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[Darvi]

Reals are, by definition, also complex numbers (but with a nil imaginary part), so yes on that I believe.

[ed boy]

Reals are, by definition, also complex numbers (but with a nil imaginary part), so yes on that I believe.

True, but your reasoning (by definition) is wrong.

The standard definition is to define the reals first, and then define the complex numbers as an ordered pair of reals (the real part and the imaginary part). One can then define a subset of the complex numbers to be the complex numbers with a zero imaginary part (the real complex numbers).

One can create a bijective map between the real real numbers and the complex real numbers, but they are not the same thing.

[Darvi]

Oh? Afaik a complex number z is any number a+bi where a and b are real. If b=0 (which is totally is real, aside from philosophical banter), z is real, so any real number is also a complex number. Also, I don't understand what you mean by "real complex numbers" and "real real numbers".

[ed boy]

There are plenty of ways to define the various number systems - I'm not going to claim that one specific one is correct and another is not, but the normal order of things is:
-Define the natural numbers (often using sets)
-Define the integers (normally a pair of natural numbers or a natural number with a boolean)
-Define a subset of the integers that correspond to the natural numbers
-Define the rationals (as a pair of integers)
-Define a subset of the rationals to correspond to the integers (and hence a subset of the rationals to correspond to the natural numbers)
-Define the reals (normally as a set of rational numbers)
-Define a subset of the real numbers to correspond to the rationals
-Define the complex numbers (a pair of natural numbers)
-Define a subset of the complex numbers to correspond to the real numbers.

By 'real complex numbers', I was referring to the subset of complex numbers that correspond to the real numbers. By 'real real numbers', I was referring to the reals, which I called that to compare with the 'real complex numbers'.

[Darvi]

Except that "Define a subset of the complex numbers to correspond to the real numbers" pretty much equals "Define the reals" because reals are a subset of complexes.

[Soadreqm]

So, uh... Are you saying that the set of real numbers defined upwards from rational numbers and the set of real numbers defined as a subset of complex numbers are not the same set?

That is so messed up.

[Darvi]

Nevermind that you cannot define reals as a set of rationals.

[Christes]

So I'm prepping for my linear algebra exam, and I'm preparing for all of the various questions I might end up facing.

Reading through all of my materials, one potential question entered my mind: Schur's decomposition works for complex numbers and unitary matrices. Does it also apply to Real numbers and orthogonal matrices? Looking through the proof and turning the general concepts in my head, I can't see why it shouldn't, but there is no mention of it anywhere, which leads me to believe that it isn't the case.

It's been a long time since I've done this sort of thing, but I think it should fail when the matrix has non-real eigenvalues.  The middle matrix of the decomposition is triangular so those eigenvalues should be down the diagonal, which is a problem if you want real entries.

So, uh... Are you saying that the set of real numbers defined upwards from rational numbers and the set of real numbers defined as a subset of complex numbers are not the same set?

That is so messed up.

It depends what you mean by "same set."  In the strictest technical sense, no.  But there exists a 1-1 map from the the first set of reals into the complex numbers that preserves every algebraic structure you could want.  This is as good as mathematicians usually expect from mathematical objects

Nevermind that you cannot define reals as a set of rationals.

I'm not quite sure what you mean, but you can define each real to be a "Dedekind cut" which is just a set of rational numbers.  You could also do something with sequences of rationals, I guess.

[Darvi]

Not really. Reals also includes irrational numbers which aren't rationals (duh).

    Whenever, then, we have to do with a cut produced by no rational number, we create a new, an irrational number, which we regard as completely defined by this cut ... . From now on, therefore, to every definite cut there corresponds a definite rational or irrational number ....
    —Richard Dedekind, Continuity and Irrational Numbers, Section IV

[ed boy]

So, uh... Are you saying that the set of real numbers defined upwards from rational numbers and the set of real numbers defined as a subset of complex numbers are not the same set?

Precisely. My mind was blown when I first found out.

Nevermind that you cannot define reals as a set of rationals.

Not really. Reals also includes irrational numbers which aren't rationals (duh).

Yes you can. A rational real number will correspond to a dedekind cut with a minimum element, an irrational number will correspond to a dedekind cut that has a minimum element (or maxiumum, depending on how you define it).

[Darvi]

Whatever, it's 2 am and I'm too tired to try and understand wikipedia.

[Miggy]

So I'm prepping for my linear algebra exam, and I'm preparing for all of the various questions I might end up facing.

Reading through all of my materials, one potential question entered my mind: Schur's decomposition works for complex numbers and unitary matrices. Does it also apply to Real numbers and orthogonal matrices? Looking through the proof and turning the general concepts in my head, I can't see why it shouldn't, but there is no mention of it anywhere, which leads me to believe that it isn't the case.

It's been a long time since I've done this sort of thing, but I think it should fail when the matrix has non-real eigenvalues.  The middle matrix of the decomposition is triangular so those eigenvalues should be down the diagonal, which is a problem if you want real entries.

Hmm... That makes a bit of sense, yes. In order to perform the proof you start by taking an eigenvector and making an orthonormal basis with it. I guess if a real matrix has exclusively complex eigenvectors, the proof falls apart. But then comes the question: Is that actually possible? I'm positively sure I remember something like that being impossible, but looking through my literature again I can't find out where it's mentioned.

Anywho, I subscribe to the idea that if we can perform an action for all complex numbers, we should be able to perform the same action with the real numbers, exactly since we can just imagine the real numbers as complex numbers without any imaginary part. However the literature I'm reading makes long jumps around in order to prove the spectral theorem for normal matrices in the real case, whereas in the complex case it's just a matter of making a simple deduction from Schur's decomposition.

[Another]

...
I guess if a real matrix has exclusively complex eigenvectors, the proof falls apart. But then comes the question: Is that actually possible? I'm positively sure I remember something like that being impossible, but looking through my literature again I can't find out where it's mentioned.
...

I think that this question is easy to answer by demonstration
0 1
-1 0
it's eigenvalues are i and -i and eigenvectors are (-i,1) and (i,1).

Real matrixes can be decomposed into real block-diagonal analog sometimes also called Schur decomposition. I think that you can prove that the blocks on the diagonal will not be larger than 2x2 using the fact that complex roots of equations with real coefficients come in conjugate pairs.

[Christes]

Anywho, I subscribe to the idea that if we can perform an action for all complex numbers, we should be able to perform the same action with the real numbers, exactly since we can just imagine the real numbers as complex numbers without any imaginary part.

Well we can, but the result might have nonreal numbers floating around in it.  The key property here is that the complex numbers are algebraically closed, but the reals are not.

[Vector]

This thread made me happy today.