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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 $ v$ is the complex conjugate. For real vectors it is the usual dot product $(v, w) = v · w$.

The new product has the property $(Av, w) = (v, A^Tw)$ for real matrices $A$ and $(λv, w) = λ(v, w)$ as well as $(v, λw) = λ(v, w).$

Now $λ(v, v) = (λv, v) = (Av, v) = (v, A^T v) = (v, Av) = (v, λv) = λ(v, v)$ shows that $\bar λ = λ$ 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 $\lambda = \bar \lambda$?

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    $\begingroup$ Is your question why $\bar \lambda = \lambda$ implies that $\lambda \in \mathbb R$? $\endgroup$ – Myridium Jan 4 '18 at 21:39
  • $\begingroup$ When proving that $\lambda(v,v)=\overline{\lambda}(v,v)$, you forgot a conjugate in the last equality. $\endgroup$ – C. Falcon Jan 4 '18 at 21:40
  • $\begingroup$ @Myridium yes , $\endgroup$ – Luai Ghunim Jan 4 '18 at 21:41
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    $\begingroup$ @LuaiGhunim because $a+bi=a-bi\implies b=-b\implies b=0$ $\endgroup$ – qbert Jan 4 '18 at 21:42
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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 $a = \lambda \in \mathbb {R}$

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Let's assume

$$\lambda=a+bi\iff \bar\lambda=a-bi$$

thus

$$\lambda=\bar\lambda\iff a+bi=a-bi\iff b=0$$

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