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

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.

[Bruno Luong]

• Sorry the question is meaningles: DET is defined only for square matrix.
• The rank of a (m x n) matrix is mathematically <= min(m,n) == 6 in your case. The kwown block A(:,1:8) is rank 6, meaning there are 6 independent columns in this block. For example A(:,1:6), which is eye(6). Any arbitrary column vector (x, y you want to add) must be a linear combination of this 6 vectors. So you can never increase the rank=6 by adding any column. The question is doubling-meaningless.

[Abdessami Jalled]

Yes you are true. sorry for the fault. I rectify.

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.

[Bruno Luong]

Sorry it'still a not-well defined problem.

Now the entire universe is a solution.

Any random X and Y would almost certainly meets that requirement. "Almost" is with probability == 1, (providing obvious condition i # j that you fail to mention).

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]

while true
    ij=randperm(4,2); % you can do all nchoosek(1:4,2) if you like
    i=ij(1); j=ij(2);
    Hi = A8(:,2*i+(-1:0));
    Hj = A8(:,2*j+(-1:0));
    H = randn(6,2);
    H = H ./ sqrt(sum(H.^2,1));
    A = [Hi,Hj,H];
    d = eigs(A,1,'smallestabs');
    d = abs(d)
    if d < 1e-12
        fprintf('detect likely singularity\n')
        return
    end
end

[John D'Errico]

Bruno is completely correct here.

[Abdessami Jalled]

sorry i'm new in Matlab and this is my first time. didn't inderstand. but our six matrix that we take are 6 lines 6 columns (each time we take 4 vectors of the known vectors with the two unknown vectors).

what i want is to find these (x1,x2,x3,x4,x5,x6) and (y1,y2,y3,y4,y5,y6) such that the determinant of any Ai, i=1...6 is different to zero. so we have 6 equation of 12 unknowns variables.

[Bruno Luong]

"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]

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]

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) .

Answer 1

[Bruno Luong]

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)