# Complex eigenvalues of a matrix in conjugate pairs (or not)

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.

• complex eigenvalues of a matrix with real entries come as conjugate pairs – J. W. Tanner Aug 10 at 14:21
• What are you taking about this is for real matrix. – A learner Aug 10 at 14:21

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.
$$Av = \lambda v \implies \bar A \bar v = \bar \lambda \bar v$$ and so $$\lambda$$ eigenvalue of $$A$$ implies $$\bar\lambda$$ eigenvalue of $$\bar A$$.
Thus, when $$A$$ is real, its eigenvalues come in conjugate pairs.