# If $M$ is p.d with $M^{-1}=\pmatrix{A&B\\C&D}^{-1}=\pmatrix{P&Q\\R&S}$, then $P-A^{-1}$ is n.n.d.

Let $M$ be a real positive definite matrix such that $M^{-1}=\pmatrix{A&B\\C&D}^{-1}=\pmatrix{P&Q\\R&S}$, where $A$ is a square matrix. Then show that $P-A^{-1}$ is non-negative definite.

As $M$ is p.d, it is invertible. As principal minors of $M$ should be positive we have,

$\det(A)>0$ and $\det(M)=\det(A)\det(D-CA^{-1}B)>0$ (assuming all matrices have orders so that $D-CA^{-1}B$ is square), thus implying $A$ and $D-CA^{-1}B$ are both invertible.

Now using the formula for inverse of block matrices,

$P=A^{-1}+A^{-1}BF^{-1}CA^{-1}$ where $F=D-CA^{-1}B$.

$\implies P-A^{-1}=A^{-1}BF^{-1}CA^{-1}$.

I don't see why this has to be n.n.d. Possibly I am missing something obvious.

• This proof only works if positive definite implies symmetric. Can we assume in this context that $M$ is meant to be a symmetric matrix? – Omnomnomnom Mar 1 '18 at 7:50
• Notably, the trick with the principal minors only works if $M$ is symmetric – Omnomnomnom Mar 1 '18 at 7:54
• @Omnomnomnom Can't we always take $M$ to be symmetric without loss of generality? – StubbornAtom Mar 1 '18 at 8:32
• It depends on what exactly "without loss of generality" means in this context. If you're using positive definite matrices to represent a quadratic form (or symmetric bilinear form), then it is indeed common to assume that $M$ is symmetric, without loss of generality. However, different matrices that induce the same quadratic form can have drastically different properties. – Omnomnomnom Mar 1 '18 at 17:11

If $M$ is a symmetric matrix, then we have $$P-A^{-1}=A^{-1}BF^{-1}CA^{-1} = [A^{-1}B]F^{-1}[A^{-1}B]^T$$ Since $F^{-1}$ is (symmetric and) positive definite, the matrix $QF^{-1}Q^T$ (where $Q = A^{-1}B$) must be (symmetric and) non-negative definite.
• You seem to take $A$ to be symmetric. Is $A$ also p.d? – StubbornAtom Mar 1 '18 at 9:54
• @StubbornAtom if $M$ is symmetric and p.d. then all of its principal submatrices are symmetric and p.d. – Omnomnomnom Mar 1 '18 at 17:08
• Doesn't this show that $P-A^{-1}$ is positive definite to be precise? Also, what would you recommend doing if $M$ is not symmetric? – StubbornAtom Mar 10 '18 at 15:44
• We can only say that $P - A^{-1}$ is positive definite (i.e. invertible) if we are given that $A^{-1}B$ is invertible, which notably can only happen if $B$ is square. – Omnomnomnom Mar 10 '18 at 16:36
• @StubbornAtom I don't think that the statement holds if we're not given that $M$ is symmetric – Omnomnomnom Mar 10 '18 at 16:38
the positive definite feature is defined for symmetric matrices in most texts. I guess for your problem too. In order M to be symmetric $A$ and $D$ should be symmetric and you should have $C=B^T$.So you can write $M$ as follows: $$\begin{bmatrix} A & B\\ B^T & D\\ \end{bmatrix}$$ You should study Schur complement.The matrix M is positive definite if and only if the schur complement of $M$ relative to $A$ is positive definite. This schur complement is defined as follows: $X/A=D-B^TA^{-1}B$ Since M is positive definite$X/A=D-B^TA^{-1}B$ is positive definie. Now it is easy to answer your question. as you have mentioned
$$P-A^{-1}=A^{-1}BF^{-1}CA^{-1}=A^{-1}BF^{-1}B^{T}A^{-1}$$ in which $F=D-CA^{-1}B=D-B^{T}A^{-1}B$ is positive definite. Here you should know that if a matrix is positive definite its inverse is positive definite too. So $F^{-1}$ is positive definite.Now consider the vector $x$. By constituting quadratic form for $P-A^{-1}$ we have: $$x^{T}(P-A^{-1})x=x^{T}A^{-1}BF^{-1}B^{T}A^{-1}x=(B^{T}A^{-1}x)^{T}F^{-1}(B^{T}A^{-1}x)\ge 0$$.The $\ge$ sign is written by using the positive definite property of $F^{-1}$ Take care that $A^{-1}$ is symmetric too.So $P-A^{-1}$ is positive semi-definite.