# A Hermitian matrix $(\textbf{A}^\ast = \textbf{A})$ has only real eigenvalues - Proof Strategy [Lay P397 Thm 7.1.3c]

Would someone please explain the proof strategy at Need verification - Prove a Hermitian matrix $(\textbf{A}^\ast = \textbf{A})$ has only real eigenvalues? I brook the algebra so I'm not asking about formal arguments or proofs. For example, $1.$ How would you determine/divine/previse to take the Hermitian conjugate and to right-multiply by $\color{orangered}{\vec{v}}$?

$2.$ Since we are given that $A$ is Hermitian and has eigenvalues, why not start the proof with $A^*\mathbf{v} = \lambda^*\mathbf{v}$? Here, $\mathbf{v}$ is an eigenvector and so by definition $\neq \mathbf{0}$.

Then \begin{align} LHS = Av & = \\ \lambda v & = \end{align}

$\iff \lambda \mathbf{v} = \lambda^*\mathbf{v} \iff \mathbf{0} = (\lambda^* - \lambda )\mathbf{v} \iff \mathbf{v} \neq \mathbf{0}, so \, (\lambda^* - \lambda )=0.$

• ..."I brook the algebra"? – Omnomnomnom Apr 24 '14 at 16:03

Intuitively, it comes down to the fact that we need to prove the fact that $\lambda \in \mathbb{R}$ using the facts that $A = A^*$ and $Av = \lambda v$ for some vector $v \neq 0$. In order to bring $A^*$ into play, we have to take the Hermitian conjugate of both sides at some point.
As for multiplying by $v$: well, once you see $A^*$, you note that this is equal to $A$; there's one step in the right direction. In order to get from "talking about $A$" to "talking about $\lambda$", we need to multiply by $v$ on the right.
It is not generally the case that for a matrix $A$ with eigen-pair $\lambda,v$ $$A^*v = \lambda^*v \tag{NOT GENERALLY TRUE}$$ While we can say that $\lambda^*$ must be an eigenvalue of $A$, the associated eigenvector for $A^*$ is not related to the associated eigenvector for $A$.