$T^2v= \lambda v \implies Tv= \sqrt{\lambda} v$? Let $V$ be  finite dimensional complex vector space, and $T:V\to V$ be a linear operator. Assume that there exists $0\neq v \in V$ such that $T^2v= \lambda v$ where $\lambda \geq 0$ and $T^2 = T \circ T.$

Question: Can we say that $Tv= \sqrt{\lambda}v$? If so, how I should justify?

 A: As we have $T^2\nu = \lambda \nu$ and $\nu\neq0$ we can say that $\lambda$ is an eigenvalue of $T^2$. Hence,$det(T^2-\lambda I) = 0$. Therefore, we have:
$$det(T^2-\lambda I) = det((T-\sqrt{\lambda}I)(T+\sqrt{\lambda}I)) = 0$$
As referenced here we can say:
$$det(T-\sqrt{\lambda}I)det(T+\sqrt{\lambda}I) = 0$$
So, $det(T-\sqrt{\lambda}I) = 0$ or $det(T+\sqrt{\lambda}I) = 0$. It means one of the $\sqrt{\lambda}$ and $-\sqrt{\lambda}$ is at least eigen value of $T$.
Anyhow, you can't say necessarily $\sqrt{\lambda}$ is eigen value of $T$.
Also, If we know all eigen values of $T$ are positive and real, we can say the statement is true.
As mentioned by Jyrki, You should be aware that this analysis about $\lambda$ not about $\nu$ as eigen vector. So, if your question is also on $\nu$ you can't find anything in this analysis.
A: No. Let $T=diag(1,-1) \in \mathbb C^{2 \times 2}$. Then $T^2=I_2$. For
$v=(1,1)^t$ we have $T^2v=1*v$, but $Tv \ne \sqrt{1}v$.
A: The most general (counter-)example: When $\dim V \geq 2$ take any two independent vectors $v$ and $w$
and define $T$ so that
 $Tv = w$ and $Tw=\lambda v$. 
On the complement (if any) of ${\rm Span} \{ v,w\}$ define $T$ as you like. 
You then have a counter-example.
To avoid the occurrence of the counter-example you should distinguish between zero and non-zero values of $\lambda$:
When $\lambda>0$ the linear-combinations $(\sqrt{\lambda}\; v)\pm  w$ will have eigenvalues $\pm \sqrt{\lambda}$. Thus if you know that all eigenvalues have e.g. positive real part this example is excluded and the conclusion holds. 
When $\lambda=0$ the example corresponds to a nil-potent part of the matrix. If you know that the matrix is diagonalizable then this example is excluded and the conclusion holds.
If e.g. you assume $T$ positive semi-definite both of the above examples are excluded and the conclusion holds.
A: No, you can't: the simplest example is a one-dimensional space $V$ and the map $T\colon V\to V$ defined by $T(v)=-v$; of course this extends to any dimension.
On the other hand, if $T$ is a positive definite hermitian operator, then the statement is true, because of the spectral theorem: write
$$
T=\sum_{k=1}^{m}\lambda_k T_k
$$
where $T_k$ are idempotent hermitian and $T_h\circ T_k=0$ for $h\ne k$. The scalars $\lambda_k$ are the eigenvalues of $T$, which are positive. 
Then
$$
T^2=\sum_{k=1}^m\lambda_k^2 T_k
$$
and so $\lambda=\lambda_k^2$ for some $k$, that is, $\lambda_k=\sqrt{\lambda}$. Without loss of generality, we can assume $k=1$. Then $v=T_1(v)$ and $T_k(v)=0$ for $k>1$. Thus
$$
T(v)=\lambda_1T_1(v)=\lambda_1v=\sqrt{\lambda}v
$$
