3 ms·

They define a generic "is_singular" function and test it with a 2x2 matrix. The problem with the determinant is not about performance. It is just useless for d

by cafaxo 3y ago

They define a generic "is_singular" function and test it with a 2x2 matrix. The problem with the determinant is not about performance. It is just useless for determining if a matrix is singular. The thing that gives it away is that the determinant is influenced by a rescaling of the matrix: det(s A) = s^n det(A) where A is a n x n matrix As an example, would you say that [[1e-10, 0], [0, 1e-10]] is singular? It has condition number 1.

renonce 3y ago
It does work in theory and for integer numbers. Such an algorithm might be used in practice when numerical stability is concerned, but condition number can certainly be defined somewhere in another textbook and you just need more context to tell it to use another method.
- cafaxo 3y ago
  Edit: Sorry, I completely messed up my original answer here. A better version: Let's say we are in a setting where we only work with integers. A matrix is invertible iff its determinant is invertible in the underlying ring. The only invertible elements in Z are -1 and 1. So, the code is also incorrect in the integer setting. Here, we should not check for 0, but for -1 or 1.
  - thaumasiotes 3y ago
    If I'm reading you correctly, you'd also need to flip the response to the check: where the original test for 0 determines that a matrix is singular if it does find 0, the new test for ±1 should determine that a matrix is singular if it does not find ±1?
    - cafaxo 3y ago
      Yes, exactly.
TwentyPosts 3y ago
Saying that the determinant is useless to determine whether a matrix is incorrect, or misleading at best. This is purely a matter of numerical stability. Of course [[1e-10, 0], [0, 1e-10]] is nonsingular, and its determinant is 1e-20, which does not equal zero. Yes, when it comes to floating point issues we might want to use something else, and that's a valid complaint when it comes to NumPy code, but from a theoretical perspective the determinant is an excellent tool to determine singularity.
- cafaxo 3y ago
  Of course. Theoretically, the determinant answers the binary question "singular" or "nonsingular". Numerically, such a binary answer is pretty useless. Here, we need a measure of how singular/nonsingular a matrix is relative to the numerical precision we are working with.

by cafaxo 3y ago

renonce 3y ago

It does work in theory and for integer numbers. Such an algorithm might be used in practice when numerical stability is concerned, but condition number can certainly be defined somewhere in another textbook and you just need more context to tell it to use another method.

cafaxo 3y ago
Edit: Sorry, I completely messed up my original answer here. A better version: Let's say we are in a setting where we only work with integers. A matrix is invertible iff its determinant is invertible in the underlying ring. The only invertible elements in Z are -1 and 1. So, the code is also incorrect in the integer setting. Here, we should not check for 0, but for -1 or 1.
- thaumasiotes 3y ago
  If I'm reading you correctly, you'd also need to flip the response to the check: where the original test for 0 determines that a matrix is singular if it does find 0, the new test for ±1 should determine that a matrix is singular if it does not find ±1?
  - cafaxo 3y ago
    Yes, exactly.

TwentyPosts 3y ago

Saying that the determinant is useless to determine whether a matrix is incorrect, or misleading at best. This is purely a matter of numerical stability. Of course [[1e-10, 0], [0, 1e-10]] is nonsingular, and its determinant is 1e-20, which does not equal zero. Yes, when it comes to floating point issues we might want to use something else, and that's a valid complaint when it comes to NumPy code, but from a theoretical perspective the determinant is an excellent tool to determine singularity.

cafaxo 3y ago
Of course. Theoretically, the determinant answers the binary question "singular" or "nonsingular". Numerically, such a binary answer is pretty useless. Here, we need a measure of how singular/nonsingular a matrix is relative to the numerical precision we are working with.