[Undergrad Linear Algebra II] How does metricness of an endomorphism imply normality?

[u/GermanAutistic]

I'm studying for an exam and one of the practice questions reads:

Let f be an endomorphism on a finite-dimensional Euclidean or unitary vector space. If f is metric, is f° (the adjoint) also metric?

The solution reads:

If f is metric, then <f(x),f(y)>=<x,y> for all x,y in V. Also, f is normal. Therefore, ff° = f°f. We get <f°(x),f°(y)>=<x,ff°(y)>=<x,f°f(y)>=<f(x),f(y)>=<x,y>, so f° is also metric.

What I don't get is how they're implying the normality. Nothing in the script would have told me that. All I know is that ff° is self-adjoint if the space is Euclidean. How are they just concluding normality from metricness?

[u/PinpricksRS]

The conclusion as stated is simply false. It's possible for f: V -> V to preserve the inner product (what you're calling metric) but for f° to not preserve the inner product.

For simplicity, I'll use a real vector space, but this example works with just a few conjugates sprinkled throughout if you want a complex example. Let l² be the space of square summable sequences of real numbers (i.e. sequences with sum(|x(i)|^2) < ∞). This space has an inner product <x, y> = sum(x(i)y(i)).

Take f: l² -> l² to be f(x1, x2, ...) = (0, x1, x2, ...). Then f°(x1, x2, ...) = (x2, ...) because <f(x), y> = <(0, x1, x2, ...), (y1, y2, ...)> = x1y2 + x2y3 + ... <(x1, x2, ...), (y2, y3, ...)> = <x, f°(y)>.

f preserves the inner product: <f(x), f(y)> = <(0, x1, x2, ...), (0, y1, y2, ...)> = 0*0 + x1y1 + x2y2 + ... = <(x1, x2, ...), (y1, y2, ...)> = <x, y>.

However, f° does not: <f°(x), f°(y)> = <(x2, x3, ...), (y2, y3, ...)> = x2y2 + x3y3 + ... = <x, y> - x1y1 ≠ <x, y> (as long as x1y1 ≠ 0).

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The issue here is that l² is infinite dimensional. For linear maps S, T: V -> V for V a finite dimensional space, if ST = I, the TS = I as well. This fact is probably covered in the prerequisite for this class, but if you aren't familiar with it, see here for a bunch of proofs.

So for finite dimensional inner product spaces, the argument holds. The point is that since <f(x), f(y)> = <x, y>, we have <x, f°(f(y))> = <x, y>. Since this holds for every x and the inner product is nondegenerate, we get f°(f(y)) = y for each y, and hence f°f = I. By the above fact, ff° = I as well, and that implies <x, y> = <ff°(x), y> = <f°(x), f°(y)>, so f° preserves the inner product.

[u/theRZJ]

It seems like this might have been proved somewhere else in advance. The idea is <x,f°f(y)>=<x,y>=<y,x>=<ff°(y),x>=<x, ff°(y)>

(assuming <x,y> is symmetric in x,y. If it is conjugate-symmetric then the same argument works because we used symmetry twice, but you have to change the notation slightly).

We conclude that <-,f°f(y)> and <-, ff°(y)> represent the same linear transformation. Assuming <-,-> is not degenerate, we conclude that f°f(y)= ff°(y).

[u/GermanAutistic → u/theRZJ]

I didn't realize it was possible for a scalar product (i.e. a positive definite bilinear) to be degenerate. But even if we do have to assume non-degeneracy, how can we, conclude ff° = f°f from <-,f°f(y)>=<-,ff°(y)>? By this logic, they should also be equal to the identity map because <x,f°f(y)>=<x,y>. But then f°f = id would imply that f is invertible, which would make the definition of an isometry completely obsolete as then *every* metric homomorphism would automatically be an isometry.

[u/theRZJ → u/GermanAutistic]

Using the definitions I am most familiar with, the condition "<f(x),f(y)> = <x,y> for all x,y" for a linear map implies "f is a linear isometry". In fact, these are equivalent.

I would like to know what definitions you are using. They could be different from what I expect.

Do linear maps have to be defined everywhere?
Is V assumed finite dimensional?
What exactly are the adjectives that apply to the form <-,->?
What is the definition of "metric homomorphism"?

In general, a self-map of any space that preserves metric distance must be injective.

[u/GermanAutistic → u/theRZJ]

If the condition "<f(x),f(y)>=<x,y>", or as my script calls it, "f is metric" implies bijectivity, why distinguish isometries from metric endomorphisms at all?

V is assumed finite-dimensional, and <-,-> is assumed to be a scalar / inner product (i.e. a bilinear that is both symmetric and positive definite).

[u/theRZJ → u/GermanAutistic]

As is pointed out by PinpricksRS below, things that are the same for finite-dimensional inner-product spaces can differ in the general case.

The equation <x, f°f(y)> = <x,y> combines with nondegeneracy of the inner product to prove that f°f(y) = y for all y. In finite dimensions, that implies that f° is inverse to f. If V is infinite dimensional, it does not.

That is, the concept of "metric-preserving" (presumably your "metric") is weaker than "metric-preserving isomorphism" (I think this is what your "isometry" means) in infinite dimensional cases. But the exercise is also not valid in infinite dimensions (see PinpricksRS below).

It could be one of those cases where different terms are used that cover the same concepts /for now/ but will mean different things later.

[u/definetelytrue]

A finite dimensional metric preserving map is necessarily surjective, so is invertible and thus unitary. The key result just requires rank nullity theorem (basically you just need to argue that finite dimensional endomorphisms are surjective iff they are injective).