Hello my question is simple. Imagine we have a matrix $Q$ which has positive and negative eigenvalues. I know that a linear combination of eigenvectors with the same eigenvalue will result in the same eigenvalue.

But can I say that any linear combination of eigenvectors with POSITIVE (>=0) eigenvalues will result in a (>=0) VALUE? This result does not have to be an eigenvalue, just a positive number


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    $\begingroup$ A linear combination of eigenvectors will only yield an eigenvector when all the eigenvectors belong to the same eigenvalue. $\endgroup$ – Lord Shark the Unknown Apr 27 '17 at 16:04
  • $\begingroup$ I do not understand this formulation. A linear combination of eigenvectors is a vector not a number (unless you are working with numbers as vectors). $\endgroup$ – kyticka Apr 27 '17 at 17:11
  • $\begingroup$ Suppose that if x is the linear combination vector then the "number" would be obtained by: $x^t Q x = c$ $\endgroup$ – m33n Apr 27 '17 at 17:23

No. Take for instance the matrix

$Q = \left(\begin{array}{rrr} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \end{array}\right)$

It has eigenvalues 1, 2, 3 and eigenvectors:

$(1, 0, 0)^T, (1, 1, 0)^T, (1, 1, 1)^T$ respectively.

Now sum the first two eigenvectors and we get:

$s = v_1 + v_2 = (2, 1, 0)^T$ which is not an eigenvector as $Qs = (3, 2, 0)^T$.

  • $\begingroup$ Yes, you are right. Sorry, I will edit my question since I did not explain myself properly. $\endgroup$ – m33n Apr 27 '17 at 16:34

Suppose A is an eignvector with eigenvalue $\alpha$ and B is an eigenvector with eigenvalue $\beta$. So

$A \mapsto \alpha A$


$B \mapsto \beta B$

Now take a linear combination of A and B - say C = A+B. Then

$C \mapsto \alpha A + \beta B$

but $\alpha A + \beta B$ is not a multiple of A+B unless $\alpha = \beta$.


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