Home > Algebra, Analysis, Differential Equations, Inequalities > Agreg 2012 Analysis Part 1

Agreg 2012 Analysis Part 1

Part 1. Finite dimension

The goal is to prove the following theorem:

Theorem 1. Let {A \in M_n(\Bbb{R})} be a square matrix with non-negative coefficients. Suppose that for every {x \in \Bbb{R}^n\setminus \{0\}} with non-negative coordinates, the vector {Ax} has strictly positive components. Then

  • (i) the spectral radius {\rho = \sup \{ |\lambda | : \lambda \in \Bbb{C} \text{ is an eigenvalue for }A\}} is a simple eigenvalue for {A};
  • (ii) there exists an eigenvector {v} of {A} associated to {\rho} with strictly positive coordinates.
  • (iii) any other eigenvalue of {A} verifies {|\lambda|<\rho};
  • (iv) there exists an eigenvector of {A^T} associated to {\rho} with strictly positive components.

1. Consider {(w_1,..,w_n) \in \Bbb{C}^n} such that {|w_1+..+w_n|=|w_1|+...+|w_n|}. Prove that for distinct {j,l \in \{1,..,n\}} we have {\text{re}(\overline{w_j}w_l)=|w_j||w_l|}. Deduce that there exists {\theta \in [0,2\pi)} such that {w_j=e^{i\theta}|w_j|,\ j=1..n}.

2. Prove that the coefficients of {A} are strictly positive.

3. For {z \in \Bbb{C}^n} we denote {|z|=(|z_1|,..,|z_n|)}. Prove that {A|z|=|Az|} if and only if there exists {\theta \in [0,2\pi)} such that {z_j=e^{i\theta}|z_j|,\ j=1..n}.

4. Denote {\mathcal{C}= \{x \in \Bbb{R}^n : x_i \geq 0, i=1..n\}}. Consider {x \in \mathcal{C}} and denote {e=(1,1,..,1) \in \Bbb{R}^n}. Prove that

\displaystyle 0 \leq (Ax|e)\leq (x|e)\max_j \sum_{k=1}^n a_{kj}.

5. Denote {\mathcal{E}= \{ t \geq 0 : \text{ there exists }x \in \mathcal{C} \setminus \{0\} \text{ such that } Ax-tx \in \mathcal{C}\}}. Prove that {\mathcal{E}} is an interval which does not reduces to {\{0\}}, it is bounded and closed.

6. Denote {\rho=\max \mathcal{E}>0}. Prove that if {x \in \mathcal{C}\setminus \{0\}} verifies {Ax-\rho x \in \mathcal{C}} then we have {Ax=\rho x}. Deduce that {\rho} is an eigenvalue of {A} and that for this eigenvalue there exists an eigenvector {v} with coordinates strictly positive.

7. Consider {z \in \Bbb{C}^n}. Prove that {Az=\rho z} and {(z|v)=0} implies {z=0}. Deduce that {\ker(A-\rho I)=\text{span}\{v\}} and every other eigenvalue of {A} verifies {|\lambda | <\rho}.

8. Prove that every eigenvector of {A} which has positive coordinates is proportional to {v}.

Theorem 1 can be used in the study of the asymptotic behavior of some linear systems of differential equations. Consider {A \in M_n(\Bbb{R})} a matrix which verifies the conditions of theorem 1. Therefore, there exists a couple of vectors {v,\phi \in \Bbb{R}^n} such that they have positive coordinates, {(v|\phi)=1, Av=\rho v,A^T \phi=\rho \phi} where {\rho} is the spectral radius of {A}. Consider the Cauchy problem

\displaystyle \frac{d}{dt}y =Ay,\ y(0)=y_0 \in \Bbb{R}^n.

9. Consider {y_0\in \Bbb{R}^n} a vector with non-negative coordinates. Justify that the above system of linear equations has a unique solution defined on {\Bbb{R}} and that for {t \geq 0} the coordinates of {y(t)} are all non-negative.

10. Prove that {(y(t)|\phi)e^{-\rho t}=(y_0|\phi)} for every {t \in \Bbb{R}}.

11. Deduce that for every {j} the function {t \mapsto e^{-\rho t}y_j(t)} is bounded on {[0,\infty)}.

12. Recall that the Dunford decomposition allows us to write {A=D+N} where {D} is diagonalizable and {N} is nilpotent with {DN=ND}. If we denote {\sigma(A)} the set of eivgenvalues of {A}, deduce that there exist the polynomial functions {P_{\lambda,j}} such that

\displaystyle y_j(t)= \sum_{\lambda \in \sigma(A) } P_{\lambda,j}(t)e^{\lambda t}.

13. Prove that {P_{\rho,j}(t)} is constant.

14. Deduce that {e^{-\rho t}y(t)} admits a limit as {t} goes to {+\infty} and express that limit in terms of {y_0,\phi} and {v}.

Hints: 1. Do it for two vectors {|a+b|^2=|a|^2+2|a||b|+|b|^2} and use {|z|^2=z\overline z}. 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 {0,w_1,..,w_n} are all colinear.

2. Since {e_i=(\delta_{ij})_j} has nonnegative components and {Ae_i=c_i} (the {i}-th column of {A}) we use the definition of {A} to see that each column of {A} has strictly positive elements.

3. {A|z|=|Az|} implies

\displaystyle \sum_j a_{ij}|z_j|=|\sum_j a_{ij}z_j|,

and since all coefficients {a_{ij}} are strictly positive we use 1. to conclude.

4. {(Ax|e)=\sum_{i=1}^n \sum_{j=1}^n a_{ij}x_j} and since all the quantities are positive we have {(Ax|e) \geq 0}. On the other hand, changing the order of summation we get

\displaystyle (Ax|e)=\sum_{j=1}^n ( x_j \sum_{i=1}^n a_{ij})

from where the second inequality comes easily.

5. First note that {Ax \in \mathcal{C}} for every {x \in \mathcal{C}} (by definition), so {0 \in \mathcal{E}}. We prove now that if {t \in \mathcal{E}} then {\lambda t \in \mathcal{E}} for {\lambda \in (0,1)}. Suppose that {Ax-tx \in \mathcal{C}}. Then {Ax=c+tx} where {c \in \mathcal{C}} and {Ax-\lambda tx =c+(t-\lambda t)x} which is also in {\mathcal{C}} since we are adding positive coordinate vectors.

Note {m=\min a_{ij}>0}. Then each component of {Ax-mx} is equal to {\sum_j a_{ij}x_j-mx_j} and the positivity is obvious by the choice of {m}. This proves that {\mathcal{E}} is not trivial. Choosing {M=\max_i \sum_{j=1}^n a_{ij}} we find that {Ax-Mx} has all components negative, so {\mathcal{E}} is bounded above.

6. If {y=Ax-\rho x \neq 0} then {Ay} has all its components strictly positive, and therefore we can find {\varepsilon >0} such that {Ay-\varepsilon Ax\in \mathcal{C}}. But then we have {Ay-\varepsilon Ax=A(Ax)-(\rho+\varepsilon)Ax} and we have contradicted the maximality of {\rho}. Therefore we must have {Ax=\rho x}, and there exists {v} such that {v \neq 0} and {Av=\rho v}. If one component of {v} would be zero we would have an equality like

\displaystyle \sum_j a_{ij}v_j =0

where {a_{ij}} are strictly positive, and this would imply that {v=0}; contradiction to {v \neq 0}.

7. We have {Az=\rho z} and {(z|v)=0}. I will use the notation {x \geq y} for vectors when we have {x-y \in \mathcal{C}}. We obviously have {A|z| \geq |Az|=\rho|z|}. Therefore {A|z|-\rho|z| \in \mathcal{C}}, which implies by 6 that {A|z|=\rho |z|}. But then we have {|Az|=A|z|} and by 3. all the components of the vector {z} have the same argument {\theta}. This reduces the problem to {z \in \mathcal{C}}, and the equality {(z|v)=0} implies {z=0}. This proves that {\{v\}^\perp \cap \ker(A-\rho I)=0} which leaves us with the desired equality.

Of course, if {\lambda \in \Bbb{C}} is another eigenvalue of {A} then if we assume that {|\lambda|=\rho} then we find (as above) an eigenvector {z} with the property that {A|z|=\rho|z|} and {(z|v)=0}, which implies {z=0}, a contradiction.

8. If {w} is another eigenvector with positive components then, denoting {\phi} the vector with positive components of {A^T} (which exists; proof exactly like above) we have

\displaystyle \lambda (w|\phi)=(Aw|\phi)=(w|A^t \phi)=\rho(w|\phi)

and since {w,\phi} have strictly positive coordinates, we have {\lambda =\rho} and {w} and {v} are proportional.

9. The solution is the exponential of {A} multiplied by {y_0}.

10. Denote {f(t)=(y(t)|\phi)e^{-\rho t}} and differentiate with respect to {t} to get

\displaystyle f'(t)= (Ay(t) | \phi)e^{-\rho t}-(y(t)|\phi)\rho e^{-\rho t}=

\displaystyle = (y(t)|A^T\phi)e^{-\rho t}-(y(t)|\phi)\rho e^{-\rho t}=0

and as a consequence {f(t)=f(0)} for every {t}.

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

12. {y(t)=e^{tA}y_0=Pe^{tD}P^{-1}e^{tN}y_0} so {e^{tD}} yields the exponentials of the eigenvalues with come constant coefficients given by multiplications with {P,P^{-1}} while {e^{tN}} is a matrix consisting of polynomials of degree at most {n}.

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

14. We know that {e^{-\rho t}y_j(t)=c_j+lower\ order\ terms}. So {y'(t)=c\rho e^{\rho t}+l.o.t =Ace^{\rho t}+l.o.t}. Therefore {Ac=c\rho} so {c=\alpha v} where the constant {\alpha} comes from the equalities

\displaystyle \alpha=(\alpha v|\phi)=(y_0|\phi).

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