# 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 $A = \overline{A}^t$.

• Did you try to diagonalize $\bar{B}^tB$? Feb 24 '16 at 12:22

## 3 Answers

Hint: make use of Schur complement and matrix congruence.

To prove this we fix some notation, we put $M=B^* B$, 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 of $B$ then : $$\left(\begin{array}{c} I & B \\ B^* & I \end{array}\right)=\left(\begin{array}{c} I & USV \\ V^*S U^* & I \end{array}\right)=\left(\begin{array}{c} U & 0 \\ O & V^* \end{array}\right)\left(\begin{array}{c} I & S \\ S & I \end{array}\right)\left(\begin{array}{c} U^* & 0 \\ 0 & V \end{array}\right)$$ 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 $s_1\leq 1$

• More concisely, we can say that $$\pmatrix{I&S\\S&I} = I + \pmatrix{0&1\\1&0} \otimes S$$ where $\otimes$ denotes the Kronecker product. Knowing some standard properties of the Kronecker product then allows you to reach the result about the eigenvalues. Feb 24 '16 at 14:30
• This question is part of more general results, that can be found in the following paper: Cauchy-Schwarz Inequalities Associated with Positive Semidefinite Matrices (Roger A. Horn and Roy Mathias, Linear Algebra and its Applications, Dec. 1990). Feb 24 '16 at 14:50

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.