Let d be a positive integer, a in R^(d x (d+1)), b in R^(d+1) and 
C= (a'*a  b) in R^((d+2) x (d+2)).
   (b'    0)

Show that 

(*) det(C) = -(det(a' b))^2.

It is easy to check the identity for d=1. I tested (*) numerically using
random matrices.

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Request for Question Clarification by livioflores-ga on 31 May 2004 09:36 PDT

Can you clarify what means:
a' (is it the transpose matrix?)
a' *a (is it the product between a and its transpose?)

Thank you.

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Clarification of Question by florian22651-ga on 01 Jun 2004 03:46 PDT

Let d be a positive integer, a in R^(d x (d+1)), b in R^(d+1) and 
C= (a^t * a  b) in R^((d+2) x (d+2)).
   (b^t      0)

Show that 

(*) det(C) = -(det(a^t, b))^2.

Notation: a^t is the transpose, a^t * a is the matrix-product between a^t
and a. Example d=1, a=(1,2), b^t=(3,4)
            (1 2 3)
det(C) = det(2 4 4) = -4
            (3 4 0) 

and 
det(1 2) = - 2.
   (3 4)

In the case d=1 it can be shown quite easily that (*) holds. For
general d I tested (*) numerically using random matrices.

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Request for Question Clarification by mathtalk-ga on 02 Jun 2004 06:01 PDT

Hi, florian22651-ga:

The formula can be proven using expansion by minors.  Would you like
detailed "hints" or simply a completed proof?

regards, mathtalk-ga

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Clarification of Question by florian22651-ga on 02 Jun 2004 09:55 PDT

Thanks for your effort. Meanwhile I found the answer myself. If you
want the money, detailed "hints" will suffice. Actually it was not
homework but a question that came up in the context of barycentric
coordinates.
Thanks again, Florian

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Request for Question Clarification by mathtalk-ga on 02 Jun 2004 10:25 PDT

Hi, florian22651-ga:

If you found the Answer for yourself, I suggest that you Close
(Expire) the Question to avoid being charged.  Note that it was
another researcher, livioflores-ga, who asked earlier about the
meaning of the notation.

Personally I like to use ' for transpose - very compact!

My proof is not elegant. Expand twice by minors, along the final
column and final row of C.  The result of omitting row i and column j
from a'*a is the same as the product of omitting row i from a' and
column j from a, so the cofactors in this double expansion are
determinants of the products of two dxd matrices.  Comparing the
double expansion of C with the (single) expansions of:

det(a' | b)
      
       |  a  |
 = det | --- |
       |  b' |

proves det(C) = - (det(a' | b))^2, just as you conjectured.

regards, mathtalk-ga