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 $\lambda = \bar \lambda$?

  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ @LuaiGhunim because $a+bi=a-bi\implies b=-b\implies b=0$ $\endgroup$ – qbert Jan 4 '18 at 21:42

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 $a = \lambda \in \mathbb {R}$


Let's assume

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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.