You work out its kernel. If it's a map from V to V then you can work out its determinant which tells you if the kernel is zero or not (but not what it is).
A function, f, in general, is injective if f(x)= f(y) implies x= y. If f is linear, then f(x)= f(y) gives f(x)- f(y)= f(x-y)= 0 while x= y is the same as x- y= 0. That is why a linear function is injective if and only if its kernel is trivial: {0}, as matt grime said.
Are there known conditions under which a Markov Chain is also a Martingale? I know only that the only Random Walk that is a Martingale is the symmetric one, i.e., p= 1-p =1/2.
Hello !
I derived equations of stress tensor 2D transformation.
Some details: I have plane ABCD in two cases (see top on the pic) and I know tensor components for case 1 only. Only plane ABCD rotate in two cases (top of the picture) but not coordinate system. Coordinate system rotates only on the bottom of picture.
I want to obtain expression that connects tensor for case 1 and tensor for case 2.
My attempt:
Are these equations correct? Is there more easier expression for stress tensor...