Self-adjoint and eigenvalues properties I wondering about something.
Let $V$ be an inner product space
$T\colon V\to V$ is a linear map
$T$ is self-adjoint and all the eigenvalues of $T$ are not negative
I need to prove that for all $v$ in $V$, $(T(v),v)\ge0$. 
So I think that if all the eigenspaces of all the eigenvalues span $V$ I am done (It is very easy).
But who can promise that? I do not know any theorem like that (am I wrong ?).
So which theorem can I use? All I know is that eigenvectors from different eigenspaces are orthogonal and nothing more.
 A: There is a theorem, that says that self-adjoint maps are allways diagonalizable, so there an eigenbasis and you are done.
Let's sketch the proof: Let $\lambda\in \mathbb C$ be an eigenvalue of $T$, $E_\lambda := \ker(\lambda- T)$ the eigenspace. Then $E_\lambda$ is $T$-invariant, and by self-adjointness, its orthogonal complement $E_\lambda^\bot$ is also: For $v \in E_\lambda^\bot$, $w \in E_\lambda$, we have
$$ (Tv, w) = (v,Tw) = (v,\lambda w) = 0$$
so $Tv \in E_\lambda^\bot$. If we restrict $T$ to $T' \colon E_\lambda^\bot \to E_\lambda^\bot$, we can inductively find a $T$-eigenbasis of $E_\lambda^\bot$ and are done.
A: Since T is self adjoint,then $T=T^{*}$ and for all $v$ we have $0 \leq ||Tv||^2=(Tv,Tv)=(T^{*}Tv,v)=(T^{2}v,v)$
A: Since $T$ is self-adjoint, it has $n$ eigenvectors $\{e_i^j\}$ (i=1,...,n) corresponding to the eigenvalues $\{\lambda_j\}$ (j= 1,..,k) s.t that $\{e_i\}$ forms a orthonormal basis for $V$.
Now assume $\lambda_i >0 \quad \forall i$, and let, $x = \sum_i^n \eta_i e_i \in V$, then
$$(x, T(x)) = \sum_j^n \lambda_j (\sum_u^{r_j} \eta_u)^2$$
where $r_j$ denotes the geometric multiplicity of the eigenvalue $\lambda_j$. 
Since by our hypothesis $\lambda_i >0$, the sum $\sum_i^n \lambda_i (\eta_i)^2$ has to be positive.
