## Agreg 2012 Analysis Part 1

**Part 1. Finite dimension**

The goal is to prove the following theorem:

**Theorem 1.** Let be a square matrix with non-negative coefficients. Suppose that for every with non-negative coordinates, the vector has strictly positive components. Then

- (i) the spectral radius is a simple eigenvalue for ;
- (ii) there exists an eigenvector of associated to with strictly positive coordinates.
- (iii) any other eigenvalue of verifies ;
- (iv) there exists an eigenvector of associated to with strictly positive components.

1. Consider such that . Prove that for distinct we have . Deduce that there exists such that .

2. Prove that the coefficients of are strictly positive.

3. For we denote . Prove that if and only if there exists such that .

4. Denote . Consider and denote . Prove that

5. Denote . Prove that is an interval which does not reduces to , it is bounded and closed.

6. Denote . Prove that if verifies then we have . Deduce that is an eigenvalue of and that for this eigenvalue there exists an eigenvector with coordinates strictly positive.

7. Consider . Prove that and implies . Deduce that and every other eigenvalue of verifies .

8. Prove that every eigenvector of which has positive coordinates is proportional to .

Theorem 1 can be used in the study of the asymptotic behavior of some linear systems of differential equations. Consider a matrix which verifies the conditions of theorem 1. Therefore, there exists a couple of vectors such that they have positive coordinates, where is the spectral radius of . Consider the Cauchy problem

9. Consider a vector with non-negative coordinates. Justify that the above system of linear equations has a unique solution defined on and that for the coordinates of are all non-negative.

10. Prove that for every .

11. Deduce that for every the function is bounded on .

12. Recall that the Dunford decomposition allows us to write where is diagonalizable and is nilpotent with . If we denote the set of eivgenvalues of , deduce that there exist the polynomial functions such that

13. Prove that is constant.

14. Deduce that admits a limit as goes to and express that limit in terms of and .

**Hints:** 1. Do it for two vectors and use . The equality condition will translate into the fact that all the vectors have the same trigonometric argument. Note that a geometrical interpretation also works, and for the equality we need that are all colinear.

2. Since has nonnegative components and (the -th column of ) we use the definition of to see that each column of has strictly positive elements.

3. implies

and since all coefficients are strictly positive we use 1. to conclude.

4. and since all the quantities are positive we have . On the other hand, changing the order of summation we get

from where the second inequality comes easily.

5. First note that for every (by definition), so . We prove now that if then for . Suppose that . Then where and which is also in since we are adding positive coordinate vectors.

Note . Then each component of is equal to and the positivity is obvious by the choice of . This proves that is not trivial. Choosing we find that has all components negative, so is bounded above.

6. If then has all its components strictly positive, and therefore we can find such that . But then we have and we have contradicted the maximality of . Therefore we must have , and there exists such that and . If one component of would be zero we would have an equality like

where are strictly positive, and this would imply that ; contradiction to .

7. We have and . I will use the notation for vectors when we have . We obviously have . Therefore , which implies by 6 that . But then we have and by 3. all the components of the vector have the same argument . This reduces the problem to , and the equality implies . This proves that which leaves us with the desired equality.

Of course, if is another eigenvalue of then if we assume that then we find (as above) an eigenvector with the property that and , which implies , a contradiction.

8. If is another eigenvector with positive components then, denoting the vector with positive components of (which exists; proof exactly like above) we have

and since have strictly positive coordinates, we have and and are proportional.

9. The solution is the exponential of multiplied by .

10. Denote and differentiate with respect to to get

and as a consequence for every .

11. If a sum of positive functions is bounded, then each of the functions are bounded.

12. so yields the exponentials of the eigenvalues with come constant coefficients given by multiplications with while is a matrix consisting of polynomials of degree at most .

13. This is a direct consequence of 11. and 12.

14. We know that . So . Therefore so where the constant comes from the equalities