michael_nielsen

That's one way. There are many other useful equivalent ways of thinking of the det: (1) As the product of the eigenvalues (for normal matrices); (2) as the product of the singular values (up to sign) for any matrix; (3) as the value of the wedge product of the columns c1^c2^ ...

It's the product of the eigenvalues for any matrix. Not just normal ones.

The 2 by 2 matrix [ [2, 0], [1, 2] ] has determinant 4, and just a single eigenvalue, 2.

Of course, if you want to raise the eigenvalues to the power of their algebraic multiplicities, then you do get the determinant. That's another nice way of thinking about it. But it wasn't what my tweet was referring too (geometric multiplicities).

Well we don't disagree then except on semantics. I think of n x n matrix always having n eigenvalues and Det is their product. I think that's very standard. But you're using a different convention.

The matrix I described above is 2 by 2, but has just a single eigenvector. So you're wrong about n x n matrices having n eigenvalues.

That's why algebraic multiplicities (2, in my example) are needed, rather than geometric multiplicities (1, in my example). This is true of any non-normal matrix, except in the trivial case with eigenvalue 1, where raising to different powers makes no difference.

No, the problem of missing an eigvec, and hence by your semantics missing an eigval, is not true of any non-normal matrix, only special non-generic ones (an infinitesimal perturbation will give full basis of eigvecs). e.g. [[2, epsilon][1,2]] has two eigvecs for ep \neq 0.