exp(A) the exponential function of a matrix
In a previous post I was going thru a section of marsden
in that pages 283-292 part of chapter 9, and it mentioned
the exponential function defined by the power series
exp(x) = 1 + x + x
2/2! +...
and gave a case of where you plug a matrix A in for x
and get a matrix exp(A)
this has always seemed to me like a cool thing to do
and I see it as illustrating a kind of umbilical connection between Lie algebra and Lie group.
The algebra element A is what gets plugged into exp() to give exp(A) which is in the group.
Or in more cagey differential geometry style----exp(tA) for t running from 0 to 1 gives you a curve down in the Lie group (the manifold) which starts out at the identity point and travels along in the manifold and reaches exp(A) as its destination. Indeed exp(tA) for t running forever gives a one-dimensional subgroup--but this is a bit too abstract for this time of morning.
What I always think is so great is that if A is 3x3 skew sym
matrix, meaning A
T = -A
then plugging A into that good old exp() power series gives a rotation matrix, one of the SO(3) Lie group.
More wonderful still, exp(A) is the rotation by exactly |v| radians about the vector v = (v1, v2, v3) as axis where A is
given by
+0 -v3 +v2
+v3 +0 -v1
-v2 +v1 +0
any skew symmetric matrix would have such a form for some
v1,v2,v3
And we may be able to convince ourselves of this, or prove it a bit, without much effort, just by looking at the power series in A.
If I stands for the identity matrix,
B = exp(A) = I + A + A
2/2! +...
Now consider that since A
T = - A, we can take the transpose of this whole power series and it will be as if we put a minus sign in front of A.
B
T = exp(A)
T = exp(- A)
But multiplying exp(x) and exp( -x) always gives one. When you multiply the two power series there is a bunch of cancelation and it boils down to the identity. So exp (-A) is the matrix INVERSE of exp(A).
B
T = exp(A)
T = exp(- A) = exp(A)
-1 = B
-1
B
T = B
-1 means that B is orthogonal
BTW one reason to think about paths exp(tA) from the identity to the endpoint exp(A) is to see clearly that exp(A) is in the same connected component of the group. O(3) is split into two pieces, one with det = 1 and one with det = -1.
The latter kind turn your shirt inside out as well as rotating it, so they are bad mothers and it is generally safer to work with the det = 1 kind which are called "special" or SO(3).
this curve going t = 0 to 1 shows that exp(A) is in the same connected component as the identity, because how could the curve ever leap the chasm between the two components?
So it shows det A = 1. But that is just mathematical monkeyshines, of course the determinant is one!
All this stuff can be written with an n sitting in for 3, but
as an inveterate skeptic I often suspect that
dimensions higher than 3 do not exist and prefer to write 3 instead of n. It looks, somehow, more definite and specific that way.
We should check out the elementary fact that [A,B] works with
skew sym matrices A and B! Why not! Maybe later today, unless someone else has already done it.
I will bring along this earlier post with an extract from pages 289-291 of the book
**************************************
SO(3) is a compact Lie group of dimension 3.
Its Lie algebra so(3) is the space of real skew-symmetric 3x3 matrices
with bracket [A,B] = AB - BA.
The Lie algebra so(3) can be identified with R
3
the 3-tuples of real numbers by a vectorspace isomorphism
called the"hat map"
v = (v1,v2,v3) goes to v-hat, which is a skew-symmetric matrix
meaning its transpose its its NEGATIVE, and you just stash the three numbers into such a matrix like:
+0 -v3 +v2
+v3 +0 -v1
-v2 +v1 +0
v-hat is a matrix and apply it to any vector w and
you get vxw.
Everybody in freshman year got to play with v x w
the cross product of real 3D vectors
and R
3 with ordinary vector addition and cross product v x w is kind of the ancestral Lie algebra from whence all the others came.
And the hat-map is a Lie algebra isomorphism
EULER'S THEOREM
Every element A in SO(3) not equal to the identity is a rotation
thru an angle φ about an axis w.
SO SO(3) IS JUST THE WAYS YOU CAN TURN A BALL---it is the group of rotations
THE EIGENVALUE LEMMA is that if A is in SO(3) one of its
eigenvalues has to be equal to 1.
The proof is just to look at the characteristic polynomial which is of degree three and consider cases.
Proof of Euler is just to look at the eigenvector with eigenvalue one----pssst! it is the axis of the rotation. Marsden takes three sentences to prove it.
A CANONICAL MATRIX FORM to write elements of SO(3) in
is
+1 +000 +000
+0 +cosφ -sinφ
+0 +sinφ cosφ
For typography I have to write 0 as +000
to leave space for the cosine and sine under it
maybe someone knows how to write handsomer matrices?
EXPONENTIAL MAP
Let t be a number and w be a vector in R
3
Let |w| be the norm of w (sqrt sum of squares)
Let w^ be w-hat, the hat-map image of w in so(3), the Lie algebra. Then:
exp(tw^) is a rotation about axis w by angle t|w|
It is just a recipe to cook up a matrix giving any amount of rotation around any axis you want.
