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find a column vector such that the determinant of a matrix A is non-zero?

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Let H1 = Span_IR{(1; 0; 0; 0; 0; 0); (0; 1; 0; 0; 0; 0)}
H2 = Span_IR{(0; 0; 1; 0; 0; 0); (0; 0; 0; 1; 0; 0)}
H3 = Span_IR{(0; 0; 0; 0; 1; 0); (0; 0; 0; 0; 0; 1)}
H4 = Span_IR{(1; 0; 1; 0; 1; 0); (0; 1; 0; 1; 0; 1)}
i want to find two vectors (x1,x2,x3,x4,x5,x6) and (y1,y2,y3,y4,y5,y6) such that
H= span{(x1,x2,x3,x4,x5,x6) ,(y1,y2,y3,y4,y5,y6)} and
SpanR(Hi ;Hj ;H ) = R^6. where i,j in {1,2,3,4}.
so we have six matrixs that must not be zero . every matrix is composed by the two vector (x1,x2,x3,x4,x5,x6) and (y1,y2,y3,y4,y5,y6) and 4 vectors among the eight vectors of H1, H2 H3 and H4.
for example
A1 = [1 0 0 0 x1 y1;0 1 0 0 x2 y2;0 0 1 0 x3 y3;0 0 0 1 x4 y4;0 0 0 0 x5 y5;0 0 0 0 x6 y6] then DET(A) must be non zero.
hope this is clear.

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Bruno Luong
Bruno Luong on 18 Sep 2019
"such that the determinant of any Ai, i=1...6 is different to zero."
This is not an equation but a constraint. What you present is strictly speaking 0 equation (and not 6 equations).
Let me give you a simple example: If I tell you find a scalar x such that
x*3.1415926 different 0.
What is x?
x can be any number different than 0. Do you think this is one equation one unknown and well defined system?
However if I ask x such that
x*3.1415926 = 2.718281828
Now this is an equation, and the answer is
x = 0.865255994045822.
You do not need to know MATLAB to see the difference, this is more about (linear) algebra.
Abdessami Jalled
Abdessami Jalled on 18 Sep 2019
Yes sorry for the misunderstanding.
if we calculate the determinant of Ai, we obtain six expressions in xi and yi. is there any method to find (x1,x2,x3,x4,x5,x6) and (y1,y2,y3,y4,y5,y6) such that DET(A) is non zero?
Bruno Luong
Bruno Luong on 19 Sep 2019
If it's not clear for you the test in my answer below
eigs(A,1,'smallestabs')
abs(d) >= 1e-12
is the robust numerical test of non-zero det(A) .

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Accepted Answer

Bruno Luong
Bruno Luong on 18 Sep 2019
Edited: Bruno Luong on 18 Sep 2019
A8 = [1 0 0 0 0 0 1 0;
0 1 0 0 0 0 0 1;
0 0 1 0 0 0 1 0;
0 0 0 1 0 0 0 1 ;
0 0 0 0 1 0 1 0;
0 0 0 0 0 1 0 1]
ij=nchoosek(1:4,2);
binaryflag = true; % put it TRUE if you want X and Y binary, even if it's not stated in the question
% This method is O(1) for binaryflag = false
while true
if binaryflag
H = rand(6,2) > 0.5;
else
H = randn(6,2);
H = H ./ sqrt(sum(H.^2,1));
end
for k=1:size(ij,1)
i=ij(k,1);
j=ij(k,2);
Hi = A8(:,2*i+(-1:0));
Hj = A8(:,2*j+(-1:0));
A = [Hi,Hj,H];
d = eigs(A,1,'smallestabs');
OK = abs(d) >= 1e-12;
if ~OK
% for binaryflag=FALSE, likely never get here !
% fprintf('detect possible singularity\n');
break
end
end
if OK
% for binaryflag=FALSE likely get here the first iteration
break;
end
end
X = H(:,1)
Y = H(:,2)

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