# Why is the conjugate of an eigenpair also an eigenpair

This is an exercise on the book fundamentals of matrix computations 1st. edition.

It asks to show that for $$A \in R^{nxn}$$, if $$(\lambda, u)$$ eigenpair, then $$(\overline{\lambda}, \overline{u})$$ is also an eigenpair.

• It helps to note that for $A,B \in \Bbb C^{n \times n}$, we have $\overline{AB} = \bar A \bar B$ (where $\bar A$ denotes the conjugate of $A$). – Omnomnomnom Nov 1 '18 at 16:50
• @Omnomnomnom I am sorry, could you give me some more? I am staring at it, but I do not see it still. – MTLaurentys Nov 1 '18 at 17:26
• If $Au = \lambda u$ then $\overline{A} \overline{u} = \overline{A u} = ?$ – Connor Harris Nov 1 '18 at 17:49