Simplifying the positive semidefinite constraint if closed-form expressions of the eigenvalues are found I came across an article and I am not sure how the simplification is done. Please find my question and related information below.
[Information]
In the paper,
$$\begin{aligned} A_1 &= I_N-\lambda_1 h_1 h_1^H+\lambda_2 r_2 h_1 h_1^H+\lambda_3 r_2h_2 h_2^H \\ A_2 &= I_N-\lambda_2 h_1 h_1^H-\lambda_3h_2 h_2^H\end{aligned}$$
where $\lambda_1,\lambda_2,\lambda_3 \geq0$, $r_1,r_2 \geq 0$, $h_1,h_2 \in \mathbb{C}^N$, and $I_N$ is the $N \times N$ identity matrix.
Furthermore there is a lemma: For $h$, $g \in \mathbb{C}^N$, define $\cos^2(\theta)=\frac{h^Hgg^Hh}{\|{h}\|^2\|g\|^2}$. Then $h h^H+g g^H$ has at most 2 non-zero eigenvalues: $$\frac{1}{2}\left(\|{h}\|^2+\|g\|^2\right) \pm \sqrt{\left(\|{h}\|^2+\|g\|^2\right)^2-4 \|{h}\|^2\|g\|^2 \sin^2\left(\theta\right)}, \tag{a}$$
and similarly $h h^H-g g^H$ has at most two nonzero eigenvalues:
$$\frac{1}{2}\left(\|{h}\|^2-\|g\|^2\right) \pm \sqrt{\left(\|{h}\|^2-\|g\|^2\right)^2+4 \|{h}\|^2\|g\|^2 \sin^2\left(\theta\right)}. \tag{b}$$
(P.s. one may assume the lemma is correct. Ref: Eigenvalues of a sum of rank-one matrices? )
[Question]
Based on the given information, the authors stated that the constraints $A_1\geq0$ and $A_2\geq0$ can respectively be reduced to the following:
$$\|{k_1}\|^2-\|{k_2}\|^2 \leq \text{min} \{ 1-\|k_1\|^2 \|k_2\|^2 \sin^2(\theta),2 \}\tag{1}$$
$$\|{l_1}\|^2+\|{l_2}\|^2 \leq \text{min} \{ 1+\|l_1\|^2 \|l_2\|^2 \sin^2(\theta),2 \} \tag{2}$$
where
$$\begin{aligned} k_1 &= \sqrt{\lambda_1-\lambda_2 r_2} h_1\\ k_2 &= \sqrt{\lambda_3 r_3}h_2\\ l_1 &= \sqrt{\lambda_2}h_1\\ l_2 &= \sqrt{\lambda_3}h_2\\ \cos^2(\theta) &= \frac{h_1^Hh_2 h_2^Hh_1}{\|{h_1}\|^2\|h_2\|^2}\end{aligned}$$
My question is: given the eigenvalues as expressed in (a) and (b), how can we reduce the constraints $A_1\geq0$ and $A_2\geq0$ to (1) and (2), respectively.
[Extension of Rodrigo de Azevedo's answer]
Many thanks to Rodrigo de Azevedo's answer. As requested, I would extend his answer to fully address my question. To avoid confusion, the eigenvalues used in his answer are denoted as $\phi_1,\phi_2$. WLOG, I will start from the 2+1 inequalities. By further exploiting orthogonality, we thus obtain
$$\begin{aligned} 
\phi_1 & \leq 1\\\\ 
\phi_2 & \leq 1\\\\ 
\phi_1 + \phi_2 & \leq 1 + \phi_1 \phi_2
\end{aligned}$$
Hence, we immediately have $A_1 \geq 0 \iff (1)$ and $A_2 \geq 0 \iff (2)$.
 A: Let
$$\begin{aligned} {\rm  A} &:= {\rm  I}_n - \begin{bmatrix} | & | \\ {\rm  v}_1 & {\rm  v}_2\\ | & | \end{bmatrix} \begin{bmatrix} \lambda_1 & 0\\ 0 & \lambda_2 \end{bmatrix} \begin{bmatrix} | & | \\ {\rm  v}_1 & {\rm  v}_2\\ | & | \end{bmatrix}^\top\\\\ &\,= {\rm  I}_n - \begin{bmatrix} | & | \\ \sqrt{\lambda_1} \, {\rm  v}_1 & \sqrt{\lambda_2} \, {\rm  v}_2\\ | & | \end{bmatrix} \begin{bmatrix} | & | \\ \sqrt{\lambda_1} \, {\rm  v}_1 & \sqrt{\lambda_2} \, {\rm  v}_2\\ | & | \end{bmatrix}^\top = {\rm  I}_n - {\rm  W} {\rm  W}^\top\end{aligned}$$
and ${\rm  w}_i := \sqrt{\lambda_i} \, {\rm  v}_i$. If $\color{blue}{{\rm  A} \succeq {\rm  O}_n}$, then
$$\begin{bmatrix} {\rm  I}_n & {\rm  W}\\ {\rm  W}^\top & {\rm  I}_2 \end{bmatrix} \succeq {\rm  O}_{n+2}$$
From the complementary Schur complement,
$${\rm  I}_2 - {\rm  W}^\top {\rm  W} = \begin{bmatrix} 1 - \| {\rm  w}_1 \|_2^2 & - \langle {\rm  w}_1, {\rm  w}_2 \rangle \\ - \langle {\rm  w}_1, {\rm  w}_2 \rangle & 1 - \| {\rm  w}_2 \|_2^2 \end{bmatrix} \succeq {\rm  O}_2$$
Using Sylvester's criterion for positive semidefiniteness, since both diagonal entries and the determinant must be non-negative, we obtain $2+1$ inequalities
$$\begin{aligned} \| {\rm  w}_1 \|_2^2 & \leq 1\\\\ \| {\rm  w}_2 \|_2^2 & \leq 1\\\\ \left( 1 - \| {\rm  w}_1 \|_2^2 \right) \left( 1 - \| {\rm  w}_2 \|_2^2 \right) - \left| \langle {\rm  w}_1, {\rm  w}_2 \rangle \right|^2 &\geq 0\end{aligned}$$
From which we conclude that
$$\begin{aligned} \| {\rm  w}_1 \|_2^2 + \| {\rm  w}_2 \|_2^2 &\leq 2\\\\ \| {\rm  w}_1 \|_2^2 + \| {\rm  w}_2 \|_2^2 &\leq 1 + \| {\rm  w}_1 \|_2^2  \| {\rm  w}_2 \, \|_2^2 \, \sin^2(\theta)\end{aligned}$$
and, finally,
$$\color{blue}{\boxed{\\ \,\\\qquad \| {\rm  w}_1 \|_2^2 + \| {\rm  w}_2 \|_2^2 \leq \min\left( 1 + \| {\rm  w}_1 \|_2^2  \| {\rm  w}_2 \, \|_2^2 \, \sin^2(\theta), 2\right) \qquad \\}}$$
