No.
In my question A\dagger was conjugate transpose and conditions 3 and 4 in Penrose's definition are just the requirement that AX and XA both should be hermitian, where X is the generalized inverse of A (in my case which is A\dagger and so 3-4 follows trivially).
Hi,
Does the equation AA^\dagger=I force A^\dagger to be the generalized inverse of A? That is: AA^\dagger=I\Rightarrow A^\dagger\text{ is the generalized inverse of } A? A is any rectangular matrix over the field of complex numbers. It is very easy to verify the first three properties, but...
Hello mscudder3,
Take any unit column vector a (i.e., nx1 matrix with sum of square of moduli of its elements being 1, where n is the number of your choice). Then the matrix m=aa^\dagger will satisfy m^2=m.
Well, I thought (and still think) Landau is a specialist in Quantum Mechanics (which is sometimes called the mathematical modelling of nature and uses lots of linear operators:P). Particularly, the whole matrix theory is a basic tool in finite dimensional QM and its modern branch Quantum...
Yes, your explanation is correct too.
By the way, these are very much elementary results and there is no confusion that these can be found in all textbooks of matrix theory/linear algebra :)
Surely I am missing some simple points, but what about say [.11, .12]? Every real within this interval must start with 1 and so does not contain any such number?
I don't know what is the function of JCF in it...it simply follows from the well known Caley-Hamilton Theorem (Every square matrix satisfies its own characteristic equation) and the result holds for any square matrix.
Oh, finally I got an answer to this question...the proof is a handy one (according to me, of course!:smile:). The answer is indeed correct.
Since \cos\theta_1\cos\theta_2\sin\theta_1\sin\theta_2\ne0, \cos\theta_1,\cos\theta_2,\sin\theta_1,\sin\theta_2<1.
so...
Plotting in mathematica with different values, I can not find an exception.
But the reasoning of closed, compact domain can not be applied, because though the domain is convex, the function is itself not a convex one. However, we can split the domain such that in each part, the function...