OK, boss. Umm, let's see. Let me start my justification for including the Lie groups here with a
Definition; A Lie group is a differentiable manifold with the algebraic structure of a group.
Yes, quite, not very illuminating, is it? So what follows, initially, is to guide our intuition
only, and should not be taken too seriously (we'll get pretty abstract soon enough!).
So, Lie groups describe natural symmetries. Suppose, for example, I hand you an object, and you want to discover if it is symmetrical. You have two options, right? You can rotate it around in all available dimensions, making sure you don't blink (smooth operation), or you can walk around it in all available directions, again completely smoothly. And both these operations must be reversible (invertible) right?
So the act of rotating the object, or, equivalently, walking around it, are referred to as
linear transformations, for which I shall give a definition shortly. But first note that, "shape symmetry" may not be all that I am interested in, so there is a class of such transformations, each of which give you information about different sorts of continuous symmetries.
I guess I can turn this around, sort of; there is a class of smooth linear transformations, each of which preserves some symmetry properties of some abstract object. Each set in this class has the algebraic structure of a group, which in due course, I'll explain.
So, enough intuition, let's get dirty.
Let $$ V$$ be a
vector space over the field $$\mathbb{F}$$. I define the gadget $$A:V \to V$$ to be a
linear transformation aka linear operator on $$V$$ iff, for all
generic $$v,\;w \in V$$ and any $$\alpha,\;\beta \in \mathbb{F}$$ that $$ Av = w$$ and
$$A(\alpha v + \beta w) = \alpha Av + \beta Aw$$.
(Note that, since $$v,\;w$$ are discrete elements in $$V$$, in general $$A$$ is not a smooth transformation, that is of class $$C^{\infty}$$; shortly I shall insist on this).
Hmm, enough for now; what's next? Oh yes, groups I think