# Intuition for the exponential of a matrix

I'm trying to understand an algorithm that tries to map points from a lie group to its lie algebra using the exponential map. The background is the representation of 3d coordinate transformations as a lie group.

I can imagine, how one can map points on a manifold (eg a sphere) to the tangent space (tangent plane) at some point x on the sphere. I can imagine to create the tangent space by computing derivatives at that point x. But I cannot imagine how the exponential of some matrix is giving me that tangent space? How can I envision the exponential of a matrix? Is there some picture or analogy I can use as intuition?

• The exponential map goes from the Lie algebra to the Lie group, not the other way around... You need to invert the exponential map – co9olguy Mar 30 '15 at 18:24
• And in general, you'd only be able to invert it locally, say in a neighborhood of the identity. – PeterJL Mar 30 '15 at 19:08

A few things that might be helpful:

1) The exponential of a matrix is always invertible.

2) Diagonal matrices are easy to exponentiate: the result is a diagonal matrix whose entries are the exponentials of the original diagonal entries.

3) Fact 2 suggests trying to diagonalize the input matrix before exponentiating it. And in fact: $\mathrm{exp}(gXg^{-1}) = g\mathrm{exp}(X)g^{-1}$.

4) Working with concrete examples is helpful! $G = SO(2)$ is a great place to start.

Also, you can find a really nice investigation of these ideas in Artin's "Algebra".

Good luck!

Let's focus on $G = SL_n(\mathbb{R})$ so we have something concrete in mind. In fact, let's choose $n = 2$ so that $G$ is the space of $2 \times 2$ real matrics whose determinant is $1$. Perhaps the first thing to note is that we can think of $G$ as living inside the space $M_2(\mathbb{R})$ of all $2 \times 2$ matrices (this includes the non-invertible ones too.) And we can clearly identify $M_2(\mathbb{R})$ with $\mathbb{R}^4$ by the map: $$A = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \mapsto (a, b, c, d).$$ The good news here is that we can now think of tangent vectors in terms of ordinary Euclidean space (and elementary vector calculus.)

Now, since $\det(A) = ad - bc$, we can actually think of $G$ as nothing more than the surface sitting inside $\mathbb{R}^4$ defined by $xw - yz = 1$. Cool! This is not much different from thinking of the sphere being defined by the equation $x^2 + y^2 + z^2 + w^2 = 1$. So, if you can find tangent vectors on spheres, then you can find tangent vectors on Lie groups!

In our present setting, we have a very nice way to find tangent vectors: Derivatives of curves. Let $A$ be some point in $SL_2(\mathbb{R})$, which we'll think of as a point $(a,b,c,d) \in \mathbb{R}^4$ with $ad - bc = 1$. If we could find a curve $\gamma(t)$ lying entirely in $SL_2(\mathbb{R})$ with $\gamma(0) = A$, then $\gamma'(0)$ would be a tangent vector based at $A$. And then we could ask: What's the relationship between the tangent vector $\gamma'(0)$ and the value of the exponential map $\exp(\gamma'(0))$?

OK, so before I go any further, is this helpful information? Does it get you closer to an answer to your question?

• Thanks for your help, unfortunately I still don't get the intuition :-/. I hope to get this one day though. What is the connection between exponentiating a matrix (even after diagonalisation) and moving to a tangent space? Is there some intuition the taking the exponent or logarithm is doing something similar than taking derivatives? – mcExchange May 24 '17 at 13:20

When I first learned about cyclic groups, the picture that I always had in my head was of the unit circle in the complex plane -- imagine my shock when I realised it wasn't a cyclic group at all! But I really wanted it to be cyclic, because it shared some really interesting properties with cyclic groups (see my post Intuition, analogies and abstraction).

The solution to the problem can be seen directly from the quickest proof that the unit circle isn't cyclic -- the fact that it isn't countable (while the integers are). So here's an idea: let's admit real powers on groups!

Ok, but how? We know the construction of integer powers on an arbitrary group, and we know how real powers work on the unit circle, or the real line (which is also real-power cyclic*, by the way), and it's conventionally equal to $$x^r=\exp(r\log x)$$ with $$\exp$$ given by its power series expansion.

But sticking just to our intuition for now, it would seem like the natural way to define a real power is to introduce a real-number parameterisation to our group -- for example, the circle group can be parameterised by $$\theta$$ and each element of the group is given by some $$g(\theta)$$. Then real powers would look like $$g(\theta)^r=g(r\theta)$$. In the case of a one-parameter group, we also have $$g(\theta_1+\theta_2)=g(\theta_1)g(\theta_2)$$, but don't get too attached to this.

If you think about it, we've now just given some additional structure to our group -- a geometric structure in addition to the group structure.

But frankly, introducing a parameterisation in this way is a bit hand-wavy. We knew what parameterisation to introduce for the circle group because we already have a picture of its geometry in our heads, but in principle, we could've introduced really any kind of ridiculous parameterisation and given it a really ugly structure and an ugly real-power. What we need is a sensible, systematic way to introduce this parameterisation -- i.e. to think about what this parameter space really is.

The answer to the question comes from Euler's formula, which relates addition on the imaginary line to multiplication on the unit circle.

$$\exp(i\theta)=g(\theta)$$

What significance does the imaginary line have to the unit circle? Well, something interesting is that the tangent to the unit circle at 1 is parallel to the imaginary line, i.e. all its elements are of the form $$1+it$$. So an idea for the parameterisation is that the parameter space is the tangent space at the identity of the group -- this is the Lie algebra of the group.

(You still need to prove that this actually works in general -- this has to do with proving that all derivatives of the exponential map at the identity can be recovered as $$g^{(k)}(0)=(g'(0))^k$$ -- this is a property of exponential functions of the form $$g(t)=e^{bt}$$, and is part of the "exponential structure" of the Lie Algebra/Lie Group correspondence.)

This is not too bad! It's not completely absurd to think about the "vicinity of the identity" of at least matrix groups, so it's not absurd to think about tangent spaces to these groups. This is where you see arguments like $$(1+\varepsilon t)^T(1+\varepsilon t)=1+\varepsilon(t+t^T)$$ implying the tangent space to an Orthogonal Group is an algebra of antisymmetric matrices, etc. -- if you have some notion of perturbing an element in your group, you can construct a Lie algbera parameterisation of it.

*To the best of my knowledge, "real-power cyclic" is not a real word -- the conventional term is "one-parameter Lie group".

See my post Introduction to Lie groups for a more complete treatment.