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I have given a nxn upper triangular matrix R and I want to solve the system of equations Rx=0 where x is a vector of size n. Moreover the lowermost diagonalement of R is 0 (R(n,n)=0). Therefore I want to set x(n)=1.

I tried some loops but I do not know how to solve it.

Thank you for your help.

fYpsE
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1 Answers1

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It's guaranteed that R has a zero eigenvalue, and the solution you want is a multiple of the eigenvector corresponding to that eigenvalue. Let's create some matrix R first:

>> R = triu(rand(3, 3));
>> R(3, 3) = 0;
>> R

R = 

    0.8147    0.9134    0.2785
         0    0.6324    0.5469
         0         0         0

Now let's get the eigenvalues and eigenvectors:

>> [V, E] = eig(R)

V =

    1.0000   -0.9806    0.4289
         0    0.1958   -0.5909
         0         0    0.6833


E =

    0.8147         0         0
         0    0.6324         0
         0         0         0

The eigenvectors are the diagonal elements of E

>> E = diag(E);
>> index = find(abs(E) < 1e-16); %// NB don't use find(E==0) because of fp problems...

Now pull out the correct eigenvector

>> v = V(:, index);

and make sure its final element is equal to 1

>> v = v / v(end)

v =

    0.6277
   -0.8648
    1.0000

You can check that this is the solution you want

>> R * v

ans =

     0
     0
     0
Chris Taylor
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  • I tried your example with triu(rand(6,6)) and defined the last 3 diagonal elements to 0 (R(4,4)=0, R(5,5)=0. R(6,6)=0) and your solution isn`t working there. – fYpsE Nov 29 '14 at 16:40
  • That's because you asked me for a solution where the last diagonal entry R(6,6) was zero. There's no reason it should also work when there is more than one zero diagonal entry. – Chris Taylor Nov 30 '14 at 15:59