Throughout this part $A$ and $B$ denote real symmetric matrices of $\mathcal { M } _ { 2 } ( \mathbb { R } )$. We set $$A = \left( \begin{array} { l l } a _ { 1 } & b _ { 1 } \\ b _ { 1 } & d _ { 1 } \end{array} \right) \quad B = \left( \begin{array} { l l } a _ { 2 } & b _ { 2 } \\ b _ { 2 } & d _ { 2 } \end{array} \right)$$ We assume $A \geqslant 0$ and $B \geqslant 0$, $\operatorname { det } A \operatorname { det } B \neq 0$ and $b _ { 1 } b _ { 2 } \neq 0$. II.F.1) Prove that we have equality in the formula of question II.E.2 if and only if the vectors $( a _ { 1 } , d _ { 1 } )$ and $( a _ { 2 } , d _ { 2 } )$ are linearly dependent, as well as the vectors $( b _ { 1 } , \sqrt { \operatorname { det } A }$ ) and $( b _ { 2 } , \sqrt { \operatorname { det } B }$ ). II.F.2) Prove then that we have equality in the formula of question II.E.2 if and only if the matrices $A$ and $B$ are proportional ($A = \lambda B$ for some $\lambda \in \mathbb { R }$, $\lambda > 0$).
Throughout this part $A$ and $B$ denote real symmetric matrices of $\mathcal { M } _ { 2 } ( \mathbb { R } )$. We set
$$A = \left( \begin{array} { l l } a _ { 1 } & b _ { 1 } \\ b _ { 1 } & d _ { 1 } \end{array} \right) \quad B = \left( \begin{array} { l l } a _ { 2 } & b _ { 2 } \\ b _ { 2 } & d _ { 2 } \end{array} \right)$$
We assume $A \geqslant 0$ and $B \geqslant 0$, $\operatorname { det } A \operatorname { det } B \neq 0$ and $b _ { 1 } b _ { 2 } \neq 0$.
II.F.1) Prove that we have equality in the formula of question II.E.2 if and only if the vectors $( a _ { 1 } , d _ { 1 } )$ and $( a _ { 2 } , d _ { 2 } )$ are linearly dependent, as well as the vectors $( b _ { 1 } , \sqrt { \operatorname { det } A }$ ) and $( b _ { 2 } , \sqrt { \operatorname { det } B }$ ).
II.F.2) Prove then that we have equality in the formula of question II.E.2 if and only if the matrices $A$ and $B$ are proportional ($A = \lambda B$ for some $\lambda \in \mathbb { R }$, $\lambda > 0$).