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I am trying to get an intuitive understanding of the meanings of right and left eigenvectors. I guess the best thing you can do is to provide examples of application. (Examples from the field of biology would be especially welcome).

Example:

A population have individuals that are classified in three classes, class A, B and C. Na, Nb, Nc represents the number of individuals in each class.

$$Na(t+1) = m*Na(t) + n*Nb(t) + o*Nc(t)$$ $$Nb(t+1) = p*Na(t) + q*Nb(t) + r*Nc(t)$$ $$Nc(t+1) = s*Na(t) + t*Nb(t) + u*Nc(t)$$

$$v(t+1) = A * v(t)$$

What is the long term equilibrium of this system of equations? Do we use left or right leading eigenvector of A. Why would be the biological meaning (if any) of the other (left/right) eigenvector?

The current intuitive sense I have is that an eigenvector of a matrix is a measure of how oriented is the distortion caused by the multiplication by this matrix. The eigenvalue is the strength of this distortion. Therefore the eigenvector linked with the biggest eigenvalue determines the long term behaviour of a system.

Here are two links on Stack exchange that did not help me answering my question. How to intuitively understand eigenvalue and eigenvector? question on left and right eigenvectors

Community
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Remi.b
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  • I think the literal definition is by far the most intuitive meaning for eigenvector -- I think what you're interested in is "What phenomena can be intuitively explained in terms of eigenvectors?" –  Aug 17 '13 at 13:35
  • I think you're right @Hurkyl! I'm looking for an understanding on when I should use the right and when I should use the left eigenvector for various mathematical modelling in ecology. – Remi.b Aug 17 '13 at 13:50
  • I can't give an intuitive explanation of both, but: when modeling age-structured population growth, the "stable age distribution" is a right eigenvector of the Leslie Matrix, and the "stable reproductive value distribution" is a left eigenvector of it. See the TREE article "Projection Matrices in Population Biology". – David Bahry May 03 '18 at 19:02
  • Intuitively for the right eigenvector. So the Leslie matrix contains entries representing to the age-specific fecundities (in the top row), and the age-specific survival probabilities (on the sub-diagonal). If you have a vector for the population's age structure this year (how many 0 year olds are there? How many 1 year olds? Etc.), then multiplying it by the Leslie Matrix gives you the population's age structure for next year. If the population is in stable age distribution, then the age classes might grow or shrink, but they'll stay in the same ratio to reach other; the vector won't rotate. – David Bahry May 03 '18 at 19:05

2 Answers2

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I suppose in the question $v(t)$ just is an abbreviation for the column vector with components $Na(t), Nb(t), Nc(t)$, which describes the distribution among classes in the population. As you can see the equation lets the matrix $A$ operate from the left on this column vector (as it should; on the other side a row vector would be required), so the interesting case would be when $v(t)$ is a right eigenvector for $A$. In that case the operation by$~A$ multiplies all components by the same scalar, which means one gets the same relative distribution in classes as before; all classes have grown by the same factor. This means it is a relative equilibrium situation (though maybe an unstable one). It would seem to me that the situation can only be sustainable if the eigenvalue equals$~1$, since otherwise one would have exponential growth, which is very hard to sustain (unless it is by a factor less than$~1$, in which case it leads to extinction).

In this setting I can see no obvious use for left eigenvectors, since they do not correspond to particular states of the system (distributions into classes) like equilibrium states, but rather to linear functions on the set of all possible states. Such a function might measure something like the amount of certain resources that would be required by the population, as depending on the state the population.

Marc van Leeuwen
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As was pointed out by others, the normalized right eigenvector (with the largest positive eigenvalue) represents the steady-state distribution of the population.

The normalized left eigenvector represents "individual reproductive value", which means how much does an individual of class $i$ contribute to the long-term population contents many generations into the future.

One can look at the population, and ask, where did an individual of class $i$ come from, one generation ago, two generations ago, and so on, deep into the past. The left eigenvector tells the asymptotic ratio of probability of coming from the different classes for any generation deep enough in the past.

See this paper by François Bienvenu

Michael Lachmann
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