We return to the example of subsection II.B with $\mu = 1$, i.e. $$A(1) = \begin{pmatrix} 1 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix}$$ If $1 \leqslant p \leqslant n$, we denote by $\mathcal{G}_{n,p}(\mathbb{R})$ the set of matrices in $\mathcal{M}_{n,p}(\mathbb{R})$ with rank equal to $p$.
How should we choose $N' = \begin{pmatrix} N_{1}' & N_{2}' \end{pmatrix}$ so that $\operatorname{det}(N'^{\top}A(1)N') = 0$?

We return to the example of subsection II.B with $\mu = 1$, i.e. $$A(1) = \begin{pmatrix} 1 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix}$$ A matrix $K \in \mathcal{M}_{n}(\mathbb{R})$ is called $F$-singular if there exists a non-zero $X \in F$ such that $\forall Z \in F, Z^{\top}KX = 0$.
Determine a vector subspace $F$ of $E_{3}$ such that $\dim F = 1$ and such that $A(1)$ is $F$-singular.

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. A matrix $K \in \mathcal{M}_{n}(\mathbb{R})$ is called $F$-singular if there exists a non-zero $X \in F$ such that $\forall Z \in F, Z^{\top}KX = 0$. If $1 \leqslant p \leqslant n$, we denote by $\mathcal{G}_{n,p}(\mathbb{R})$ the set of matrices in $\mathcal{M}_{n,p}(\mathbb{R})$ with rank equal to $p$. Let $F$ be a vector subspace of $E_{n}$ of dimension $n-p$, where $1 \leqslant p \leqslant n-1$.
Show that $A$ is $F$-singular if $\operatorname{det}(N'^{\top}AN') = 0$ for a matrix $N' \in \mathcal{G}_{n,p}(\mathbb{R})$ that one will define.

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. If $1 \leqslant p \leqslant n$, we denote by $\mathcal{G}_{n,p}(\mathbb{R})$ the set of matrices in $\mathcal{M}_{n,p}(\mathbb{R})$ with rank equal to $p$. Let $F$ be a vector subspace of $E_{n}$ of dimension $n-p$, where $1 \leqslant p \leqslant n-1$. Let $N' \in \mathcal{G}_{n,p}(\mathbb{R})$ be a matrix whose columns form a basis of $F^{\perp}$. We now assume that $A_{s} \in \mathcal{S}_{n}^{++}(\mathbb{R})$.
Show that if $X \in \mathcal{M}_{p,1}(\mathbb{R})$ is non-zero then $X^{\top}N'^{\top}AN'X > 0$.

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. If $1 \leqslant p \leqslant n$, we denote by $\mathcal{G}_{n,p}(\mathbb{R})$ the set of matrices in $\mathcal{M}_{n,p}(\mathbb{R})$ with rank equal to $p$. Let $F$ be a vector subspace of $E_{n}$ of dimension $n-p$, where $1 \leqslant p \leqslant n-1$. Let $N' \in \mathcal{G}_{n,p}(\mathbb{R})$ be a matrix whose columns form a basis of $F^{\perp}$. We assume that $A_{s} \in \mathcal{S}_{n}^{++}(\mathbb{R})$.
Deduce that the real eigenvalues of $N'^{\top}AN'$ are strictly positive.

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. If $1 \leqslant p \leqslant n$, we denote by $\mathcal{G}_{n,p}(\mathbb{R})$ the set of matrices in $\mathcal{M}_{n,p}(\mathbb{R})$ with rank equal to $p$. Let $F$ be a vector subspace of $E_{n}$ of dimension $n-p$, where $1 \leqslant p \leqslant n-1$. Let $N' \in \mathcal{G}_{n,p}(\mathbb{R})$ be a matrix whose columns form a basis of $F^{\perp}$. We assume that $A_{s} \in \mathcal{S}_{n}^{++}(\mathbb{R})$.
Deduce that $\operatorname{det}(N'^{\top}AN') > 0$.

We denote $E_{n} = \mathcal{M}_{n,1}(\mathbb{R})$ equipped with the inner product $(X \mid Y) = X^{\top}Y$. A matrix $K \in \mathcal{M}_{n}(\mathbb{R})$ is called $F$-singular if there exists a non-zero $X \in F$ such that $\forall Z \in F, Z^{\top}KX = 0$. Let $F$ be a vector subspace of $E_{n}$ of dimension $n-p$, where $1 \leqslant p \leqslant n-1$. We assume that $A_{s} \in \mathcal{S}_{n}^{++}(\mathbb{R})$.
Deduce that $A$ is $F$-regular for every non-zero vector subspace $F$ of $E_{n}$.

In all that follows, $\mathbf{K = C}$. We set $E = \mathbf{C}^n$. We are given $A \in \mathcal{M}_n(\mathbf{C})$. For every eigenvalue $\lambda$ of the matrix $A$, we denote by $m_\lambda$ its multiplicity, and we introduce the vector subspace $$F_\lambda = \operatorname{Ker}\left((A - \lambda I_n)^{m_\lambda}\right) = \operatorname{Ker}\left((u - \lambda \operatorname{Id}_E)^{m_\lambda}\right).$$
$\mathbf{20}$ ▷ Show that $\mathbf{C}^n = \bigoplus_{\lambda \in \operatorname{Sp}(A)} F_\lambda$.

We denote by $\mathcal { C } _ { J } = \left\{ M \in \mathcal { M } _ { n } ( \mathbb { R } ) \mid J M = M J \right\}$ the centralizer of $J = \left( \begin{array}{cc} 0 & -I_m \\ I_m & 0 \end{array} \right)$, and every $M \in \mathcal{C}_J$ has the form $M = \left( \begin{array}{cc} U & -V \\ V & U \end{array} \right)$ for some $U, V \in \mathcal{M}_m(\mathbb{R})$. Deduce that, for every matrix $M \in \mathcal { C } _ { J } , \operatorname { det } ( M ) \geqslant 0$.
One may consider the product of block matrices $\left( \begin{array} { c c } I _ { m } & 0 \\ \mathrm { i } I _ { m } & I _ { m } \end{array} \right) \left( \begin{array} { c c } U & - V \\ V & U \end{array} \right) \left( \begin{array} { c c } I _ { m } & 0 \\ - \mathrm { i } I _ { m } & I _ { m } \end{array} \right)$.

We denote by $\mathcal { C } _ { J } = \left\{ M \in \mathcal { M } _ { n } ( \mathbb { R } ) \mid J M = M J \right\}$ the centralizer of the matrix $J$, and it has been shown that $M \in \mathcal{C}_J$ if and only if $M = \left( \begin{array} { c c } U & - V \\ V & U \end{array} \right)$ for some $U, V \in \mathcal{M}_m(\mathbb{R})$. Deduce that, for every matrix $M \in \mathcal { C } _ { J } , \operatorname { det } ( M ) \geqslant 0$.
One may consider the product of block matrices $\left( \begin{array} { c c } I _ { m } & 0 \\ \mathrm { i } I _ { m } & I _ { m } \end{array} \right) \left( \begin{array} { c c } U & - V \\ V & U \end{array} \right) \left( \begin{array} { c c } I _ { m } & 0 \\ - \mathrm { i } I _ { m } & I _ { m } \end{array} \right)$.

In all that follows, $\mathbf{K = C}$. We set $E = \mathbf{C}^n$. We are given $A \in \mathcal{M}_n(\mathbf{C})$. For every eigenvalue $\lambda$ of the matrix $A$, we denote by $m_\lambda$ its multiplicity, and we introduce the vector subspace $$F_\lambda = \operatorname{Ker}\left((A - \lambda I_n)^{m_\lambda}\right).$$
$\mathbf{21}$ ▷ Deduce from question 20) the existence of three matrices $P, D$ and $N$ in $\mathcal{M}_n(\mathbf{C})$ such that $A = P(D + N)P^{-1}$, where $D$ is diagonal, $N$ is nilpotent, $DN = ND$, and $P$ is invertible.

Show that, if $A$ and $B$ belong to $S_n^{++}(\mathbf{R})$, then: $$\forall t \in [0,1], \quad \operatorname{det}((1-t)A + tB) \geq \operatorname{det}(A)^{1-t} \operatorname{det}(B)^t$$ Justify that this inequality remains valid for $A$ and $B$ only in $S_n^+(\mathbf{R})$.

Show that, if $A$ and $B$ belong to $S _ { n } ^ { + + } ( \mathrm { R } )$, then:
$$\forall t \in [ 0,1 ] , \quad \operatorname { det } ( ( 1 - t ) A + t B ) \geq \operatorname { det } ( A ) ^ { 1 - t } \operatorname { det } ( B ) ^ { t }$$
Justify that this inequality remains valid for $A$ and $B$ only in $S _ { n } ^ { + } ( \mathbf { R } )$.

What can be deduced about the function $\ln \circ \det$ on $S_n^{++}(\mathrm{R})$?

What can we deduce about the function $\ln \circ \det$ on $S _ { n } ^ { + + } ( \mathbf { R } )$ ?

Let $A \in S_n^{++}(\mathbf{R})$ and let $g : t \in \mathbf{R} \mapsto \operatorname{det}(I_n + tA)$. Express, for all $t \in \mathbf{R}$, $g(t)$ using the eigenvalues of $A$. Deduce that $g$ is of class $C^\infty$ on $\mathbf{R}$.

Let $A \in S _ { n } ^ { + + } ( \mathbf { R } )$ and let $g : t \in \mathbf { R } \mapsto \operatorname { det } \left( I _ { n } + t A \right)$. Express, for all $t \in \mathbf { R } , g ( t )$ using the eigenvalues of $A$. Deduce that $g$ is of class $C ^ { \infty }$ on $\mathbf { R }$.

Let $A \in S_n^{++}(\mathbf{R})$ and $M \in S_n(\mathbf{R})$. Let $\varepsilon_0 > 0$ be such that for all $t \in ]-\varepsilon_0, \varepsilon_0[, A + tM \in S_n^{++}(\mathrm{R})$. Let $\alpha \in ]-\frac{1}{n}, +\infty[\backslash\{0\}$ and $\varphi_\alpha(t) = \frac{1}{\alpha} \operatorname{det}^{-\alpha}(A + tM)$. Show that $\varphi_\alpha$ is twice differentiable at 0 and that $$\varphi_\alpha''(0) = \operatorname{det}^{-\alpha}(A)\left(\alpha \operatorname{Tr}^2(A^{-1}M) + \operatorname{Tr}\left((A^{-1}M)^2\right)\right).$$

Let $A \in S _ { n } ^ { + + } ( \mathbf { R } )$ and $M \in S _ { n } ( \mathbf { R } )$, and let $\varepsilon_0 > 0$ be such that for all $t \in ] - \varepsilon _ { 0 } , \varepsilon _ { 0 } [ , A + t M \in S _ { n } ^ { + + } ( \mathbf { R } )$. Let $\alpha \in ] - \frac { 1 } { n } , + \infty \left[ \backslash \{ 0 \} \right.$ and $\varphi _ { \alpha } ( t ) = \frac { 1 } { \alpha } \operatorname { det } ^ { - \alpha } ( A + t M )$. Show that $\varphi _ { \alpha }$ is twice differentiable at 0 and that
$$\varphi _ { \alpha } ^ { \prime \prime } ( 0 ) = \operatorname { det } ^ { - \alpha } ( A ) \left( \alpha \operatorname { Tr } ^ { 2 } \left( A ^ { - 1 } M \right) + \operatorname { Tr } \left( \left( A ^ { - 1 } M \right) ^ { 2 } \right) \right) .$$

Let $A \in S_n^{++}(\mathbf{R})$ and $M \in S_n(\mathbf{R})$. Let $\alpha \in ]-\frac{1}{n}, +\infty[\backslash\{0\}$ and $\varphi_\alpha(t) = \frac{1}{\alpha} \operatorname{det}^{-\alpha}(A + tM)$. Show that, if $\varphi_\alpha''(0) > 0$, then there exists $\eta > 0$ such that for all $t \in ]-\eta, \eta[$, $$\frac{1}{\alpha} \operatorname{det}^{-\alpha}(A + tM) \geq \frac{1}{\alpha} \operatorname{det}^{-\alpha}(A) - \operatorname{Tr}(A^{-1}M) \operatorname{det}^{-\alpha}(A) t.$$

Let $A \in S _ { n } ^ { + + } ( \mathbf { R } )$ and $M \in S _ { n } ( \mathbf { R } )$, and let $\alpha \in ] - \frac { 1 } { n } , + \infty \left[ \backslash \{ 0 \} \right.$. Show that, if $\varphi _ { \alpha } ^ { \prime \prime } ( 0 ) > 0$, then there exists $\eta > 0$, such that for all $t \in ] - \eta , \eta [$,
$$\frac { 1 } { \alpha } \operatorname { det } ^ { - \alpha } ( A + t M ) \geq \frac { 1 } { \alpha } \operatorname { det } ^ { - \alpha } ( A ) - \operatorname { Tr } \left( A ^ { - 1 } M \right) \operatorname { det } ^ { - \alpha } ( A ) t$$

We consider $M \in \mathscr { D } _ { \rho } \left( S _ { n } ( \mathbb { R } ) \right)$. We fix a real eigenvalue $\lambda$ of $M _ { \mid t = 0 }$ and denote by $d$ its multiplicity as a root of $\chi _ { \mid t = 0 }$ where $\chi = \det(XI_n - M)$. We set $A = F ( M )$, $B = G ( M )$, $Q = ( B U \mid A V ) \in \mathscr { D } _ { \rho _ { 2 } } \left( \mathscr { M } _ { n } ( \mathbb { R } ) \right)$.
Show that, for all $a \in U _ { \rho _ { 2 } }$, the direct sum $\operatorname { im } \left( B _ { a } U \right) \oplus \operatorname { im } \left( A _ { a } V \right) = \mathbb { R } ^ { n }$ of question 22a is orthogonal for the standard inner product on $\mathbb { R } ^ { n }$.

We consider $M \in \mathscr { D } _ { \rho } \left( S _ { n } ( \mathbb { R } ) \right)$. We fix a real eigenvalue $\lambda$ of $M _ { \mid t = 0 }$ and denote by $d$ its multiplicity as a root of $\chi _ { \mid t = 0 }$ where $\chi = \det(XI_n - M)$. We set $Q = ( B U \mid A V ) \in \mathscr { D } _ { \rho _ { 2 } } \left( \mathscr { M } _ { n } ( \mathbb { R } ) \right)$ and $Q^{-1} \cdot M \cdot Q = \operatorname{Diag}(M_1, M_2)$.
Show that there exists $\rho _ { 3 } \in \mathbb { R } _ { + } ^ { * }$ such that $\rho _ { 3 } \leqslant \rho _ { 2 }$ and matrices $R _ { 1 } \in \mathrm { GL } _ { d } \left( \mathscr { D } _ { \rho _ { 3 } } ( \mathbb { R } ) \right) , R _ { 2 } \in \mathrm { GL } _ { n - d } \left( \mathscr { D } _ { \rho _ { 3 } } ( \mathbb { R } ) \right)$ such that the matrix $Q \cdot \operatorname { Diag } \left( R _ { 1 } , R _ { 2 } \right)$ is orthogonal. (One may use the result of question 17.)

Prove Theorem 2: Let $M \in \mathscr { D } _ { \rho } \left( S _ { n } ( \mathbb { R } ) \right)$. Then there exists $r \in \mathbb { R } _ { + } ^ { * }$ such that $r \leqslant \rho$ and an orthogonal matrix $P \in \mathscr { D } _ { r } \left( \mathscr { M } _ { n } ( \mathbb { R } ) \right)$ such that $P ^ { \mathrm { T } } \cdot M \cdot P$ is diagonal.

Show that, for all natural integers $p$ and $q$ greater than or equal to 2, for any matrix $M = (M(i,j))_{1 \leqslant i,j \leqslant q} \in \mathcal{M}_q(\mathbb{R})$ and for all $(i,j) \in \llbracket 1,q \rrbracket^2$, the coefficient with index $(i,j)$ of the matrix $M^p$ is $$\sum_{(k_2,\ldots,k_p) \in \llbracket 1,q \rrbracket^{p-1}} M(i,k_2)\left(\prod_{r=2}^{p-1} M(k_r, k_{r+1})\right) M(k_p, j),$$ the product being equal to 1 in the case where $p = 2$.