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 $| \alpha | = 1$.

We fix a symplectic form $\omega$ on $E$. Show by induction that there exists a basis $\widetilde { \mathcal { B } }$ of $E$ such that $$\operatorname { Mat } _ { \widetilde { \mathcal { B } } } ( \omega ) = \left( \begin{array} { c c c c } J _ { 2 } & 0 & \cdots & 0 \\ 0 & J _ { 2 } & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & J _ { 2 } \end{array} \right)$$

For all $n$ in $\mathbb{N}$, determine the degree of $T_n$, then show that $\left(T_k\right)_{0 \leqslant k \leqslant n}$ is a basis of $\mathbb{C}_n[X]$.
The sequence of polynomials $\left(T_n\right)_{n \in \mathbb{N}}$ is defined by $T_0 = 1, T_1 = X$ and $\forall n \in \mathbb{N}, T_{n+2} = 2X T_{n+1} - T_n$.

Show by induction that
$$\forall N \in \mathbf { N } ^ { * } , \forall z \in D , \prod _ { k = 1 } ^ { N } \frac { 1 } { 1 - z ^ { k } } = \sum _ { n = 0 } ^ { + \infty } p _ { n , N } z ^ { n }$$

Let $I$ be an interval of $\mathbf{R}$. Let $f : I \rightarrow \mathbf{R}$ be a convex function. Show that, for all $p \in \mathbf{N}^\star$, for all $(\lambda_1, \ldots, \lambda_p) \in (\mathbf{R}_+)^p$ such that $\sum_{i=1}^p \lambda_i = 1$ and for all $(x_1, \ldots, x_p) \in I^p$, we have: $$f\left(\sum_{i=1}^p \lambda_i x_i\right) \leq \sum_{i=1}^p \lambda_i f(x_i)$$ Hint: You may proceed by induction on $p$.

Let $M \in S_n^+(\mathbf{R})$ be a non-zero matrix. Show the inequality $\frac{\operatorname{Tr}(M)}{n} \geq \operatorname{det}^{1/n}(M)$.
Hint: You may show that $x \mapsto -\ln(x)$ is convex on $\mathbf{R}_+^\star$.

Let $I$ be an interval of $\mathbf { R }$. Let $f : I \rightarrow \mathbf { R }$ be a convex function. Show that, for all $p \in \mathbf { N } ^ { \star }$, for all $\left( \lambda _ { 1 } , \ldots , \lambda _ { p } \right) \in \left( \mathbf { R } _ { + } \right) ^ { p }$ such that $\sum _ { i = 1 } ^ { p } \lambda _ { i } = 1$ and for all $\left( x _ { 1 } , \ldots , x _ { p } \right) \in I ^ { p }$, we have:
$$f \left( \sum _ { i = 1 } ^ { p } \lambda _ { i } x _ { i } \right) \leq \sum _ { i = 1 } ^ { p } \lambda _ { i } f \left( x _ { i } \right)$$
Hint: You may proceed by induction on $p$.

Let $M \in S _ { n } ^ { + } ( \mathbf { R } )$ be a non-zero matrix. Show the inequality $\frac { \operatorname { Tr } ( M ) } { n } \geq \operatorname { det } ^ { 1 / n } ( M )$.
Hint: You may show that $x \mapsto - \ln ( x )$ is convex on $\mathbf { R } _ { + } ^ { \star }$.

143. In the sequence $\{U_n\}$, with initial conditions $U_1 = U_2 = 1$ and recurrence $U_{n+1} = U_n + U_{n-1}$, using induction, the expression $(U_n^2 - U_{n+1} \times U_{n-1})$ equals which number?
(1) $-1$ (2) $1$ (3) $(-1)^n$ (4) $(-1)^{n+1}$