The partially prescribed inverse problem

Now we consider a type of problem known as a matrix completion problem. We pick one suggestion by Hugo Woerderman and we are grateful to him for discussions.

Given matrices a, b, c and d, we wish to determine under what conditions there exists matrices x, y, z and w such that the block two by two matrices

are inverses of each other. Also, we wish to find formulas for x, y, z and w.

This problem was solved in a paper by W.W. Barrett, C.R. Johnson, M. E. Lundquist and H. Woerderman [BJLW] where they showed it splits into several cases depending upon which of a, b, c and d are invertible. In our next example, we assume that a, b, c and d are invertible and derive the result which they obtain. If one runs NCProcess1 on the polynomial equations which state that a, b, c and d are invertible together with the eight polynomial equations which come from the two matrices above being inverses of each other, one gets the spreadsheet:

THE ORDER IS NOW THE FOLLOWING: a < a < b < b < c < c < d < d « z « x « y « w

	YOUR SESSION HAS DIGESTED
	THE FOLLOWING RELATIONS

THE FOLLOWING VARIABLES HAVE BEEN SOLVED FOR:
{w, x, y}
The corresponding rules are the following:

w a d z b d

x d - d z b

y c - b z c

The expressions with unknown variables {}
and knowns {a, b, c, d, a, b, c, d}

a a 1

a a 1

b b 1

b b 1

c c 1

c c 1

d d 1

d d 1

	USER CREATIONS APPEAR BELOW

	SOME RELATIONS WHICH APPEAR BELOW
		MAY BE UNDIGESTED

THE FOLLOWING VARIABLES HAVE NOT BEEN SOLVED FOR:
{z} The expressions with unknown variables {z}
and knowns {a, b, c, d}

z b z z + d a c

This spreadsheet shows that, if a, b, c and d are invertible, then one can find x, y, z and w such that the matrices in (§) are inverses of each other if and only if z b z = z + d a c. The spreadsheet also gives formulas for x, y and w in terms of z.

In [BJLW] they also solve the problem in the case that a is not invertible -- the answer is more complicated and involves conditions on ranks of certain matrices. It is not clear whether or not these can be derived in a purely algebraic fashion.