Suppose we have a real, symmetric matrix $A(x_1,x_2,x_3)$ given by \begin{pmatrix} a_{1,1} & a_{1,2} & x_1 & x_2 \\ a_{2,1} & a_{2,2} & a_{2,3} & x_3 \\ x_1 & a_{3,2} & a_{3,3} & a_{3,4} \\ x_2 & x_3 & a_{3,4} & a_{4,4} \end{pmatrix}

We would like to complete this matrix into a positive definite matrix by choosing appropriate values for $x_1,x_2, x_3$. (Assume that the $a$ values permit this by themselves). Many completions are possible.

However, only one unique completion is possible such that its inverse is of the form \begin{pmatrix} b_{1,1} & b_{1,2} & 0 & 0 \\ b_{2,1} & b_{2,2} & b_{2,3} & 0 \\ 0 & b_{3,2} & b_{3,3} & b_{3,4} \\ 0 & 0 & b_{3,4} & b_{4,4} \end{pmatrix} i.e it has zeroes in the positions corresponding to $x_1, x_2, x_3$. Furthermore, apparently these values of $x_1,x_2,x_3$ end up maximizing the determinant of $A$.

How do we prove this, and what are the corresponding values of $x_1, x_2, x_3$?

The relationship between the completion and maximizing determinant seemed clear since the conditions on the inverse are equivalent to the partial derivatives of the determinant function being zero. But I am unable to establish uniqueness.

  • $\begingroup$ scgroup.hpclab.ceid.upatras.gr/class/SCII/Various/… looks like it might be relevant. $\endgroup$ Sep 9, 2012 at 1:29
  • $\begingroup$ Can you please give me the values of the parameters $a_{ij}$? It's easier to solve an instance of the problem before thinking of all the instances of the problem! $\endgroup$ Sep 9, 2012 at 3:30
  • $\begingroup$ Sure. Assume the matrix is, say \begin{pmatrix} 3 & 2 & x_1 & x_2 \\ 2 & 2 & 1 & x_3 \\ x_1 & 1 & 2 & 1 \\ x_2 & x_3 & 1 & 1 \end{pmatrix} $\endgroup$
    – BharatRam
    Sep 9, 2012 at 5:28
  • $\begingroup$ In the above case, writing out the Sylvester's conditions, and the specific minors being $0$, we get $(x_1,x_2,x_3)=(1,1/2,1/2)$, but I obtained this through brute computation of polynomial roots, rather than in any insightful algebraic way. $\endgroup$
    – BharatRam
    Sep 9, 2012 at 5:37
  • $\begingroup$ @Gerry: The following paper is most relevant. But it attacks the problem in general, and has many borrowed analytic notions. I was hoping it could be explicitly demystified in the simple special case which I asked. orion.math.uwaterloo.ca/~hwolkowi/henry/reports/… $\endgroup$
    – BharatRam
    Sep 9, 2012 at 5:56

1 Answer 1


Let $f:x=(x_1,x_2,x_3)\rightarrow \det(A(x))$.

$\textbf{Proposition}.$ If $A(x)$ is invertible, then

$Df_{x}=0$ iff $A(x)^{-1}$ is a tridiagonal matrix.

$\textbf{Proof}.$ "$Df_x=0$" and "$A(x)^{-1}$ is a tridiagonal matrix" can each be written in the form of an algebraic system of $3$ algebraic equations. These $2$ systems have same Grobner basis. $\square$

If $A$ is symm. $>0$, then considering the determinants of the submatrices $[1..3],[1.3]$ and $[2.4],[2..4]$ of $A$, we deduce that $x_1,x_3$ are bounded by functions of the $(a_{i,j})$. Then $\det(A)=(-a_{2,2}a_{3,3}+a_{3,2}^2)x_2^2+\cdots=(<0)x_2^2+\cdots$ must be $>0$ and $x_2$ is also bounded. Thus the max of $\det(A)$ is reached for at least one solution of the above system.

For the OP's example (cf. his comment above), the system has $5$ real solutions. We must choose the one that realizes $\max(\det(A))$.


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