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Given a set of distinct non-real eigenvalues $\lambda_1, \dots, \lambda_N$, so that $\lambda_{2n} = \overline{\lambda_{2n+1}}$. Accordingly given a set of non-real orthonormal eigenvectors $v_1, \dots, v_N$, so that $v_{2n} = \overline{v_{2n+1}}$. (N is even.)

We define $V = [v_1, \dots, v_N]$ and $\Lambda = \textrm{diag}(\lambda_1, \dots, \lambda_N)$.

Is the matrix $V \Lambda V^{-1}$ real?

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First, one minor observation. I believe you meant $v_{2n}=\overline{v_{2n-1}}$ (instead of $\overline{v_{2n+1}}$) otherwise $v_1$ is not conjugated of any vector.

Now, write $v_{2n-1}=a_{2n-1}-ia_{2_n}$ and $v_{2n}=a_{2n-1}+ia_{2_n}$, where $a_{2n-1},a_{2n}\in\mathbb{R}^n$.

Notice tha $a_{2n-1}=\dfrac{v_{2n-1}+v_{2n}}{2}$ and $a_{2n}=\dfrac{i(v_{2n-1}-v_{2n})}{2}$.

Now $V\Lambda V^{-1}a_{2n-1}=\dfrac{\lambda_{2n-1}v_{2n-1}+\lambda_{2n}v_{2n}}{2}=\dfrac{\lambda_{2n-1}v_{2n-1}+\overline{\lambda_{2n-1}v_{2n-1}}}{2}\in\mathbb{R}^n$

$V\Lambda V^{-1}a_{2n}=\dfrac{\lambda_{2n-1}iv_{2n-1}-\lambda_{2n}iv_{2n}}{2}=\dfrac{\lambda_{2n-1}v_{2n-1}+\overline{i\lambda_{2n-1}v_{2n-1}}}{2}\in\mathbb{R}^n$.

Notice that $v_1,\ldots,v_n$ are linear combinations of $a_1,\ldots,a_n$. Thus, $\{a_1,\ldots,a_n\}$ generates $\mathbb{C}^n$. Therefore, they are linear independent over the complex numbers. So they are linear independent over the real numbers and they form a basis of $\mathbb{R}^n$.

Let $e_1,\ldots,e_n$ be the canonical basis of $\mathbb{R}^n$. Thus, $e_i=\sum_{j=1}^n\beta_{ij}a_j$, where $\beta_j\in\mathbb{R}$.

Thus, $V\Lambda V^{-1}e_i=\sum_{j=1}^n\beta_jV\Lambda V^{-1}a_j\in\mathbb{R}^n$.

Finally notice that $V\Lambda V^{-1}e_i$ is the column $i$ of $V\Lambda V^{-1}$. Thus, $V\Lambda V^{-1}$ is a real matrix.

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