## Sunday, September 26, 2010

### My first published book

 This year I published my first book Linear Mappings of Free Algebra: First Steps in Noncommutative Linear Algebra Book was published by Lambert Academic Publishing.

## Tuesday, March 9, 2010

Recently playing with complex numbers I discovered that transformation caused by matrix $$\left( \begin{array}{cc} a_0 & -a_1 \\ a_1 & a_0 \end{array} \right)$$ is equivalent to multiplication over complex number $z=a_0+a_1i$. The Caushy-Riemann equation follows from this matrix.

Initially I assumed similar transformation for quaternion. However appropriate matrix for quaternion looks too restrictive. I assumed less restrictive condition for derivative of quaternion function $$\frac {\partial y^0} {\partial x^0} = \frac{\partial y^1}{\partial x^1} = \frac{\partial y^2}{\partial x^2} = \frac{\partial y^3}{\partial x^3}$$ $\frac{\partial y^i}{\partial x^j} =- \frac{\partial y^j}{\partial x^i} \quad i\ne j$ I see that such functions like $y=ax$, $y=xa$, $y=x^2$ satisfy to this equations. The same time conjugation does not satisfy it. A lot of questions should be answered to understand if this is set of function that generalize set of complex functions in case of quaternions. You can find details in my paper eprint arXiv:0909.0855 Quaternion Rhapsody, 2010

## Tuesday, February 2, 2010

### Matrix of Mappings

I recently published new paper
eprint arXiv:1001.4852 The Matrix of Linear Mappings, 2010

In this paper, I consider change in notation of mapping. Instead of notation
y=f(x)
I use notation
y = f ° x
Let us consider ring of mappings of R-algebra A with product f ° g
This point of view allows me to consider ring R as representation in algebra A. If f is additive mapping of algebra A then I can write value of map in form f ° h. In particular I can write derivative of mapping of algebra in form
∂f(x)(h)=∂f(x) ° h
This allow me to separate derivative and increment of argument. In this case I need to use tensor notation for derivative. For instance, if f ° x = x2 then
$$\partial f(x)\circ dx=(x\otimes 1+1\otimes x)\circ dx=x\ dx+dx\ x$$