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This is also largely a consequence of the fact that most linear algebra courses are computational by nature. They'll ask you to compute lots of things as a mean
by daniel-levin 10y ago
This is also largely a consequence of the fact that most linear algebra courses are computational by nature. They'll ask you to compute lots of things as a means of assessment. Work out this determinant. Find the eigenvalues of that matrix. "Matrices are grids of numbers". Take this matrix and write it in terms of this other basis. I only really perceived the impact of linear transformations being vector-space-structure preserving mappings long after I'd seen their analogues in the form of homomorphisms, homeomorphisms, diffeomorphisms etc. Nobody told us that determinants are just the alternating k-tensor on real k-space (up to a constant factor). At some level, this is because most students taking linear algebra don't have the mathematical maturity to stomach a course in finite dimensional vector spaces. I was indignant when I found out that a matrix was not just a grid of numbers, but rather a manifestation of a linear transformation, with respect to a particular basis.
I only really started to understand linear algebra when I was forced to in a differential geometry class. The opening chapters were a review intended to fix my university's notoriously broken linear algebra training. As I said, it comes with the territory. You can't design a course based on Halmos' FDVS and expect students coming in, that is, students who've scraped through calculus 1, to manage. So, the recipe book / cookbook style abounds. Granted, there were inklings of mathematics in my linear algebra course. I don't think anyone really appreciated it however. It's hard to grok "vector space over a field" when you've never been introduced to the abstract concept of a field.
When my second year stats lecturer told me that a determinant of a 2x2 matrix was an area, I almost didn't believe him.
Fun exercise I was told about just the other day. Every invertible matrix with integer coefficients has determinant +-1. I would never have known how to solve that after my linear algebra course.
- abecedarius 10y agoI'm confused: [42] is an invertible matrix with integer coefficients, right?
- taejo 10y agoI guess daniel-levin meant to say "invertible matrix with integer coefficients whose inverse also has invertible coefficients". The fact isn't too hard to see: * If M has integer coefficients, then det(M) is an integer. * det(inv(M)) = 1/det(M) * Since M has integer coefficients, det(M) is an integer * Since inv(M) has integer coefficients, det(inv(M)) is an integer * So det(M) and 1/det(M) are both integers, so det(M) is either 1 or -1
- taejo 10y agos/invertible coefficients/integer coefficients/
- sixo 10y agoYes, the claim is false. It's not obvious to me what it's been confused for. Edit: oh, the other commenter figured it out.