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I have learnt that in a matrix, if there are complex eigenvalues, they should come as conjugate pairs. Also, I know that, in a diagonal matrix, eigenvalues are the diagonal elements.

So how about the following matrix?

$$\begin{pmatrix} i & 0\\ 0& 2 \end{pmatrix}$$

Shouldn't the eigenvalues be $i$ and $2$, where it doesn't have a conjugate pair?!

I appreciate your help to clarify my mistake.

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    $\begingroup$ complex eigenvalues of a matrix with real entries come as conjugate pairs $\endgroup$ – J. W. Tanner Aug 10 at 14:21
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    $\begingroup$ What are you taking about this is for real matrix. $\endgroup$ – A learner Aug 10 at 14:21
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Recall that the eigenvalues of a matrix $A$ are the zeroes of its characteristic polynomial $\chi_A(x) = \det (x I - A)$. Of course it is entirely possible for the roots of $\chi_A$ to not occur in pairs of complex conjugates as shown by your example. However, if we restrict the coefficients of $\chi_A$ to be real (e.g. if your matrix $A$ is real) then we will find that any complex roots occur in pairs of conjugates by the complex conjugate root theorem.

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Complex eigenvalues of matrices with real entries come as conjugate pairs.

This is not necessarily the case for matrices with complex entries.

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$Av = \lambda v \implies \bar A \bar v = \bar \lambda \bar v $ and so $\lambda$ eigenvalue of $A$ implies $\bar\lambda$ eigenvalue of $\bar A$.

Thus, when $A$ is real, its eigenvalues come in conjugate pairs.

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