Post History
#2: Post edited
by
Trevor
·
2024-02-22T05:44:24Z (9 months ago)
simplified equation
- I found a solution for the case I am considering, which is when $A$ is a $3{\times}3$ cross product matrix of vector $a \in \mathbb{R}^3$:
- $$A = \begin{bmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{bmatrix}, ~~~~~ a = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} .$$
- Using the explicit equation for the inverse of a $3{\times}3$ matrix, the inverse can be derived as:
$$( I + A )^{-1} = \frac{1}{1+\|a\|^2} (I - A + \|a\|^2 I + A^2) .$$- I'm not sure if this can be extended to more general cases.
- I found a solution for the case I am considering, which is when $A$ is a $3{\times}3$ cross product matrix of vector $a \in \mathbb{R}^3$:
- $$A = \begin{bmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{bmatrix}, ~~~~~ a = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} .$$
- Using the explicit equation for the inverse of a $3{\times}3$ matrix, the inverse can be derived as:
- $$( I + A )^{-1} = \frac{1}{1+\|a\|^2} ((1 + \|a\|^2)I - A + A^2) .$$
- I'm not sure if this can be extended to more general cases.
#1: Initial revision
by
Trevor
·
2024-02-22T04:48:12Z (9 months ago)
I found a solution for the case I am considering, which is when $A$ is a $3{\times}3$ cross product matrix of vector $a \in \mathbb{R}^3$:
$$A = \begin{bmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{bmatrix}, ~~~~~ a = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} .$$
Using the explicit equation for the inverse of a $3{\times}3$ matrix, the inverse can be derived as:
$$( I + A )^{-1} = \frac{1}{1+\|a\|^2} (I - A + \|a\|^2 I + A^2) .$$
I'm not sure if this can be extended to more general cases.