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Home/ Questions/Q 7612607
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Editorial Team
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Editorial Team
Asked: May 31, 20262026-05-31T01:57:23+00:00 2026-05-31T01:57:23+00:00

The problem i am trying to understand is easy but i cant seem to

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The problem i am trying to understand is easy but i cant seem to get the correct result in matlab. The actual problem is that i want to get the weight vectors of a 2 hidden layer input RBF using just the plain distance as a function, i.e. no Baysian or Gaussian function as my φ. I will use the function with 2 centres let’s say 0,0 and 1,1. So this will give me a Matrix φ of:

[0 sqrt(2) ; 1 1; 1 1; sqrt(2) 0] *[w1; w2] = [0;1;1;0] As defined my the XOR function.

When i apply the pseudoinverse of the Φ in matlab * [0;1;1;0] though i get [0.33 ; 0.33] which is not the correct value which would allow me to get the correct output values [0;1;1;0].

i.e. .33 * sqrt(2) != 0 .

Can someone explain to me why this is the case?

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  1. Editorial Team
    Editorial Team
    2026-05-31T01:57:25+00:00Added an answer on May 31, 2026 at 1:57 am

    I’ll take a swag at this. The matrix, I’ll call A, A = [0 sqrt(2) ; 1 1; 1 1; sqrt(2) 0] has full column rank, but not full row rank, i.e. rank(A) = 2. Then you essentially solve the system Ax = b, where x is your weighting vector. You could also just do x = A\b in Matlab, which is supposedly a much more accurate answer. I get the same answer as you. This is a very rough explanation, when your system can not be solved for a certain solution vector, it means that there exists no such vector x that can be solved for Ax = b. What Matlab does is try to estimate the answer as close as possible. I’m guessing you used pinv, if you look at the Matlab help it says:

    If A has more rows than columns and is not of full rank, then the overdetermined least squares problem

    minimize norm(A*x-b)
    

    does not have a unique solution. Two of the infinitely many solutions are

    x = pinv(A)*b 
    

    and

    y = A\b 
    

    So, this appears to be your problem. I would recommend looking at your φ matrix if possible to come up with a more robust system. Hope this is useful.

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