Block matrix and spectral radius I have a problem about a positive definite matrix. I cannot prove this.

Let $B= [b_{ij}]$ be a $m \times m$ matrix. Let $\overline{B}^t$ be the conjugate transpose of $B$. If we have a strict inequality on the spectral radius $$\rho(\overline{B}^tB) < 1$$ show that the block matrix $$\begin{bmatrix}
I_n & B \\ 
\overline{B}^t & I_n 
\end{bmatrix}$$ is positive definite.

If you don't mind, help me to solve this problem.
Remake $\rho(A) = \max_i \lvert \lambda_i \rvert $ where $\lambda_i$ is eigenvalue of $A$.
A matrix $A$ is positive definite if $\overline{x}^t A x >0$ for all $x\in \mathbb{C}^n$ and $ A = \overline{A}^t $.
 A: Directly consider the eigenvalue equation:
$$\begin{bmatrix}I & B \\ \overline{B}^T & I\end{bmatrix}\begin{bmatrix}u \\ v\end{bmatrix} = \lambda \begin{bmatrix}u \\ v\end{bmatrix}.$$
Bring the identity portion to the right hand side, to get:
$$\begin{bmatrix} & B \\ \overline{B}^T & \end{bmatrix}\begin{bmatrix}u \\ v\end{bmatrix} = (\lambda-1) \begin{bmatrix}u \\ v\end{bmatrix}.$$
This is the same as the following two equations,
\begin{align}
Bv &= (\lambda-1) u \\
\overline{B}^T u &= (\lambda-1) v.
\end{align}
Solving the first equation for $u$ and substituting it into the second yields,
$$\overline{B}^T B v = (\lambda-1)^2 v.$$
In other words, for $\lambda$ to be an eigenvalue of your big matrix, $(\lambda-1)^2$ must be an eigenvalue of $B^TB$. When the eigenvalues of $B^TB$ are less than 1, as presupposed by the question, this means,
$$(\lambda-1)^2 < 1$$
and so the eigenvalues of the big matrix, $\lambda$, lie in the interval 
$$\lambda \in (0,2).$$
In particular this implies that the eigenvalues are strictly positive, so the big matrix is positive definite.
A: Hint: make use of Schur complement and matrix congruence.
A: Say $M=B^* B$, then we have :
$$
\|B^2\|=\|B^*B\| = \rho(B^*B) = s_1^2
$$
The first equality is a propriety of matricial norm, the second because $M$ is Hermitian, and $s_1$ is the first singular values of $B$ (the eigenvalues of $(BB^*)^{1/2}$, ordered $s_1\geq s_2 \geq ...\geq s_n$)
So the condition  $\rho(M)\leq 1$ is equivalent to say that $B$ is a contraction ($\|B\|\leq 1$)
If $n=m=1$ then the condition is obvious, in general let $B=USV$ the [Singular value decomposition][1] of $B$ then :
$$
\left(\begin{array}{c}
I & B \\
B^* & I
\end{array}\right)=\left(\begin{array}{c}
I & USV \\
V^*S U^* & I
\end{array}\right)=\left(\begin{array}{c}
U & 0 \\
0 & V^*
\end{array}\right)\left(\begin{array}{c}
I & S \\
S & I
\end{array}\right)\left(\begin{array}{c}
U^* & 0 \\
0 & V
\end{array}\right)
$$
So the matrix $\left(\begin{array}{c}
I & B \\
B^* & I
\end{array}\right)$ is unitarily equivalent to $\left(\begin{array}{c}
I & S \\
S& I
\end{array}\right)$, which in turn unitarily equivalent to the direct sum :
$$
\left(\begin{array}{c}
1 & s_1 \\
s_1 & 1
\end{array}\right)\oplus \left(\begin{array}{c}
1 & s_2 \\
s_2 & 1
\end{array}\right)\oplus\dots\oplus \left(\begin{array}{c}
1 & s_n \\
s_n & 1
\end{array}\right)
$$
and these $2\times2$ matrices are positive if and only if $s_1\leq 1$
[1]: https://en.wikipedia.org/wiki/Singular_value_decomposition
