# Positive semidefiniteness of a block matrix of positive semidefinite matrices

Given any symmetric matrix $\mathbf{M} = \begin{pmatrix} \mathbf{A} & \mathbf{B}\\ \mathbf{B}^\mathrm{T}& \mathbf{C} \end{pmatrix}$, the following conditions are equivalent:

(1) $\mathbf{M}\succeq 0$ ($\mathbf{M}$ is positive semidefinite)

(2) $\mathbf{A}\succeq 0$, $\;\;(\mathbf{I}-\mathbf{A}\mathbf{A}^T )\mathbf{B}= 0$, $\;\;\mathbf{C} - \mathbf{B}^T\mathbf{A}^{\dagger }\mathbf{B} \succeq 0$.

(3) $\mathbf{C}\succeq 0$, $\;\;(\mathbf{I}-\mathbf{C}\mathbf{C}^T )\mathbf{B}= 0$, $\;\;\mathbf{A} - \mathbf{B}\mathbf{C}^{\dagger }\mathbf{B}^T \succeq 0$.

Then, if we are given that $\mathbf{A}\succeq 0$, $\mathbf{B}\succeq 0$ and $\mathbf{C}\succeq 0$, is there any way to verify (2) or (3) and therefore conclude that $\mathbf{M}$ will be positive semidefinite as well?

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Just because $M$ is positive semidefinite, it does not follow that $A^{-1}$ exists. – Chris Godsil May 14 '13 at 22:35
Sorry, I should have used the pseudo-inverse for the case of semipositive definite matrices... – María Pérez May 15 '13 at 10:05

The conditions$\;\;(\mathbf{I}-\mathbf{A}\mathbf{A}^T )\mathbf{B}= 0$, $\;\;(\mathbf{I}-\mathbf{C}\mathbf{C}^T )\mathbf{B}= 0$, and $\mathbf{B} \succeq0$, do not follow from anything. W/out them the equivalence follows from the Sylvester criteria for positive definiteness and determinant formula for Schur complement. What is the source of the problem?

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What do you mean that these conditions do not follow from anything? The source of the problem is that I have to use an algorithm which requires a positive semidefinite matrix, but what I have is a set of positive semidefinite matrices A, B, C, which compose M, and I will need to verify if M is positive semidefinite as well so I can use it. – María Pérez May 15 '13 at 12:08
@MaríaPérez Put it simply, (1) does not imply (2) or (3), and (2), (3) do not imply each other. Each of (2) and (3) does imply (1), provided that the $A$ in (2) or the $C$ in (3) are invertible, and the error in (3) is corrected ($A-B^TC^{-1}B$ should read $A-BC^{-1}B^T$). – user1551 May 15 '13 at 12:14

Hint By definition a matrix $M\in\mathbb{R}^{n\times n}$ is positive semidefinite if $v^TMv\geq 0$ for all $v\in\mathbb{R}^n$. Use the fact $\det M= \det\Big(\mathbf{C} - \mathbf{B}^T\mathbf{A}^{-1}\mathbf{B} \Big)$

and proof by inductin in $n\in\mathbb{N}$ for $n\geq 1$.

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Your claim is false, take the $3\times3$ matrix $M$ with all entries 1 and then set $M_{3,3}$ equal to $-1$. This satisfies your claim - the three leading minors of the modified matrix are 1, 0, 0 - but it is not positive semidefinite. – Chris Godsil May 14 '13 at 22:47
@ChrisGodsil, I corrected my mistake. But it is a mistake that does not compromise the idea of proof.Thanks – MathOverview May 14 '13 at 23:35
What if $A$ is not invertible? And $\det(M)=\det(A)\det(C-B^TA^{-1}B)$, if $A$ is invertible. – Chris Godsil May 14 '13 at 23:54
@ChrisGodsil Well, this is a hint to solve the exercise. It is clear that during the resolution will be issues to be considered. – MathOverview May 15 '13 at 10:59