In this part, we study the case $n = 2$. For two real numbers $u$ and $v$ such that $u \geqslant v$, we denote: $$S(u, v) = \left\{M \in \mathcal{S}_{2}(\mathbb{R}) \mid s^{\downarrow}(M) = (u, v)\right\}.$$ We fix $a_{1} \geqslant a_{2}$ and $b_{1} \geqslant b_{2}$, four real numbers satisfying the relation $a_{1} - a_{2} \geqslant b_{1} - b_{2}$, and $$\Sigma = \left\{s^{\downarrow}(A + B) \mid A \in S\left(a_{1}, a_{2}\right),\, B \in S\left(b_{1}, b_{2}\right)\right\}.$$ Show that $$\Sigma = \left\{s^{\downarrow}(A + B) \,\middle|\, A = \left(\begin{array}{cc} a_{1} & 0 \\ 0 & a_{2} \end{array}\right),\, B \in S\left(b_{1}, b_{2}\right)\right\}.$$
In this part, we study the case $n = 2$. For two real numbers $u$ and $v$ such that $u \geqslant v$, we denote:
$$S(u, v) = \left\{M \in \mathcal{S}_{2}(\mathbb{R}) \mid s^{\downarrow}(M) = (u, v)\right\}.$$
We fix $a_{1} \geqslant a_{2}$ and $b_{1} \geqslant b_{2}$, four real numbers satisfying the relation $a_{1} - a_{2} \geqslant b_{1} - b_{2}$, and
$$\Sigma = \left\{s^{\downarrow}(A + B) \mid A \in S\left(a_{1}, a_{2}\right),\, B \in S\left(b_{1}, b_{2}\right)\right\}.$$
Show that
$$\Sigma = \left\{s^{\downarrow}(A + B) \,\middle|\, A = \left(\begin{array}{cc} a_{1} & 0 \\ 0 & a_{2} \end{array}\right),\, B \in S\left(b_{1}, b_{2}\right)\right\}.$$