# If $\Re(x^*Mx)>0\forall x\neq0$ for a block matrix $M$, is a sub-element matrix invertible?

Let $M = \begin{pmatrix} A & B \\C & D\end{pmatrix}$ be a block matrix. Define $a_n:=n^t A n$ for some real vector $n$ and analogously $b_n, c_n, d_n$. Does $\Re(x^*Mx)>0\forall x\neq0$ imply invertibility of $\begin{pmatrix} a_n & b_n\\ c_n & d_n\end{pmatrix}$?

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I presume then that all the blocks are of the same size? – J. M. Aug 31 '11 at 14:59
@J.M.: yes, I should have mentioned that – Tobias Kienzler Aug 31 '11 at 15:01

I'm assuming that your $*$ means Hermitian conjugate (= conjugate transpose), and $n \ne 0$.
Note that for scalars $r$ and $s$, $(\overline{r} n^t\ \overline{s} n^t) M \pmatrix{rn \cr sn} = r \overline{r} a_n + \overline{r} s b_n + r \overline{s} c_n + s \overline{s} d_n = (\overline{r} \ \overline{s}) \pmatrix{a_n & b_n\cr c_n & d_n\cr} \pmatrix{r\cr s\cr}$. If $\pmatrix{a_n & b_n\cr c_n & d_n\cr}$ was not invertible, you could take $\pmatrix{r\cr s\cr}$ to be a vector in its null space and get 0.
Ok, so that would mean we found a vector $x^* = (\bar r n^t, \bar s n^t)$ for which $x^*Mx=0$ which would contradict $\Re(x^*Mx)>0$. Thanks! – Tobias Kienzler Sep 1 '11 at 8:02