Leading positive eigenvalues with negative eigenvectors?

Question

So, here it is my problem. I have a non-negative matrix 15x15 and it's a discrete-time model. I also have 5 parameters - I can't get, analytically, the value of the eigenvalue. So, I'm assigning values to these parameters (using the software Mathematica). However, in some circumstances, I have a leading eigenvalue superior to 1 that has a negative eigenvector. Is this mathematically possible? I know that, in my model, this would be meaningless and, therefore, should be ignored, but my concern is that I might be doing something wrong. Example below.

{{0.04112903225806455`, 0, 0.023951612903225825`, 
0.04112903225806455`, 0.050000000000000044`, 0.003951612903225813`, 
0.025000000000000022`, 0, 0.003951612903225813`, 0, 0, 
0.023951612903225825`, 0.025000000000000022`, 0, 0}, {0, 0.375`, 
0.23951612903225805`, 0, 0, 0.039516129032258096`, 0, 0.375`, 
0.039516129032258096`, 0.5`, 0.25`, 0.23951612903225805`, 0, 0.25`, 
0}, {0, 0, 0.00266129032258065`, 0, 0, 0.03556451612903228`, 
0.025000000000000022`, 0, 0.03556451612903228`, 0, 
0.025000000000000022`, 0.00266129032258065`, 0.025000000000000022`, 
0.025000000000000022`, 0.050000000000000044`}, {0.4838709677419355`,
0, 0.30483870967741933`, 0.4838709677419355`, 0.9090909090909091`, 
0.04677419354838713`, 0.45454545454545453`, 0, 0.04677419354838713`,
0, 0, 0.30483870967741933`, 0.45454545454545453`, 0, 
0}, {0.01612903225806453`, 0, 0.013064516129032266`, 
0.01612903225806453`, 0.09090909090909098`, 0.00016129032258064557`,
0.022727272727272745`, 0, 0.00016129032258064557`, 0, 0, 
0.013064516129032266`, 0.022727272727272745`, 0, 0}, {0, 0, 
0.13064516129032255`, 0, 0, 0.0016129032258064544`, 0, 0, 
0.0016129032258064544`, 0, 0, 0.13064516129032255`, 0, 0, 0}, {0, 0,
0.0014516129032258087`, 0, 0, 0.0014516129032258087`, 
0.022727272727272745`, 0, 0.0014516129032258087`, 0, 0, 
0.0014516129032258087`, 0.022727272727272745`, 0, 0}, {0, 0.375`, 
0.00047181586699280805`, 0, 0, 0.00007239502721053675`, 0, 0.375`, 
0.00007239502721053675`, 0.5`, 0.0004992760497278392`, 
0.00047181586699280805`, 0, 0.0004992760497278392`, 0}, {0, 0, 
0.40439337959958055`, 0, 0, 0.00499251085925409`, 0, 0, 
0.00499251085925409`, 0, 0, 0.40439337959958055`, 0, 0, 0}, {0, 
0.125`, 0.00020220680013977485`, 0, 0, 2.4963802486392006`*^-6, 0, 
0.125`, 2.4963802486392006`*^-6, 0.5`, 0.0002496380248639196`, 
0.00020220680013977485`, 0, 0.0002496380248639196`, 0}, {0, 0, 
0.04493259773328677`, 0, 0, 0.04493259773328677`, 0, 0, 
0.04493259773328677`, 0, 0.49925108592540823`, 0.04493259773328677`,
0, 0.49925108592540823`, 0}, {0, 0, 0.000052423985221423164`, 0, 0,
0.0006515552448948302`, 0.0002498875506022013`, 0, 
0.0006515552448948302`, 0, 0.0004992760497278392`, 
0.000052423985221423164`, 0.0002498875506022013`, 
0.0004992760497278392`, 0.0004997751012044027`}, {0, 0, 
0.04493259773328677`, 0, 0, 0.04493259773328677`, 
0.24987505622469894`, 0, 0.04493259773328677`, 0, 0, 
0.04493259773328677`, 0.24987505622469894`, 0, 0}, {0, 0, 
0.00002246742223775278`, 0, 0, 0.00002246742223775278`, 0, 0, 
0.00002246742223775278`, 0, 0.0002496380248639196`, 
0.00002246742223775278`, 0, 0.0002496380248639196`, 0}, {0, 0, 
0.00499251085925409`, 0, 0, 0.40439337959958055`, 
0.24987505622469894`, 0, 0.40439337959958055`, 0, 
0.49925108592540823`, 0.00499251085925409`, 0.24987505622469894`, 
0.49925108592540823`, 0.9995002248987958`}}

The leading eigenvalue is 1.01535 and the associated eigenvector:

{-0.00298633, -0.302466, -0.0452828, -0.0373315, -0.00146164, 
-0.00592459, -0.000197951, -0.289254, -0.0183387, -0.143545, 
-0.00609804, -0.000462362, -0.00417405, -3.04917*10^-6, -0.894628}

The most important question is: is there something wrong? Am I missing something? If I'm not, can I just ignored it, giving the fact that negative values are, in my model, simply meaningless?

PS: I'm a biologist, taking that into account, if you can :)

Answer 1

All multiples (except 0) of an eingenvector are eigenvectors with the same eigenvalue

Let $v$ be an eigenvector of the matrix $A$ with eigenvalue $\lambda$. Then for any $r\ne 0$, $rv$ satisfies

$$A(rv)=r(Av)=r(\lambda v)=\lambda(rv),$$

which is the definition of an eigenvector.

So in particular, $-v$ is an eigenvector.

It does not make much sense to think of vectors with negative components to be as negative vectors, as there (probably) is no (meaningful) ordering, which makes all these vectors smaller than the zero vector. Changing the sign of a vector means reflecting it at the origin.