grandes-ecoles 2016 QIV.B.1

grandes-ecoles · France · centrale-maths1__mp 3x3 Matrices Matrix Algebraic Properties and Abstract Reasoning

For any primitive matrix $A$ in $\mathcal{M}_n(\mathbb{R})$, we denote $r$ the spectral radius of $A$, $L = xy^\top$ where $x > 0$ is a direction vector of $\operatorname{Ker}(A - rI_n)$ and $y > 0$ is a direction vector of $\operatorname{Ker}(A^\top - rI_n)$ with $y^\top x = 1$. We set $B = A - rL$.
Let $\lambda$ be a nonzero eigenvalue of $B$ and let $z$ be an associated eigenvector.
Show that $Lz = 0$, then $Az = \lambda z$. Deduce $\rho(B) \leqslant r$.

For any primitive matrix $A$ in $\mathcal{M}_n(\mathbb{R})$, we denote $r$ the spectral radius of $A$, $L = xy^\top$ where $x > 0$ is a direction vector of $\operatorname{Ker}(A - rI_n)$ and $y > 0$ is a direction vector of $\operatorname{Ker}(A^\top - rI_n)$ with $y^\top x = 1$. We set $B = A - rL$.

Let $\lambda$ be a nonzero eigenvalue of $B$ and let $z$ be an associated eigenvector.

Show that $Lz = 0$, then $Az = \lambda z$. Deduce $\rho(B) \leqslant r$.

Paper Questions

QI.A.1 QI.A.2 QI.B.1 QI.B.2 QII.A QII.B QII.C QIII.A QIII.B.1 QIII.B.2 QIII.B.3 QIII.B.4 QIII.B.5 QIII.C.1 QIII.C.2 QIII.C.3 QIII.D.1 QIII.D.2 QIV.A.1 QIV.A.2 QIV.A.3 QIV.A.4 QIV.B.1 QIV.B.2 QIV.B.3 QIV.C QV.A.1 QV.A.2 QV.A.3 QV.A.4 QV.A.5 QV.B.1 QV.B.2 QV.C.1 QV.C.2 QVI.A QVI.B.1 QVI.B.2 QVI.B.3 QVI.C.1 QVI.C.2 QVI.D