# Conditions for a (2x2) block matrix to be Positive definite

Let us consider a given symmetric matrix $$M=\begin{bmatrix}0 & B\\ B^{T} & C \end{bmatrix}.$$ 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.

• 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. Jun 30, 2017 at 14:04
• If the (1,1) block is zero, then also $B$ must be zero. Jun 30, 2017 at 14:11
• 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. Jun 30, 2017 at 14:20
• 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. Jun 30, 2017 at 14:20
• Also have a look at math.stackexchange.com/q/2280671/321264. Jun 30, 2017 at 14:25

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.