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Let us consider a given symmetric matrix \begin{equation} M=\begin{bmatrix}0 & B\\ B^{T} & C \end{bmatrix}. \end{equation} My question may seem very simple but what conditions must the block matrices follow in order to make the big matrix positive definite please? A, B and C are matrices. Thanks.

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    $\begingroup$ Not with a zero subblock because $$\begin{bmatrix} v^* & 0 \end{bmatrix}M\begin{bmatrix} v \\ 0 \end{bmatrix} = v^*0v = 0.$$ Positive semi-definite is still a possibility. $\endgroup$ Jun 30, 2017 at 14:04
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    $\begingroup$ If the (1,1) block is zero, then also $B$ must be zero. $\endgroup$ Jun 30, 2017 at 14:11
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    $\begingroup$ It's the same thing: $$ \begin{bmatrix} 0 & v^* \end{bmatrix} M \begin{bmatrix} 0 \\ v \end{bmatrix} = v^*0v = 0. $$ Also you can conjugate by $$ \begin{bmatrix} 0 & I \\ I & 0 \end{bmatrix} $$ to get back to the first form. $\endgroup$ Jun 30, 2017 at 14:20
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    $\begingroup$ A matrix with zero(s) on the diagonal can never ever be positive definite. It can only be positive semi-definite and, if $M$ has zeros on the diagonal, that can happen only if for every diagonal zero entry the whole corresponding row and column is zero. That is a consequence of this fact. $\endgroup$ Jun 30, 2017 at 14:20
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    $\begingroup$ Also have a look at math.stackexchange.com/q/2280671/321264. $\endgroup$ Jun 30, 2017 at 14:25

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I'm creating an answer from the comments.


If $M$ has a zero subblock in the (1,1) position then $M$ cannot be positive definite because $$ \begin{bmatrix} v^T & 0 \end{bmatrix}M\begin{bmatrix} v \\ 0 \end{bmatrix} = v^T0v = 0. $$

Nor can it have a zero subblock in the (2,2) position because then it is conjugate to a matrix with a zero subblock in the (1,1) position: $$ \begin{bmatrix} 0 & I \\ I & 0 \end{bmatrix} \begin{bmatrix}A & B\\ B^{T} & 0 \end{bmatrix} \begin{bmatrix} 0 & I \\ I & 0 \end{bmatrix}^{-1} = \begin{bmatrix} 0 & B^T \\ B & A \end{bmatrix}. $$ You can also see this directly $$ \begin{bmatrix} 0 & v^T \end{bmatrix} \begin{bmatrix} A & B\\ B^{T} & 0 \end{bmatrix} \begin{bmatrix} 0 \\ v \end{bmatrix} = v^T0v = 0. $$


If the (1,1) subblock is $0$ and $M$ is positive semidefinite then the $(1,2)$ and $(2,1)$ must be zero as a consequence of this result. This leaves us with

$$ \begin{bmatrix}0 & 0\\0 & C\end{bmatrix} $$

which is positive semidefinite iff $C$ is positive semidefinite.

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