@r9m I just present them the reality. Some are shocked to realize they aren't able to solve a problem for kids. I'm not aggressive, but I hate the way they try to show off.
Some are decent, and simply ask good questions you'd love to answer.
@Chris'ssis So you squeal interviewers during interviews ? :P .. thought its the task of interviewers to squeal the candidates .. not the other way round :P ..
Hi @DanielFischer! May I havea minute of your time? If so, that's my question: Am I possible to define a function that given an $n\times n$ matrix (let it be symmetric, positive-definite) it returns the orthogonal matrix after applying SVD to it? I mean, I would like to present it in a rigorous way, so, could I find a function, or operator, or something? Thanks a lot!
:16527700 Yeah, but I think you didn't get my point. I wanted to say that people should accept you're right when you're right and when you can prove this. You cannot say you're wrong just because you have a discussion with your boss.
@nullgeppetto I'm not sure which orthogonal matrix you mean. The orthogonal matrices in an SVD as well as in an orthogonal diagonalization of a symmetric matrix are not unique. So you'd have to make a somewhat arbitrary choice. But in principle, you can make that choice and define such a function. You can even describe such a function explicitly for invertible symmetric matrices $A$. Then $A^TA$ has a unique positive definite square root, and $\sqrt{A^TA}^{-1}A$ is orthogonal.
However, computing $\sqrt{A^TA}$ is not entirely trivial.
@DanielFischer, thanks! I mean that, if $A$ is a symmetric positive definite $n\times n$ matrix, then its SVD is $A=VD^TV$, where $D=\text{diag}(\lambda_1,\ldots,\lambda_n)$ is the diagonal matrix containing the positive eigenvalues of $A$, and $V$ is an orthogonal matrix. Isn't $V$ unique?
@nullgeppetto That is $VD({}^TV)$, is it? Then the columns of $V$ are eigenvectors of $A$. If all eigenvalues of $A$ are distinct, then you only can multiply any subset of the columns by $-1$. But if $A$ has eigenspaces of dimension greater than $1$, there is more freedom in the choice of $V$.
@nullgeppetto You can define a function that returns one possible $V$. But that involves an arbitrary choice (that's not necessarily a bad thing). Mathematically, you can also define a function returning the set of all possible $V$.
@PedroTamaroff hint: the norm can only take certain values. so, as last time, if N(x) is the product of two rationally prime values N() can't take, then x must be irreducible.
@robjohn Is it okay(by community rules) If I put sth like 'I'll reward a +100 bounty for a nice answer' in a question that I asked (without actually putting the bounty .. coz it expires in 7 days and I loose the points for nothing :P ) ? :D
@r9m There was a meta question about this, and I think the consensus was that circumventing the bounty system was not good. Let me look for the thread.
