Aug 29, 2013 · Given two square matrices A, B A, B with same dimension, what conditions will lead to this result? Or what result will this condition lead to? I thought this is a quite simple question, but I can find little information about it. Thanks.

I know that there are three important results when taking the Determinants of Block matrices $$\\begin{align}\\det \\begin{bmatrix} A & B \\\\ 0 & D \\end ...

Jun 23, 2022 · Question: prove Γ(a)Γ(b) = Γ(a + b)B(a, b) Γ (a) Γ (b) = Γ (a + b) B (a, b) using polar transformation: my attempt: L.H.S =

Can there be said anything about det(A + B)? If A/B are symmetric (or maybe even of the form λI) - can then things be said?

Jan 11, 2015 · there is a similar thread here Coordinate-free proof of $\operatorname {Tr} (AB)=\operatorname {Tr} (BA)$?, but I'm only looking for a simple linear algebra proof.

Personally, I think the NFA construction for union is straight-forward. It also leads to smaller automata (which blow up while determinising, of course). "I can easily verify my own answer by some arbitrary inputs" -- this statement is wrong, but equally valid for both DFA and NFA. The only difference is that you might have to check many runs in the NFA as opposed to one run in …

Given two square matrices $A$ and $B$, how do you show that $$\det (AB) = \det (A) \det (B)$$ where $\det (\cdot)$ is the determinant of the matrix?

As a remark, it is actually legitimate to assume that A A and B B are simultaneously diagonalisable (surprise, surprise!), so the proposition is trivial. But obviously, the reason why we can make such an assumption is way beyond the scope of undergraduate (or even graduate) linear algebra courses.

Mar 28, 2017 · We now have a word u u such that w = au w = a u and u u has one more b b than it has a a 's. Hence it can be written as s1bs2 s 1 b s 2 in such a way that each si s i has the same number of a a 's and b b 's (and is possibly empty). Both s1 s 1 and s2 s 2 are strictly shorter than w w; hence the induction hypothesis applies to each si s i and we are done.

First of all, am I being crazy in thinking that if λ is an eigenvalue of AB, where A and B are both N × N matrices (not necessarily invertible), then λ is also an eigenvalue of BA? If it's not true, then under what conditions is it true or not true? If it is true, can anyone point me to a citation? I couldn't find it in a quick perusal of Horn & Johnson. I have seen a couple proofs that the ...