I am trying to understand parts of the authors solution given to the following question:
The generators of $\mathrm{SO}(3)$ can be chosen as $$t^1=\begin{pmatrix}0 & 0 & 0\\ 0 & 0 & -i \\\ 0 & i & 0 \\ \end{pmatrix},\ t^2=\begin{pmatrix}0 & 0 & i\\ 0 & 0 & 0 \\ -i & 0 & 0 \\ \end{pmatrix},\ t^3=\begin{pmatrix}0 & -i & 0\\ i & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}$$ Diagolizing $t^3$, find the $\mathrm{SO}(3)$ group element $\mathrm{R}(\theta)=\exp\left(-i\theta t^3\right)$.
The author's solution is:
To diagonalize $t^3$, we first have to find its eigenvalues by solving $$\det\left(t^3-\mathbb{I}\lambda\right)=\det\begin{pmatrix}-\lambda & -i & 0\\ i & -\lambda & 0 \\ 0 & 0 & -\lambda \\ \end{pmatrix}=(\lambda^2-1)\times(-\lambda)=0$$ The eigenvalues are therefore $\lambda=0$ and $\lambda=\pm 1$. The eigenvector corresponding to $\lambda=0$ is clearly $\left(0,0,1\right)^T$ and the eigenvectors corresponding to $\lambda=\pm 1$ are found by solving $$\begin{pmatrix}0 & -i & 0\\ i & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}\begin{pmatrix}a\\ b \\ 0\\ \end{pmatrix}=\begin{pmatrix}-ib\\ ia \\ 0\\ \end{pmatrix}=\pm \begin{pmatrix}a\\ b \\ 0\\ \end{pmatrix}\tag{1}$$ which gives $$\frac{1}{\sqrt{2}}\left(1,\pm i,0\right)\tag{2}$$ Therefore we can write $$t^3=U^\dagger \hat t U\tag{3}$$ where $$U=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & -i & 0\\ 1 & i & 0 \\ 0 & 0 & \sqrt{2} \\ \end{pmatrix},\tag{4}$$ $$\hat t=\begin{pmatrix}+1 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix},\tag{5}$$ $$[....]$$
I don't need to type any more of the solution as I don't understand eqns. $(2)-(4)$.
Should eqn. $(2)$ actually read $$\frac{1}{\sqrt{2}}\left(1,\pm i,0\right)^\dagger=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\ \mp i \\ 0\\ \end{pmatrix}, \, \text{for} \ \lambda=\pm 1?$$ If true then $t^3$ has eigenvectors $$\frac{1}{\sqrt{2}}\begin{pmatrix}1\\ -i \\ 0\\ \end{pmatrix} \text{if}\,\,\lambda = +1\,\quad\ \frac{1}{\sqrt{2}}\begin{pmatrix}1\\ i \\ 0\\ \end{pmatrix} \text{if}\,\, \lambda = -1,\quad\ \frac{1}{\sqrt{2}}\begin{pmatrix}0\\ 0 \\ \sqrt{2}\\ \end{pmatrix} \text{if}\,\, \lambda = 0\tag{a}$$
Then writing the matrix of eigenvectors, $U$, with the eigenvectors of $(\mathrm{a})$ as columns in the same order as the eigenvalues in $(5)$ should yield $$U=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1 & 0\\ -i & i & 0 \\ 0 & 0 & \sqrt{2} \\ \end{pmatrix}=\begin{pmatrix}1/\sqrt2 & 1/\sqrt2 & 0\\ -i/\sqrt2 & i/\sqrt2 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$$ which is not the same as eqn. $(4)$. In fact, this is the transpose of eqn. $(4)$.
For eqn. $(3)$, $\det U = i$, $U$ is unitary (as the rows are orthonormal), so $U^\dagger=U^{-1}$.
In order to find the inverse of $U$, I first take its transpose $$U^T=\begin{pmatrix}1/\sqrt2 & -i/\sqrt2 & 0\\ 1/\sqrt2 & i/\sqrt2 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}\tag{b}$$ then replacing each element of $(\mathrm{b})$ with its cofactor with the associated signature $\begin{pmatrix}+ & - & +\\ - & + & - \\\ + & - & + \\ \end{pmatrix}$, I find that $$U^{-1}=\frac{1}{\det U}\begin{pmatrix}+\begin{vmatrix}i/\sqrt2 & 0\\ 0 & 1 \\ \end{vmatrix} & -\begin{vmatrix}1/\sqrt2 & 0\\ 0 & 1 \\ \end{vmatrix} & 0\\ -\begin{vmatrix}-i/\sqrt2 & 0\\ 0 & 1 \\ \end{vmatrix} & +\begin{vmatrix}1/\sqrt2 & 0\\ 0 & 1 \\ \end{vmatrix} & 0 \\ 0 & 0 & +\begin{vmatrix}1/\sqrt2 & -i/\sqrt2\\ 1/\sqrt2 & i/\sqrt2 \\ \end{vmatrix} \\ \end{pmatrix}$$ $$=-i\begin{pmatrix}i/\sqrt2 & -1/\sqrt2 & 0\\ i/\sqrt2 & 1/\sqrt2 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}=\frac{1}{\sqrt2}\begin{pmatrix}1 & i & 0\\ 1 & -i & 0 \\ 0 & 0 & \sqrt2 \\ \end{pmatrix}$$
Putting this altogether and omitting the calculation details,
$$U^\dagger \hat t U=\frac12\begin{pmatrix}1 & i & 0\\ 1 & -i & 0 \\ 0 & 0 & \sqrt2 \\ \end{pmatrix}\begin{pmatrix}+1 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}\begin{pmatrix}1 & 1 & 0\\ -i & i & 0 \\ 0 & 0 & \sqrt{2} \\ \end{pmatrix}$$$$=\begin{pmatrix}0 & 1 & 0\\ 1 & 0 & 0 \\\ 0 & 0 & 0 \\ \end{pmatrix}\ne t^3\tag{c}$$
But using the version for $U$ as given in eqn. $(4)$ of the authors solution and again omitting the calculation details,
$$U^\dagger \hat t U=\frac12\begin{pmatrix}1 & 1 & 0\\ i & -i & 0 \\ 0 & 0 & \sqrt2 \\ \end{pmatrix}\begin{pmatrix}+1 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}\begin{pmatrix}1 & -i & 0\\ 1 & i & 0 \\ 0 & 0 & \sqrt{2} \\ \end{pmatrix}$$$$=\begin{pmatrix}0 & -i & 0\\ i & 0 & 0 \\\ 0 & 0 & 0 \\ \end{pmatrix} = t^3\tag{d}$$
Interestingly, if I write (with the omission of the calculation details), $$U \hat t U^\dagger=\frac12\begin{pmatrix}1 & 1 & 0\\ -i & i & 0 \\ 0 & 0 & \sqrt{2} \\ \end{pmatrix}\begin{pmatrix}+1 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}\begin{pmatrix}1 & i & 0\\ 1 & -i & 0 \\ 0 & 0 & \sqrt2 \\ \end{pmatrix}$$$$=\begin{pmatrix}0 & i & 0\\ -i & 0 & 0 \\\ 0 & 0 & 0 \\ \end{pmatrix}=-t^3\tag{e}$$
I've stared at this for sometime now but I just cannot understand why the author's solution, $(\mathrm{d})$, gives the correct result, but my attempt in eqn. $(\mathrm{c})$ $\big(\text{or}\, (\mathrm{e})\big)$ does not.
Can someone please explain where I am going wrong here?
(Sorry for the lengthy post, I've trying to regain some linear algebra skills so needed to show a lot of working).