I'm reading through this book. Which is full of linear algebra, and there're some aspects I'm not quite familiar with. Lemma 5.6. states the following:
Here given $T \in \mathbb{R}^3$ the hat operator is defined as
$$ \widehat{T} = \left( \begin{array}{lll} 0 & -t_z & t_y \\ t_z & 0 & -t_x \\ -t_y & t_x & 0 \end{array} \right) $$
The proof is below:
I don't understand rigorously the following snippet:
My attempt in understanding why is based on the exponential matrix, firstly because of eigevalues/eigenvector argument we have
$$ e^{\widehat{\omega} \theta} \widehat{T} \omega = \widehat{T} \omega $$
we can represent the exponential as $$ e^{\widehat{\omega} \theta} = I + \sin (\theta) \widehat{\omega} + (1 - \cos(\theta) \widehat{\omega}^2 \Rightarrow e^{\widehat{\omega} \theta} \widehat{T}\omega = \widehat{T}\omega + \left( \sin (\theta) \widehat{\omega} + (1 - \cos(\theta) \widehat{\omega}^2 \right) \widehat{T}\omega \Rightarrow e^{\widehat{\omega} \theta} \widehat{T}\omega - \widehat{T}\omega = \left( \sin (\theta) \widehat{\omega} + (1 - \cos(\theta) \widehat{\omega}^2 \right) \widehat{T}\omega $$
Therefore I have the equality
$$ \left( \sin (\theta) \widehat{\omega} + (1 - \cos(\theta) \widehat{\omega}^2 \right) \widehat{T}\omega = 0 $$
From here I got stuck I don't know what to do to infer that $\widehat{T}\omega = 0$
Any thoughts about it? I would much appreciate your help.
The key is that the eigenspace of $e^{\widehat\omega\theta}$ corresponding to $1$ is one-dimensional and spanned by $\omega$. The equation $e^{\widehat\omega\theta}\widehat T\omega=\widehat T\omega$ says that either $\widehat T\omega$ is an eigenvector of $e^{\widehat\omega\theta}$ with eigenvalue $1$ or is equal to $0$, which means that $\widehat T\omega$ must be a scalar multiple of $\omega$. On the other hand, you know from the properties of $\widehat T$ that $\widehat T\omega$ is orthogonal to $\omega$, but the only scalar multiple of any vector $v$ that is orthogonal to $v$ is $0$: $cv\cdot v=c\,(v\cdot v)=0$ iff $c=0 \lor v=0$.
Your approach works, too. Picking up where you left off, observe that for all $v$, $\widehat\omega v\perp\widehat\omega^2v$. Thus, the expression $$\sin(\theta)\widehat\omega\widehat T\omega+(1-\cos\theta)\widehat\omega^2\widehat T\omega$$ is a linear combination of two orthogonal vectors. For this to be zero, either $\sin\theta=1-\cos\theta=0$, or $\widehat T\omega=0$. The former possibility is eliminated by hypothesis.