Recent content by D.K.

Well, that's quite surprising to hear. Would you be so kind as to recommend a quantum mechanic textbook that in your opinion is the best when it comes to math rigor?

So, I am about to read Landau's and Lifschitz's textbook on Quantum Mechanics. What kind of mathematics I should be already familiar with in order to completely understand the above mentioned material? Thanks for all the advice.

So, I am about to read Landau's and Lifschitz's textbook on Classical Mechanics. What kind of mathematics I should be already familiar with in order to completely understand the above mentioned material? Would real-variable calculus and linear algebra be sufficient for the task? Thanks for all...

Nevermind, it turned out to be rather easy.

Is there an easy way to prove that for any irrational \xi the set: \{x \in \mathbb{R}: x = p + q\xi, \ p, q \in \mathbb{Z}\} is dense in \mathbb{R}? I know a proof involving notions from measure theory of which I unfortunately know nothing about. Any help would be very appreciated.

I found it easy to prove: null(A) + null(B) >= null(AB), using maps instead of matrices. Indeed, take /phi_A, /phi_B and /phi_AB defined by matrices A, B and AB respectively. If we restrict the domain of /phi_AB to just those vectors X of which images \phi_B(X) lies in the Ker(A), it's...

For sure the inequality holds: max(null(A), null(B)) <= null(AB). Oh, and null(A) + null(B) >= null(AB) turns out to be simply equivalent to the statement we were trying to prove from the beginning.

Would you be so kind as to explain how did you get Rank (AB) + Null(AB) – Rank(B) >= Rank(B) – Rank(B) + Null(AB)? The rest - including Rank(AB) <= min(Rank(A),Rank(B)) - is perfectly clear to me.

Sorry, I've meant to write it the way you did. I'm aware that it'll do the trick but don't know why is it true.

Well, I see a way to use the rank-nullity theorem here successfully provided it is known that: for any A, B, AB it is the case that: nullity A + nullity B <= nullity AB. But is the above true?

I'm afraid that doesn't suffice. For suppose that m <= n < = s, e.g. m = 4, n = 4, s = 5. Then: rank(A) <= 4, rank(B) <= 4, rank(AB) <= 4. That, however, doesn't make rule out as impossible following: rank(A) = 4, rank(B) = 4, rank(AB) = 2, which would make the statement...

Homework Statement Prove that for any m x s matrix A and any s x n matrix B it holds that: rank(A) + rank(B) - s is less or equal to: rank(AB) The Attempt at a Solution Obviously, the following are true: - rank(A) is less or equal to s, - rank(B) is less or equal to s, -...