4 ms·

It sounds like you're assuming diagonalizability, which is stronger than "eigen values defined over the field". Even over the complex numbers there are matrice

It sounds like you're assuming diagonalizability, which is stronger than "eigen values defined over the field". Even over the complex numbers there are matrices that aren't diagonalizable, like [[2,1],[0,2]].

xyzzyz 3y ago
In practice, all matrices are diagonalizable, because the set of non-diagonalizable matrices is very small. Concretely, a random matrix is almost surely diagonalizable (ie. with probability 1).
- FooBarBizBazz 3y ago
  This "with zero probability" statement hides that there's a probability distribution assumed. It's really saying that any matrix can be perturbed to be diagonalizable. Which is true. But matrices in that set have a way of showing up a lot in models that the human mind conceives. For example, if you model a point mass accelerating without drag, you will end up with a matrix that isn't diagonalizable. If all matrices were diagonalizable, every linear ODE would be solved by a bunch of exponentials and sinusoids; there'd be no polynomials. But we see things like "1/2 a t^2" all the time.
  - henrydark 3y ago
    Also "any matrix can be perturbed" assumes a normed field, though those equations work over other fields as well
JadeNB 3y ago
> It sounds like you're assuming diagonalizability, which is stronger than "eigen values defined over the field". > Even over the complex numbers there are matrices that aren't diagonalizable, like [[2,1],[0,2]]. While that's completely true, and I'm glad someone said it, it's not much of an issue for 2×2 matrices. You can tell quickly when it's happened, because the characteristic polynomial is a square (itself easily tested by computing the gcd with the derivative … or just because it's easy to recognize when a quadratic is a perfect square) but the matrix isn't scalar; and then we have that the semisimple part S is scalar, the nilpotent part N satisfies N^2 = 0, and so (S + N)^n equals S^n + n S N^{n - 1} for n > 1.
deleted 3y ago
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FooBarBizBazz 3y ago
Thank you! This why the Jordan Form exists at all. (All real symmetric, or more generally Hermitian matrices are diagonalizable, which is the Spectral Theorem.)

xyzzyz 3y ago

In practice, all matrices are diagonalizable, because the set of non-diagonalizable matrices is very small. Concretely, a random matrix is almost surely diagonalizable (ie. with probability 1).

FooBarBizBazz 3y ago
This "with zero probability" statement hides that there's a probability distribution assumed. It's really saying that any matrix can be perturbed to be diagonalizable. Which is true. But matrices in that set have a way of showing up a lot in models that the human mind conceives. For example, if you model a point mass accelerating without drag, you will end up with a matrix that isn't diagonalizable. If all matrices were diagonalizable, every linear ODE would be solved by a bunch of exponentials and sinusoids; there'd be no polynomials. But we see things like "1/2 a t^2" all the time.
- henrydark 3y ago
  Also "any matrix can be perturbed" assumes a normed field, though those equations work over other fields as well

JadeNB 3y ago

> It sounds like you're assuming diagonalizability, which is stronger than "eigen values defined over the field". > Even over the complex numbers there are matrices that aren't diagonalizable, like [[2,1],[0,2]]. While that's completely true, and I'm glad someone said it, it's not much of an issue for 2×2 matrices. You can tell quickly when it's happened, because the characteristic polynomial is a square (itself easily tested by computing the gcd with the derivative … or just because it's easy to recognize when a quadratic is a perfect square) but the matrix isn't scalar; and then we have that the semisimple part S is scalar, the nilpotent part N satisfies N^2 = 0, and so (S + N)^n equals S^n + n S N^{n - 1} for n > 1.

deleted 3y ago

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FooBarBizBazz 3y ago

Thank you! This why the Jordan Form exists at all. (All real symmetric, or more generally Hermitian matrices are diagonalizable, which is the Spectral Theorem.)