$f$ symmetric with positive eigenvalues $ \Longleftrightarrow\langle f(v),v\rangle>0~~v\neq0$

Question

I need to demonstrate the previous.

What i said is... ...symmetrical endomorphism operator with all the eigenvalues positive means that he represents an "expansion" of the space $W$. this means that if i write every vector $v$ as:

$$v=\lambda_1 e_1 + \lambda_2 e_2 +...+\lambda_n e_n$$

with $e_1,...e_n$ indipendent vectors from the autospaces of $f$. Then $f(v)$ will be like:

$$f(v)=\Lambda_1e_1+\Lambda_2e_2+...+\Lambda_ne_n$$

But sign of $(e_i,e_i)$ will be $(+,+)$ or $(-,-)$

(because i said that $f$ is an "expansion" of the space $W$!) and $\Lambda_i$ will be all positive numbers!

So the euclidean scalar product of $\Omega(f(v),v)$ will be always $>0$ (if $v$ is different form $0$).

Im wrong? If yes, why?

Answer 1

Since $f$ is symmetric, there is an orthonormal basis $\{e_1,...,e_n\}$ of $W$, consisting of eigenvectors of $f$, such that $f$ has the representation (spectral theorem !):

$f(v)= \sum_{k=1}^n \lambda_k <v,e_k>e_k,$

where

(1) $f(e_k)=\lambda e_k$.

It follows:

(2) $<f(v),v>=\sum_{k=1}^n \lambda_k |<v,e_k>|^2$.

Can you take it from here ?

Answer 2

If $f$ is symmetric with positive eigenvalues then is an orthonormal basis $(e_i)_i$ such that each $e_i$ is an eigenvalue of $f$ and every vector $v$ has a unique the decompostion of the form $$v= \sum_i \langle e_i,v\rangle e_i~~\Longrightarrow f(v)=\sum_i \langle e_i,v\rangle f(e_i) = \sum_i \lambda_i\langle e_i,v\rangle e_i$$

Since $f(e_i) =\lambda_ie_i$

Hence

$$\langle f(v),v\rangle=\Big\langle\sum_i \lambda_i\langle e_i,v\rangle e_i,v\Big\rangle= \sum_i \lambda_i\langle e_i,v\rangle \langle e_i,v\rangle>0 $$

Conversely, if $$\langle f(v),v\rangle>0~~v\neq 0$$

Then , $$0< \langle f(e_i),e_i\rangle= \lambda_i\langle e_i,e_i\rangle =\lambda_i$$

that is $$\lambda_i>0$$