# Proving $A=0$ and $B$ is positive definite given $||A||<1$ and block matrix is positive semidefinite

Let $A$ be an $m\times n$ matrix with $\|A\|<1$. Suppose that the matrix,

\begin{bmatrix} 0 & A \\ A^T & B \end{bmatrix}

is positive semidefinite. Show that $A=0$ and that $B$ is positive semidefinite.

## 1 Answer

Since the matrix is positive semidefinite, we get $$0\le\pmatrix{0\\x}^T \pmatrix{ 0 & A\\A^T&B} \pmatrix{0\\x} = x^TBx,$$ hence $B$ is positive semidefinite. Similarly, it follows $$0\le\pmatrix{y\\x}^T \pmatrix{ 0 & A\\A^T&B} \pmatrix{y\\x} = x^TBx + 2y^TAx$$ for all $x,y$.

Suppose there is $x$ with $Ax\ne0$. Then we set $y:= \lambda \cdot Ax$. using this in the above inequality implies $$0 \le x^TBx + 2 \lambda \|Ax\|^2$$ for all $\lambda\in \mathbb R$. Sending $\lambda\to-\infty$ yields a contradiction. Hence $Ax=0$ for all $x$. We can conclude $A=0$ by choosing $x$ to be $e_1, e_2, ..., e_m$, where the $e_i$'s are $n \times 1$ unit vectors with $1$ in the $ith$ position and $0$ elsewhere.

• Did you use $$||A|| < 1$$? If so, where? – BCLC Oct 11 '16 at 15:35
• @BCLC no, this assumption does not help anything. – daw Oct 12 '16 at 6:14