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.

  • 1
    $\begingroup$ This proof only works if positive definite implies symmetric. Can we assume in this context that $M$ is meant to be a symmetric matrix? $\endgroup$ – Omnomnomnom Mar 1 '18 at 7:50
  • $\begingroup$ Notably, the trick with the principal minors only works if $M$ is symmetric $\endgroup$ – Omnomnomnom Mar 1 '18 at 7:54
  • $\begingroup$ @Omnomnomnom Can't we always take $M$ to be symmetric without loss of generality? $\endgroup$ – StubbornAtom Mar 1 '18 at 8:32
  • 1
    $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ You seem to take $A$ to be symmetric. Is $A$ also p.d? $\endgroup$ – StubbornAtom Mar 1 '18 at 9:54
  • $\begingroup$ @StubbornAtom if $M$ is symmetric and p.d. then all of its principal submatrices are symmetric and p.d. $\endgroup$ – Omnomnomnom Mar 1 '18 at 17:08
  • $\begingroup$ 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? $\endgroup$ – StubbornAtom Mar 10 '18 at 15:44
  • 1
    $\begingroup$ 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. $\endgroup$ – Omnomnomnom Mar 10 '18 at 16:36
  • $\begingroup$ @StubbornAtom I don't think that the statement holds if we're not given that $M$ is symmetric $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.