Rotation matrix arbitrary axis?

[Yoran91]

Hello everyone.

I'm having some trouble with rotation matrices. I'm given three matrices

$J_1, J_2, J_3$

which form a basis for the set of skew-symmetric matrices ($\mathfrak{so}_3$).
Further, the matrix exponent function is such that

$exp[\alpha J_i]=R_i(\alpha)$,

so taking the exponent of the matrix $J_i$ yields precisely the rotation matrix around the i-th standard basisvector, specified by the angle $\alpha$.

Now I wish to show that any rotation matrix $R(n,\alpha)$ specified by an axis (unit vector) $n$ and angle $\alpha$, can be written as

$R(n,\alpha)= exp(\alpha \sum n_i J_i )$.

Here, $n_i$ are the components of the unit vector n.
I know I will need the surjectivity of exp for this. Given the surjectivity of exp, how does one show the above is true?

 

[eljose79]

Hello there,

As a fellow scientist, I understand the difficulties you may be facing with rotation matrices. Let me try to explain how you can prove the given statement using the surjectivity of the matrix exponent function.

Firstly, let's recall that the matrix exponent function is defined as exp(A) = I + A + (1/2)A^2 + (1/6)A^3 + ... + (1/n!)A^n + ... where I is the identity matrix. Since J_1, J_2, and J_3 form a basis for \mathfrak{so}_3, we can express any skew-symmetric matrix A as a linear combination of these matrices, i.e., A = \sum\limits_{i=1}^3 a_iJ_i where a_i are scalar coefficients.

Now, let's consider the rotation matrix R(n,\alpha) specified by an axis n and angle \alpha. We can express this matrix as R(n,\alpha) = I + \alpha A where A is a skew-symmetric matrix. Since n is a unit vector, we can write A = \sum\limits_{i=1}^3 n_iJ_i where n_i are the components of n.

Using the definition of the matrix exponent function, we can write R(n,\alpha) as R(n,\alpha) = exp(\alpha A) = I + \alpha A + (1/2)(\alpha A)^2 + (1/6)(\alpha A)^3 + ... + (1/n!)(\alpha A)^n + ...

Substituting A = \sum\limits_{i=1}^3 n_iJ_i, we get R(n,\alpha) = I + \alpha \sum\limits_{i=1}^3 n_iJ_i + (1/2)(\alpha \sum\limits_{i=1}^3 n_iJ_i)^2 + (1/6)(\alpha \sum\limits_{i=1}^3 n_iJ_i)^3 + ... + (1/n!)(\alpha \sum\limits_{i=1}^3 n_iJ_i)^n + ...

Now, using the properties of skew-symmetric matrices, we can simplify this expression to R(n,\alpha) = I + \alpha \sum\limits_{i=1}^3 n_iJ_i - (