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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!

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  • $\begingroup$ Assuming that you mean $j^2 = -1$ and $\sigma, \omega$ real in $\Lambda_1,$ you have a matrix with trace $2 \sigma$ and determinant $\sigma^2 + \omega^2.$ Can you guess a matrix of reals with matching trace and determinant? $\endgroup$ – Will Jagy Nov 16 '15 at 21:26
  • $\begingroup$ @WillJagy Yes, it's $$ \begin{bmatrix} \sigma & -\omega \\ \omega & \sigma \end{bmatrix} $$, what's the idea? Thanks for helping! $\endgroup$ – space_voyager Nov 16 '15 at 21:34
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    $\begingroup$ good. This is similar to your $\Lambda_1,$ call this one $A,$ there is a complex matrix $P$ with $P^{-1} A P = \lambda_1;$ uses eigenvalues, eigenvectors, but a certain amount of care and checking correctness. Off to grocery store $\endgroup$ – Will Jagy Nov 16 '15 at 21:39
  • $\begingroup$ @WillJagy I think I got it, tell me if this is correct: Basis: $\{\frac{j}{2}v-\frac{j}{2}v^*,\frac{1}{2}v+\frac{1}{2}v^*,v_3,...,v_n\}$ Transformation matrix: $$ T=\begin{bmatrix} P & 0 \\ 0 & I \end{bmatrix} $$ Where $P=\begin{bmatrix}-j & j\\ 1 & 1\end{bmatrix}$ and $I$ is the identity matrix $\in\mathbb R^{(n-2)\times(n-2)}$ $\endgroup$ – space_voyager Nov 16 '15 at 22:12

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