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.


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.

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

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