Finding a basis so that a Matrix is real-valued

Question

Let $V = \mathbb C^4 , \phi_2\in \text{End(V)}$ $$D_S(\phi_2) = \begin{pmatrix} i &0&0&0\\ 1 &i & 1 &1 \\ 0 & 0&i&2i \\ 0 &0&0&-i \end{pmatrix}$$

Is there a Basis T, so that for $D_T(\phi_2) : a_{ij}\in \mathbb R $

I know that $D_T(\phi_2) = T^{-1} \cdot D_S(\phi_2) \cdot T \Longleftrightarrow D_T(\phi_2) \cdot T^{-1}= T^{-1} \cdot D_S(\phi_2)$

So let $\quad T^{-1} = \begin{pmatrix} a_1 &a_2&a_3&a_4\\ a_5 &a_6 & a_7 &a_8 \\ a_9 & a_{10}&a_{11}&a_{12} \\ a_{13} &a_{14}&a_{15}&a_{16} \end{pmatrix}$ . After multiplying it with $D_S(\phi_2)$ it follows $a_i = 0 \quad \forall i$. So there isnt such a basis.

My question is: Is there a faster way to prove this?

Answer 1

Yes, there is a faster way.

1. The trace of a matrix is invariant of the basis. (This follows from $tr(AB) = tr(BA)$.)
2. Then note that the trace of your matrix is $3i$. This is impossible if the entries where real.