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

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