I am working on a problem with the following matrix $$G=\begin{pmatrix}I&A\\ A^T&0 \end{pmatrix}$$ where $A\in\mathbb{R}^{m\times n}$ and $m>n$ with full column rank.
Then by rescaling, the $\lambda_{\min}$ of $A^TA$ is equal to one. Thus the condition number of $A^T A=\lambda_{\max}$. I want to prove, there exists a universal constant $c>0$ such that $$Cond(G)\le c\sqrt{Cond(A^TA)}=c\sqrt{\lambda_\max}$$
My Attempt
Computing the eigenvalue of $G$ and by definition $x\in\mathbb{R}^n $ and $y\in\mathbb{R}^m$, then $$\begin{cases}x+A^Ty&=\lambda x\\ A^Tx&=\lambda y \end{cases}$$ By easy observation, $\lambda =0$ implies $x,y$ are zero vectors so zero is not an eigenvalue. By simple substitution, we have $$\lambda^2x-\lambda x=AA^T x$$ By linear algebra, the eigenvalue of $AA^T$ and $A^TA$ are the same. By above equation, $x$ is an eigenvector of $AA^T$ and let $\sigma_1\le\cdots\le \sigma_m$ be the eigenvalue of $AA^T$ so $\lambda^2-\lambda -\sigma_k=0$ will be the solution of the eigenvalues of $G$.
I discover some problems, first, it may not have real roots but $G$ is symmetric so all eigenvalue is real. Second the number of roots is $2m$, it maybe too much for a $(m+n)\times (m+n)$ matrix.
Anyone can help me?