grandes-ecoles

Papers (176)
2025
centrale-maths1__official 40 centrale-maths2__official 36 mines-ponts-maths1__mp 17 mines-ponts-maths1__pc 21 mines-ponts-maths1__psi 21 mines-ponts-maths2__mp 28 mines-ponts-maths2__pc 23 mines-ponts-maths2__psi 25 polytechnique-maths-a__mp 35 polytechnique-maths__fui 9 polytechnique-maths__pc 27 x-ens-maths-a__fui 10 x-ens-maths-a__mp 18 x-ens-maths-b__mp 6 x-ens-maths-c__mp 6 x-ens-maths-d__mp 31 x-ens-maths__pc 27 x-ens-maths__psi 30
2024
centrale-maths1__official 21 centrale-maths2__official 28 geipi-polytech__maths 9 mines-ponts-maths1__mp 23 mines-ponts-maths1__psi 9 mines-ponts-maths2__mp 14 mines-ponts-maths2__pc 19 mines-ponts-maths2__psi 20 polytechnique-maths-a__mp 42 polytechnique-maths-b__mp 27 x-ens-maths-a__mp 43 x-ens-maths-b__mp 29 x-ens-maths-c__mp 22 x-ens-maths-d__mp 41 x-ens-maths__pc 20 x-ens-maths__psi 23
2023
centrale-maths1__official 37 centrale-maths2__official 32 e3a-polytech-maths__mp 4 mines-ponts-maths1__mp 14 mines-ponts-maths1__pc 21 mines-ponts-maths1__psi 21 mines-ponts-maths2__mp 21 mines-ponts-maths2__pc 13 mines-ponts-maths2__psi 22 polytechnique-maths__fui 3 x-ens-maths-a__mp 24 x-ens-maths-b__mp 10 x-ens-maths-c__mp 10 x-ens-maths-d__mp 10 x-ens-maths__pc 22
2022
centrale-maths1__mp 22 centrale-maths1__pc 33 centrale-maths1__psi 42 centrale-maths2__mp 26 centrale-maths2__pc 37 centrale-maths2__psi 40 mines-ponts-maths1__mp 26 mines-ponts-maths1__pc 20 mines-ponts-maths1__psi 23 mines-ponts-maths2__mp 22 mines-ponts-maths2__pc 9 mines-ponts-maths2__psi 18 x-ens-maths-a__mp 8 x-ens-maths-b__mp 19 x-ens-maths-c__mp 17 x-ens-maths-d__mp 47 x-ens-maths1__mp 13 x-ens-maths2__mp 26 x-ens-maths__pc 7 x-ens-maths__pc_cpge 14 x-ens-maths__psi 22 x-ens-maths__psi_cpge 26
2021
centrale-maths1__mp 34 centrale-maths1__pc 36 centrale-maths1__psi 28 centrale-maths2__mp 21 centrale-maths2__pc 38 centrale-maths2__psi 28 x-ens-maths2__mp 35 x-ens-maths__pc 29
2020
centrale-maths1__mp 42 centrale-maths1__pc 36 centrale-maths1__psi 38 centrale-maths2__mp 2 centrale-maths2__pc 35 centrale-maths2__psi 39 mines-ponts-maths1__mp_cpge 22 mines-ponts-maths2__mp_cpge 19 x-ens-maths-a__mp_cpge 10 x-ens-maths-b__mp_cpge 19 x-ens-maths-c__mp 10 x-ens-maths-d__mp 13 x-ens-maths1__mp 13 x-ens-maths2__mp 20 x-ens-maths__pc 6
2019
centrale-maths1__mp 37 centrale-maths1__pc 40 centrale-maths1__psi 38 centrale-maths2__mp 37 centrale-maths2__pc 39 centrale-maths2__psi 46 x-ens-maths1__mp 24 x-ens-maths__pc 18 x-ens-maths__psi 9
2018
centrale-maths1__mp 21 centrale-maths1__pc 31 centrale-maths1__psi 39 centrale-maths2__mp 23 centrale-maths2__pc 35 centrale-maths2__psi 30 x-ens-maths1__mp 18 x-ens-maths2__mp 13 x-ens-maths__pc 17 x-ens-maths__psi 20
2017
centrale-maths1__mp 45 centrale-maths1__pc 22 centrale-maths1__psi 17 centrale-maths2__mp 30 centrale-maths2__pc 28 centrale-maths2__psi 44 x-ens-maths1__mp 24 x-ens-maths2__mp 7 x-ens-maths__pc 17 x-ens-maths__psi 19
2016
centrale-maths1__mp 41 centrale-maths1__pc 31 centrale-maths1__psi 33 centrale-maths2__mp 25 centrale-maths2__pc 42 centrale-maths2__psi 17 x-ens-maths1__mp 10 x-ens-maths2__mp 32 x-ens-maths__pc 1 x-ens-maths__psi 20
2015
centrale-maths1__mp 18 centrale-maths1__pc 11 centrale-maths1__psi 42 centrale-maths2__mp 44 centrale-maths2__pc 1 centrale-maths2__psi 14 x-ens-maths1__mp 16 x-ens-maths2__mp 19 x-ens-maths__pc 30 x-ens-maths__psi 20
2014
centrale-maths1__mp 28 centrale-maths1__pc 26 centrale-maths1__psi 36 centrale-maths2__mp 24 centrale-maths2__pc 23 centrale-maths2__psi 29 x-ens-maths2__mp 13
2013
centrale-maths1__mp 3 centrale-maths1__pc 45 centrale-maths1__psi 20 centrale-maths2__mp 32 centrale-maths2__pc 50 centrale-maths2__psi 32 x-ens-maths1__mp 14 x-ens-maths2__mp 10 x-ens-maths__pc 22 x-ens-maths__psi 9
2012
centrale-maths1__pc 23 centrale-maths1__psi 20 centrale-maths2__mp 27 centrale-maths2__psi 20
2011
centrale-maths1__mp 27 centrale-maths1__pc 15 centrale-maths1__psi 21 centrale-maths2__mp 29 centrale-maths2__pc 8 centrale-maths2__psi 28
2010
centrale-maths1__mp 7 centrale-maths1__pc 23 centrale-maths1__psi 9 centrale-maths2__mp 10 centrale-maths2__pc 36 centrale-maths2__psi 27
2016 x-ens-maths__psi

20 maths questions

Q1 Matrices Direct Proof of an Inequality View
Let $x , y$ be strictly positive vectors of $\mathbb { R } ^ { n }$ and let $S , R$ be two sign diagonal matrices.
(a) Show that $$( S x \mid R y ) \leq ( x \mid y )$$ with equality if and only if $R = S$.
(b) Prove the uniqueness of $S$ in Broyden's theorem.
(c) Show that $$\| S x + R y \| \leq \| x + y \|$$ with equality if and only if $R = S$.
Q2 Matrices Linear System and Inverse Existence View
Let $O$ be an orthogonal matrix of $M _ { n } ( \mathbb { R } )$ and $S$ a sign diagonal matrix. Show that the equality $O x = S x$ with $x \in \mathbb { R } ^ { n }$ strictly positive is equivalent to $$( * ) \left\{ \begin{array} { c } \left( I _ { n } + O \right) x \geq 0 \\ \left( I _ { n } - O \right) x \geq 0 \\ x > 0 \end{array} \right.$$
Q3 Matrices Matrix Decomposition and Factorization View
We assume $n = 2$. We identify elements $x = \binom { x _ { 1 } } { x _ { 2 } } \in \mathbb { R } ^ { 2 }$ with vectors $\vec { x } = x _ { 1 } \vec { i } + x _ { 2 } \vec { j }$ of the Euclidean plane relative to an orthonormal frame $(\Omega , \vec { i } , \vec { j })$.
Let $O$ be the matrix of a reflection relative to a line passing through $\Omega$ and directed by a vector $\vec { v } _ { + }$. Determine a vector $x \in \mathbb { R } ^ { 2 }$ strictly positive and a sign diagonal matrix $S \in M _ { 2 } ( \mathbb { R } )$ such that $O x = S x$. Hint: begin by treating the case where $\vec { v } _ { + } \in \{ \vec { i } , \vec { j } \}$.
Q4 Matrices Matrix Decomposition and Factorization View
We assume $n = 2$. We identify elements $x = \binom { x _ { 1 } } { x _ { 2 } } \in \mathbb { R } ^ { 2 }$ with vectors $\vec { x } = x _ { 1 } \vec { i } + x _ { 2 } \vec { j }$ of the Euclidean plane relative to an orthonormal frame $(\Omega , \vec { i } , \vec { j })$.
Let $O$ be the matrix of a rotation with center $\Omega$ and angle $\theta \in ] - \pi , \pi ]$ nonzero. Using a drawing, find two vectors $x _ { + }$ and $x _ { - }$ such that $$O x _ { + } = \operatorname { diag } ( 1 , - 1 ) x _ { + } \text { and } O x _ { - } = \operatorname { diag } ( - 1,1 ) x _ { - }$$ Then discuss according to the sign of $\theta$, which of $x _ { + }$ and $x _ { - }$ is strictly positive.
Q5 Matrices Linear System and Inverse Existence View
With the notations of Broyden's theorem, we denote by $M \in M _ { 3 n } ( \mathbb { R } )$ the following block matrix $$M = \left( \begin{array} { c c c } 0 & 0 & I _ { n } + O \\ 0 & 0 & I _ { n } - O \\ - \left( I _ { n } + { } ^ { t } O \right) & - \left( I _ { n } - { } ^ { t } O \right) & 0 \end{array} \right)$$ Using Tucker's theorem, show that there exist positive vectors $x , z _ { 1 } , z _ { 2 } \in \mathbb { R } ^ { n }$ such that $$\left\{ \begin{array} { l } \left( I _ { n } + O \right) x \geq 0 \\ \left( I _ { n } - O \right) x \geq 0 \\ - \left( I _ { n } + { } ^ { t } O \right) z _ { 1 } - \left( I _ { n } - { } ^ { t } O \right) z _ { 2 } \geq 0 \\ z _ { 1 } + \left( I _ { n } + O \right) x > 0 \\ z _ { 2 } + \left( I _ { n } - O \right) x > 0 \\ x - \left( I _ { n } + { } ^ { t } O \right) z _ { 1 } - \left( I _ { n } - { } ^ { t } O \right) z _ { 2 } > 0 \end{array} \right.$$
Q6 Matrices Matrix Algebra and Product Properties View
Show that $\left\| z _ { 1 } - z _ { 2 } \right\| = \left\| z _ { 1 } + z _ { 2 } \right\|$ and that $- \left( I _ { n } + { } ^ { t } O \right) z _ { 1 } - \left( I _ { n } - { } ^ { t } O \right) z _ { 2 } = 0$.
Q7 Matrices Deduction or Consequence from Prior Results View
Deduce from question 6 that $x > 0$ and $x + O x \geq 0$ as well as $x - O x \geq 0$. Conclude.
Q8 Matrices Linear System and Inverse Existence View
Show that if $M \in M _ { n } ( \mathbb { R } )$ is antisymmetric (that is ${ } ^ { t } M = -M$) then $I _ { n } + M$ is an invertible matrix.
Q9 Matrices Matrix Group and Subgroup Structure View
Show that if $M \in M _ { n } ( \mathbb { R } )$ is antisymmetric, the matrix $$O = \left( I _ { n } + M \right) ^ { - 1 } \left( I _ { n } - M \right)$$ is orthogonal.
Q10 Matrices Deduction or Consequence from Prior Results View
Deduce from Broyden's theorem that there exist a strictly positive vector $x$ and a sign diagonal matrix $S$ such that $O x = S x$ and deduce that $u = x + S x$ is the positive vector of Tucker's theorem.
Q11 Proof Matrix Norm, Convergence, and Inequality View
We prove Broyden's theorem by induction on the dimension. We assume the result holds up to rank $n - 1$ and we write $O$ in the form of a block matrix $$O = \left( \begin{array} { l l } P & r \\ { } ^ { t } q & \alpha \end{array} \right)$$ where $P \in M _ { n - 1 } ( \mathbb { R } )$ and thus $r , q \in \mathbb { R } ^ { n - 1 }$ and $\alpha \in \mathbb { R }$.
Show that $| \alpha | \leq 1$ with equality if and only if $q = r = 0$.
Q12 Proof Prove a general algebraic or analytic statement by induction View
We prove Broyden's theorem by induction on the dimension. We assume the result holds up to rank $n - 1$ and we write $O$ in the form of a block matrix $$O = \left( \begin{array} { l l } P & r \\ { } ^ { t } q & \alpha \end{array} \right)$$ where $P \in M _ { n - 1 } ( \mathbb { R } )$ and thus $r , q \in \mathbb { R } ^ { n - 1 }$ and $\alpha \in \mathbb { R }$.
Treat the case $| \alpha | = 1$.
Q13 Proof Matrix Algebra and Product Properties View
We prove Broyden's theorem by induction on the dimension. We assume $| \alpha | < 1$ and we introduce the matrices $$Q _ { - } = P - \frac { r { } ^ { t } q } { \alpha - 1 } , \quad Q _ { + } = P - \frac { r { } ^ { t } q } { \alpha + 1 }$$ where $O = \left( \begin{array} { l l } P & r \\ { } ^ { t } q & \alpha \end{array} \right)$ with $P \in M _ { n - 1 } ( \mathbb { R } )$, $r , q \in \mathbb { R } ^ { n - 1 }$, $\alpha \in \mathbb { R }$.
Show that ${ } ^ { t } P P + q { } ^ { t } q = I _ { n - 1 }$, ${ } ^ { t } P r + \alpha q = 0$ and ${ } ^ { t } r r + \alpha ^ { 2 } = 1$.
Q14 Matrices Matrix Group and Subgroup Structure View
We prove Broyden's theorem by induction on the dimension. We assume $| \alpha | < 1$ and we introduce the matrices $$Q _ { - } = P - \frac { r { } ^ { t } q } { \alpha - 1 } , \quad Q _ { + } = P - \frac { r { } ^ { t } q } { \alpha + 1 }$$ where $O = \left( \begin{array} { l l } P & r \\ { } ^ { t } q & \alpha \end{array} \right)$ with $P \in M _ { n - 1 } ( \mathbb { R } )$, $r , q \in \mathbb { R } ^ { n - 1 }$, $\alpha \in \mathbb { R }$.
Show that the matrices $Q _ { + }$ and $Q _ { - }$ are orthogonal.
Q15 Proof Matrix Algebra and Product Properties View
We prove Broyden's theorem by induction on the dimension. We assume $| \alpha | < 1$ and we introduce the matrices $$Q _ { - } = P - \frac { r { } ^ { t } q } { \alpha - 1 } , \quad Q _ { + } = P - \frac { r { } ^ { t } q } { \alpha + 1 }$$ where $O = \left( \begin{array} { l l } P & r \\ { } ^ { t } q & \alpha \end{array} \right)$ with $P \in M _ { n - 1 } ( \mathbb { R } )$, $r , q \in \mathbb { R } ^ { n - 1 }$, $\alpha \in \mathbb { R }$.
Show that $${ } ^ { t } Q _ { + } Q _ { - } = I _ { n - 1 } - \frac { 2 } { 1 - \alpha ^ { 2 } } q { } ^ { t } q$$ and deduce that $$Q _ { - } = Q _ { + } - \frac { 2 } { 1 - \alpha ^ { 2 } } Q _ { + } q { } ^ { t } q$$
Q16 Proof Matrix Algebra and Product Properties View
We prove Broyden's theorem by induction on the dimension. We assume $| \alpha | < 1$ and we introduce the matrices $$Q _ { - } = P - \frac { r { } ^ { t } q } { \alpha - 1 } , \quad Q _ { + } = P - \frac { r { } ^ { t } q } { \alpha + 1 }$$ Using the induction hypothesis for $Q _ { + }$ (resp. for $Q _ { - }$), we denote by $x _ { + } > 0$ (resp. $x _ { - } > 0$) a vector of $\mathbb { R } ^ { n - 1 }$ and $S _ { + }$ (resp. $S _ { - }$) the sign diagonal matrix, such that $$Q _ { + } x _ { + } = S _ { + } x _ { + } , \quad \text { resp. } \quad Q _ { - } x _ { - } = S _ { - } x _ { - }$$
Show that $$\left( S _ { + } x _ { + } \mid S _ { - } x _ { - } \right) = \left( x _ { + } \mid x _ { - } \right) - \frac { 2 } { 1 - \alpha ^ { 2 } } \left( x _ { + } \mid q \right) \left( x _ { - } \mid q \right)$$
Q17 Proof Deduction or Consequence from Prior Results View
We prove Broyden's theorem by induction on the dimension. We assume $| \alpha | < 1$. Using the induction hypothesis for $Q _ { + }$ (resp. for $Q _ { - }$), we denote by $x _ { + } > 0$ (resp. $x _ { - } > 0$) a vector of $\mathbb { R } ^ { n - 1 }$ and $S _ { + }$ (resp. $S _ { - }$) the sign diagonal matrix, such that $Q _ { + } x _ { + } = S _ { + } x _ { + }$, resp. $Q _ { - } x _ { - } = S _ { - } x _ { - }$. We set
  • $\eta _ { + } = - \frac { \left( x _ { + } \mid q \right) } { \alpha + 1 } , \quad \eta _ { - } = - \frac { \left( x _ { - } \mid q \right) } { \alpha - 1 }$
  • $z _ { + } = \binom { x _ { + } } { \eta _ { + } } , \quad z _ { - } = \binom { x _ { - } } { \eta _ { - } }$
  • $S ^ { + } = \left( \begin{array} { c c } S _ { + } & 0 \\ 0 & + 1 \end{array} \right) , \quad S ^ { - } = \left( \begin{array} { c c } S _ { - } & 0 \\ 0 & - 1 \end{array} \right)$
Show using question 1(a) that in the case where $S _ { + } \neq S _ { - }$ then one of the pairs $(z _ { + } , S ^ { + })$ or $(z _ { - } , S ^ { - })$ satisfies Broyden's theorem.
Q18 Proof Existence Proof View
We now assume that $S _ { + } = S _ { - }$ and we assume that $\left( x _ { + } \mid q \right) = 0$. We denote by $z = \binom { x _ { + } } { 0 }$, $R ^ { + } = \left( \begin{array} { c c } S _ { + } & 0 \\ 0 & + 1 \end{array} \right) , R ^ { - } = \left( \begin{array} { c c } S _ { + } & 0 \\ 0 & - 1 \end{array} \right)$.
(a) Show that $O z = R^{+} z = R^{-} z$.
(b) We now write $$O = \left( \begin{array} { c c } \alpha ^ { \prime } & { } ^ { t } q ^ { \prime } \\ r ^ { \prime } & P ^ { \prime } \end{array} \right)$$ where $P ^ { \prime } \in M _ { n - 1 } ( \mathbb { R } )$. Construct then $z ^ { \prime } = \binom { \eta ^ { \prime } } { x ^ { \prime } } \in \mathbb { R } ^ { n }$ with $x ^ { \prime } \in \mathbb { R } ^ { n - 1 }$ strictly positive and $\eta ^ { \prime } \geq 0$ such that there exists a sign diagonal matrix $R ^ { \prime }$ satisfying $O z ^ { \prime } = R ^ { \prime } z ^ { \prime }$.
(c) In the case where $\eta ^ { \prime } = 0$, and using question 1(c), show that there exists a sign diagonal matrix $S$ such that $O \left( z + z ^ { \prime } \right) = S \left( z + z ^ { \prime } \right)$ and conclude.
Q19 Proof Deduction or Consequence from Prior Results View
For $A \in M _ { n , m } ( \mathbb { R } )$ and $b \in \mathbb { R } ^ { n }$ as in Farkas' lemma, we set $$B = \left( \begin{array} { c c c c } 0 & 0 & A & - b \\ 0 & 0 & - A & b \\ - { } ^ { t } A & { } ^ { t } A & 0 & 0 \\ { } ^ { t } b & - { } ^ { t } b & 0 & 0 \end{array} \right)$$ Let, by Tucker's theorem, $y = { } ^ { t } \left( z _ { 1 } , z _ { 2 } , x , t \right) \geq 0$ such that $B y \geq 0$ and $y + B y > 0$.
Show that if $t > 0$ then for $z = z _ { 1 } - z _ { 2 }$, we have $- { } ^ { t } A z \geq 0$ and $( b \mid z ) > 0$.
Q20 Proof Deduction or Consequence from Prior Results View
For $A \in M _ { n , m } ( \mathbb { R } )$ and $b \in \mathbb { R } ^ { n }$ as in Farkas' lemma, we set $$B = \left( \begin{array} { c c c c } 0 & 0 & A & - b \\ 0 & 0 & - A & b \\ - { } ^ { t } A & { } ^ { t } A & 0 & 0 \\ { } ^ { t } b & - { } ^ { t } b & 0 & 0 \end{array} \right)$$ Let, by Tucker's theorem, $y = { } ^ { t } \left( z _ { 1 } , z _ { 2 } , x , t \right) \geq 0$ such that $B y \geq 0$ and $y + B y > 0$.
If $t > 0$ show that $A x = t b$ and conclude.