I know that for a real-valued matrix, complex eigenvalues come in complex conjugate pairs. However, I'm wondering what happens for repeated complex eigenvalues (i.e. complex eigenvalues with multiplicity greater than 1). In that case, does the complex conjugate of the repeated complex eigenvalue have the same multiplicity as that eigenvalue? If that statement holds, how can we show that it's true?

  • $\begingroup$ Yes, the complex conjugate has the same multiplicity $\endgroup$ Aug 19, 2020 at 8:30

1 Answer 1


We indeed find that if a real matrix $A$ has a complex eigenvalue $\lambda$, then the conjugate eigenvalue $\bar \lambda$ has the same algebraic and geometric multiplicity. In fact, we can say a bit more: $$ (A - \lambda I)|_{\ker(A - \lambda I)} ``=" (A - \bar \lambda I)|_{\ker(A - \bar \lambda I)}, $$ which is to say that all structures associated with the eigenvalues are the same. That is, the Jordan form of $A$ has the same number and sizes of blocks for $\lambda$ and $\bar \lambda$.

As far as proving multiplicities goes, we have the following: the algebraic multiplicity is the multiplicity of the root $\lambda$ in the characteristic polynomial $p(x) = \det(xI - A)$. As is true for any polynomial with real coefficients, the multiplicity of the root $\lambda$ is this the same as the multiplicity of the root $\bar \lambda$.

For geometric multiplicity, one approach is as follows: we note that matrices satisfy $\overline{A B} = \bar A \bar B$. It follows that if $v$ is a (complex) eigenvector associated with eigenvalue $\lambda$, then we have $$ Av = \lambda v \implies \overline{Av} = \overline{\lambda v} \implies A \bar v = \bar \lambda \bar v. $$ In other words, the map $v \mapsto \bar v$ is an invertible $\Bbb R$-linear map between the eigenspaces of $A$ associated with $\lambda$ and $\bar \lambda$. It follows that these spaces have the same dimension.


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