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Re: Free Math Books

#71
post #56

Earlier quoted context omitted.

The beauty of Halmos' derivation, which is similar but not identical to exterior algebra (wedge product), is that his approach is basis independent. A determinant by his definition is scalar invariant over all bases. It is very geometrical in nature.

The determinant inherently involves a basis (or at the very least a choice of unit n-vector). Or if you like you can think of the determinant as a function of a square matrix (grid of numbers), rather than a function of a collection of vectors. When you take the basis out, that's the wedge product, which inherently includes the orientation. Conveniently, there is only one degree of freedom for n-vectors in n-dimensio…

Let me sketch a way to get the determinant basis-free:

Say we live in an n-dimensional vector space V and have an endomorphism f : V -> V. Now, we consider the pullback [1] f* : Λⁿ(V) -> Λⁿ(V) induced by f on the vector space of n-linear alternating forms Λⁿ(V) on V.

This is just an endomorphism on Λⁿ(V). However, Λⁿ(V) is one-dimensional, hence necessarily invariant under f*. This means f* has an eigenvalue (!). This eigenvalue is what we usually call the determinant of f.

This is completely independent of any choice of basis, orientation, or an inner product.

[1] That is, given an element w ∈ Λⁿ(V) and an arbitrary n-tuple v₁, ..., vₙ of vectors from V, we have (f*w)(v₁, ..., vₙ) = w(f(v₁), ..., f(vₙ))

Re: Free Math Books

#72
post #71

Earlier quoted context omitted.

The determinant inherently involves a basis (or at the very least a choice of unit n-vector). Or if you like you can think of the determinant as a function of a square matrix (grid of numbers), rather than a function of a collection of vectors. When you take the basis out, that's the wedge product, which inherently includes the orientation. Conveniently, there is only one degree of freedom for n-vectors in n-dimensio…

Let me sketch a way to get the determinant basis-free: Say we live in an n-dimensional vector space V and have an endomorphism f : V -> V. Now, we consider the pullback [1] f* : Λⁿ(V) -> Λⁿ(V) induced by f on the vector space of n-linear alternating forms Λⁿ(V) on V. This is just an endomorphism on Λⁿ(V). However, Λⁿ(V) is one-dimensional, hence necessarily invariant under f*. This means f* has an eigenvalue (!). Thi…

> and have an endomorphism f

And the "outermorphism" f̱ of your linear transformation, when limited to considering its application to an arbitrary pseudoscalar, returns another pseudoscalar which necessarily has the same orientation, making that a scaling operation.

So what we could say in that case is that f̱(p) / p = d (some scalar, the "determinant" of f), where p is any pseudoscalar p = v1 ∧ v2 ∧ ··· ∧ vn.

This turns out to be about the same as what I wrote a few comments upthread. We are just dealing with

f̱( v1 ∧ v2 ∧ ··· ∧ vn ) / ( v1 ∧ v2 ∧ ··· ∧ vn ) = d

= ( f(v1) ∧ f(v2) ∧ ··· ∧ f(vn) ) / ( v1 ∧ v2 ∧ ··· ∧ vn )

instead of ( v1 ∧ v2 ∧ ··· ∧ vn ) / ( e1 ∧ e2 ∧ ··· ∧ en ) = d

And now we are talking about a property of a linear transformation instead of a property of a collection of n vectors.

In many practical situations, an oriented quantity like v1 ∧ v2 ∧ ··· ∧ vn is more useful than a scalar ratio d though.

Re: Free Math Books

#73
post #69

Earlier quoted context omitted.

Look for Dover books, particularly from Russian authors. Cheap and good!

For linear algebra: Shilov.

I own four copies of that book :-)

Two hard copies since I moved countries, one on kindle and one on Apple Books.

Great book :D

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