I am having a little bit trouble while understanding this theorem.
It says that: A symmetric matrix has real eigenvalues.
Proof. Extend the dot product to complex vectors by $(v, w) = \sum_i \bar {v_i}{w_i}$, where $\bar {v_i}$ is the complex conjugate of $v_i$. For real vectors it is the usual dot product $(v, w) = v \cdot w$.
The new product has the property $(Av, w) = \left(v, A^{\mathrm T}w\right)$ for real matrices $A$ and $(\lambda v, w) = \lambda(v, w)$ as well as $(v, \lambda w) = \lambda(v, w)$.
Now $\lambda(v, v) = (\lambda v, v) = (Av, v) = \left(v, A^{\mathrm T} v\right) = (v, Av) = (v, \lambda v) = \lambda(v, v)$ shows that $\bar \lambda = \lambda$ because $(v, v) \neq 0$ for $v \neq 0$.
How do get this expression ($ \lambda <v,w>$) because $\lambda$ is an eigenvalue if $AX = \lambda X$ and $(A-\lambda\mathbb{I}) = 0$?
Or in other words, how do we go from $(A-\lambda\mathbb{I}) = 0$ to $ \lambda <v,w>$?
How does this prove that $\lambda$ is real by showing that $\lambda = \bar \lambda$?
Best Answer
Write $\lambda = a + bi $ where $a,b $ are real numbers. If $\lambda = \overline {\lambda} $, then $a + bi = a - bi $. Hence $2bi =0$ and $b =0$. This shows that $a = \lambda \in \mathbb {R}$