably not possible. I couldn't derive any expression and also couldn't find anything online. So, is it possible?

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    What are you given? If you are given $v_1,\dots,v_n$, then of course it's possible. If you are just given the matrix, then it's possible to find the eigenvectors (at least, numerically), so again it's possible to find the sum. So, what are you really asking? – Gerry Myerson Mar 23 '16 at 11:28
  • I am given a matrix. I want to find the sum of eigenvectors without explicitly finding the individual eigenvectors. – Goobs Mar 23 '16 at 11:30
  • what is the rank of your matrix? – Math-fun Mar 23 '16 at 11:31
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    Since any scalar multiple of an eigenvector is again an eigenvector you can't know "the eigenvectors" when all you have is the matrix. So you can't know their sum. What do you need it for? – Ethan Bolker Mar 23 '16 at 11:35
  • Note that an eigenvector is not unique. Each scalar multiple of it will be another eigenvector. For a fixed generic matrix, this sum can be anything (except zero). – Friedrich Philipp Mar 23 '16 at 11:35
  • @Ethan, note that Goobs has normalized the eigenvectors. – Gerry Myerson Mar 23 '16 at 11:36
  • @Friedrich, note that Goobs has normalized the eigenvectors. – Gerry Myerson Mar 23 '16 at 11:36
  • @GerryMyerson Then there are still $2^n$ possibilities for the sum. – Friedrich Philipp Mar 23 '16 at 11:36
  • @Friedrich, OK, so normalize it even more, by insisting the norm is 1 and the first component is positive. Then it's a well-defined problem (but I don't think there's any way to solve it that doesn't amount to finding the eigenvectors and adding them). – Gerry Myerson Mar 23 '16 at 11:39
  • Do you know the rank of your matrix? Do the eigenvectors constitute a complete basis of the full space? – seoanes Mar 23 '16 at 11:46
  • Yeah the eigen vectors constitute a complete basis – Goobs Mar 23 '16 at 11:48
  • Any of those $2^n$ possibilities is O.K. – Goobs Mar 23 '16 at 11:49
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    You need one more assumption to specify even normalized eigenvectors:: all the eigenvalues must have multiplicity one. With that assumption let's see if there's an answer to your question ... – Ethan Bolker Mar 23 '16 at 11:56
  • You only know the matrix, right? Anything more? Is it, e.g., symmetric? – Friedrich Philipp Mar 23 '16 at 12:03
  • Assume that it is symmetric if it can't be done for a more general problem. – Goobs Mar 23 '16 at 12:05

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