Change of basis help turning a sub-matrix from complex to real

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I have the following matrix:

$$ \Lambda=\begin{bmatrix} \Lambda_1 \\ & \Lambda_2 \\ & & \ddots \\ & & & \Lambda_m \end{bmatrix}\in\mathbb R^{n\times n} $$

Where $\Lambda_1\in\mathbb C^{2\times 2}$ equals the following:

$$ \Lambda_1=\begin{bmatrix} \sigma+j\omega & 0 \\ 0 & \sigma-j\omega \end{bmatrix} $$

Note that $\Lambda$ is the Jordan decomposition of matrix $A\in\mathbb R^{n\times n}$ with respect to the basis $\{v,v^*,v_3,...,v_n\}$ where $v=x+jy\in\mathbb C^n$ and $v^*=x-jy\in\mathbb C^n$ are two eigenvectors of $A$ associated with $\lambda=\sigma+j\omega$ and $\lambda^*=\sigma-j\omega$ respectively.

Question: I need to find transformation matrix, $T$, into a new basis (specify what basis it is, too) in which:

$$ T\Lambda T^{-1}=\begin{bmatrix} \widetilde\Lambda_1 \\ & \Lambda_2 \\ & & \ddots \\ & & & \Lambda_m \end{bmatrix}\in\mathbb R^{n\times n} $$

Where $\widetilde\Lambda_1\in\mathbb R^{2\times 2}$.

The trouble is, I am unable to find this new basis, nor the transformation matrix for it. My current thinking is that the new basis should be:

$$ \left\{ \begin{bmatrix} \sigma+j\omega \\ 0 \\ \vdots \\ 0 \end{bmatrix}\in\mathbb C^n, \begin{bmatrix} 0 \\ \sigma-j\omega \\ \vdots \\ 0 \end{bmatrix}\in\mathbb C^n,v_3,...,v_n \right\} $$

Which would give $\widetilde A_1=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$, however at that point (assuming the basis is correct), I am unable to find the change of basis matrix... Thanks for helping me out with this!