Lower bound of the smallest eigenvalue of block hermitian matrix Let $J$ and $L$ two squared matrix and let $A$ the matrix as follows
A=\begin{pmatrix}
         J^TJ+L^TL&J^T\\
         J        &I\\
         \end{pmatrix}
where I denote the identity matrix
I ask if there is a lower bounds of the smallest eignevalue of A in function of the eigenvalue of $J$ and $L$.
 A: Hint: Set
$$P=\pmatrix{I&0\\-J&I}$$
and investigate the eigenvalues of $P^T AP$ which are related to the eigenvalues of $A$:
$$A_1=P^T A P=\pmatrix{L^T L&0\\0&I}$$
Edit: A little more explanation
Note that the eigenvalues of $A$ and $A_2=P^{-1}A P$ are equal. Denote the smallest eigenvalue by $\underline{\lambda}$ and the largest one by $\bar{\lambda}$. We know that $\lambda_{L^T L}=\lambda_L^2$ (which can be proved by SVD). Hence
$$\underline{\lambda}(A_1)=\min\{1,\underline{\lambda}(L^T L)\}=\min\{1,{\underline{\lambda}_L}^2\}$$
And since $\Vert MN\Vert\le\Vert M\Vert\cdot\Vert N\Vert$, one can write:
$$\bar{\lambda}\left((MN)^{-1}\right)\le\bar{\lambda}(M^{-1})\,\bar{\lambda}(N^{-1})\implies\underline{\lambda}(M)\underline{\lambda}(N)\le\underline{\lambda}(MN)$$
Therefore
$$\begin{align}\underline{\lambda}(A)=\underline{\lambda}(A_2)=\underline{\lambda}\left((P^T P)^{-1}A_1\right)&\ge\underline{\lambda}\left((P^T P)^{-1}\right)\underline{\lambda}(A_1)\\&=\frac{1}{\bar{\lambda}(P^T P)}\underline{\lambda}(A_1)\\&=\frac{1}{\Vert P\Vert^2}\min\{1,{\underline{\lambda}_L}^2\}
\end{align}$$
You can see that the result is independent of $J$ because $\lambda_i(P)=1$
