Re: Determinant of a particular symmetric matrix

Alois Steindl wrote: 
Hello
sajesse <sajburguet@xxxxxxxx> writes:


Hi all,
My problem is the following.
I have a symmetric n x n matrix, which contains only positive values.
Each value on the main diagonal is equal to 1, and other values v are such that 0 < v < 1.
Numerically, it seems that the determinant of such a matrix is always different from 0, so that it always has an inverse (very interesting to me).
I'd like to know if there exist a proof of this property (if it's true).
If someone could help me... thanks. 

Its generally not true: Consider the matrix

[ 1 a b ]
[ a 1 c ]
[ b c 1 ]

with b=2/5, c=3/5 and a=6/25+4 sqrt(21)/25 approx 0.97,

then the determinant vanishes.

Alois



A similar example is the Toeplitz matrix

[ 1 a b ]
[ a 1 a ]
[ b a 1 ]

in which, with a = 1/sqrt(2), the determinant vanishes at b = 0, b = 1, though these b values are not in the open interval [0,1].
If a = 0.9 then det. vanishes at b = ~0.62 and b = 1

So the original premise seems to be false.

john


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