Lemma 5.6. of "Invitation to 3D vision" 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.
 A: 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.
