Conditions for positive definiteness: matrix inequality Let $0<\alpha<1$ and $A,B\in\mathbb{R}^{n\times n}$. I am trying to find conditions on $A$ and $B$ such that
\begin{equation}
I_n-\frac{1}{\alpha}B^{\rm T}B-\frac{1}{4\alpha(1-\alpha)}( A^{\rm T}B+A)^{\rm T}( A^{\rm T}B+A)>0.
\end{equation}
However, I do not know how to proceed. Any idea or suggestion is appreciated. 
 A: Hint: 
Rewrite it as should be positive definite as you desired:
$$Sc \triangleq  \Big (I_n - \frac{1}{\alpha}B^T B \Big )- \frac{1}{4\alpha (1-\alpha)}\Big\{(A^TB+A)^T I_n (A^TB+A)\Big \} \succ 0$$
Then, by using Schur complement, $S_c$ is positive definite if and only if the matrix $S$ defined as: 
$$S \triangleq \begin{pmatrix} \frac{1}{4\alpha (1-\alpha)}I_n & A^TB+A \\ (A^TB+A)^T & I_n - \frac{1}{\alpha}B^T B \end{pmatrix},$$
is: 
$$S \succ 0 \quad \text{with the conditions mentioned below would be satisfied.}
$$
Also, we have: 
$$ \lambda_i(S) >0; \quad \text{for all } i = 0,1, ..., n.$$
Now, it needs to check for which $A$ and $B$ the above relationship (Schur complement) can be satisfied.
Second, is to show the eigenvalues of such block matrix is positive.

Side note: Schur complement
For any symmetric matrix X, of the form: 
$$X \triangleq \begin{pmatrix} A & B \\ B^T & C \end{pmatrix},$$
if $A$ is invertible, then the following statement holds:
$X \succ 0$ if and only if $A \succ 0$ and $C-B^T A^{-1} B \succ 0.$
A: The problem reduces to showing that \begin{equation}
\lambda_{\max}\left(\frac{1}{\alpha}B^{\rm T}B+\frac{1}{4\alpha(1-\alpha)}( A^{\rm T}B+A)^{\rm T}( A^{\rm T}B+A)\right)<1.
\end{equation}
Let $\sigma_A$ and $\sigma_B$ be the largest singular values of $A$ and $B$, respectively. Then, it follows that
\begin{align}
\lambda_{\max}&\left(\frac{1}{\alpha}B^{\rm T}B+\frac{1}{4\alpha(1-\alpha)}( A^{\rm T}B+A)^{\rm T}( A^{\rm T}B+A)\right)\\&\leq\frac{1}{\alpha}\sigma_B^2+\frac{1}{4\alpha(1-\alpha)}\sigma_{\max}^2\left(A^{\rm T}B+A\right)\\
&\leq\frac{1}{\alpha}\sigma_B^2+\frac{1}{4\alpha(1-\alpha)}\left(\sigma_{\max}^2\left(A^{\rm T}B\right)+2\sigma_{\max}\left(A^{\rm T}B\right)\sigma_A+\sigma_A^2\right)\\
&\leq\frac{1}{\alpha}\sigma_B^2+\frac{1}{4\alpha(1-\alpha)}\left(\sigma_A^2\sigma_B^2+2\sigma_A^2\sigma_B+\sigma_A^2\right)\\
\end{align}
Next, assume 
\begin{equation}
\frac{1}{\alpha}\sigma_B^2+\frac{1}{4\alpha(1-\alpha)}\left(\sigma_A^2\sigma_B^2+2\sigma_A^2\sigma_B+\sigma_A^2\right)<1,
\end{equation}
and it follows that 
\begin{equation}
4\alpha^2-4\alpha(1+\sigma_B^2)+4\sigma_B^2+\sigma_A^2(\sigma_B+1)^2<0,
\end{equation}
which implies $\sigma_B<1$ and $\sigma_A<1-\sigma_B$ since $0<\alpha<1$. Thus, $\sigma_A+\sigma_B<1$ is a sufficient condition.
