grandes-ecoles 2017 QII.C.6

grandes-ecoles · France · centrale-maths1__mp 3x3 Matrices Block Matrix Multiplication and Determinant Identity

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. We assume $n \geqslant 3$. $A$ is an invertible matrix in $\mathcal{M}_{n}(\mathbb{R})$. Let $N' = \begin{pmatrix} N_{1}' & N_{2}' \end{pmatrix}$.
Show that $$\operatorname{det}(N'^{\top}AN') = \left(N_{1}'^{\top}A_{s}N_{1}'\right)\left(N_{2}'^{\top}A_{s}N_{2}'\right) - \left(N_{1}'^{\top}A_{s}N_{2}'\right)^{2} + \left(N_{1}'^{\top}A_{a}N_{2}'\right)^{2}$$

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. We assume $n \geqslant 3$. $A$ is an invertible matrix in $\mathcal{M}_{n}(\mathbb{R})$. Let $N' = \begin{pmatrix} N_{1}' & N_{2}' \end{pmatrix}$.

Show that
$$\operatorname{det}(N'^{\top}AN') = \left(N_{1}'^{\top}A_{s}N_{1}'\right)\left(N_{2}'^{\top}A_{s}N_{2}'\right) - \left(N_{1}'^{\top}A_{s}N_{2}'\right)^{2} + \left(N_{1}'^{\top}A_{a}N_{2}'\right)^{2}$$

Paper Questions

QI.A.1 QI.A.2 QI.B.1 QI.B.2 QI.B.3 QI.B.4 QI.C.1 QI.C.2 QI.C.3 QII.A.1 QII.A.2 QII.A.3 QII.A.4 QII.A.5 QII.A.6 QII.A.7 QII.A.8 QII.B.1 QII.B.2 QII.B.3 QII.C.1 QII.C.2 QII.C.3 QII.C.4 QII.C.5 QII.C.6 QII.C.7 QII.C.8 QII.D.1 QII.D.2 QII.E.1 QII.E.2 QII.E.3 QII.E.4 QII.E.5 QIII.A.1 QIII.A.2 QIII.A.3 QIII.A.4 QIII.B.1 QIII.B.2 QIII.B.3 QIII.C.1 QIII.C.2 QIII.C.3