Prove that, for any complex matrix $A$ of order $3$, there exists a unitary matrix $U$ of order $3$, such that $UAU^{-1}$ is a matrix of the form $\left( \begin{matrix} *& 0& *\\ *& *& 0\\ *& 0& *\\ \end{matrix} \right) $.
We know unitary matrix is a matrix that satisfies $UU^*=U^*U=I$. However, I think this is far from the claim above, and I don't know how to proceed further.
$A$ has some eigenvector $v_1$. Its perpendicular space $v_1^\perp$ has dimension $2$.
If $\dim(Av_1^\perp)=2$, then pick any non-zero vector $w\in v_1^\perp\cap Av_1^\perp$ so that $w=Av_2$ for some $v_2\ne0$.
If $\dim(Av_1^\perp)\le1$ then pick $v_2\in v_1^\perp$ such that $Av_2=0$.
In both cases, $v_2\in v_1^\perp$ and $Av_2\in v_1^\perp$.
Finally, add a final vector $v_3$ perpendicular to $v_1,v_2$, and normalize all vectors, so that $U:=[v_3,v_1,v_2]^*$ is unitary. Thus $U^*j=v_1$, $U^*k=v_2$, etc., and $$UAU^*=[v_3,v_1,v_2]^{-1}A[v_3,v_1,v_2]$$ is of the required form since \begin{align}UAU^*j&=UAv_1=\lambda Uv_1=\begin{pmatrix}0\\*\\0\end{pmatrix}\\UAU^*k&=UAv_2=U(\alpha v_2+\beta v_3)=\begin{pmatrix}*\\0\\*\end{pmatrix}.\end{align}