Normal matrices Does $\|Tv\|\leq\|v\|$ (for all $v \in V$) leads to $T$ is normal?
If not, when I add the additional information that every e.e of $T$ is of the absolute value 1, can I prove $T$ is unitary? 
Thanks!
 A: As Daniel Fischer has noted in the comments, the answer is no. One way to see this is as follows: on the vector space of linear operators on $V$, you can define a norm. The norm of $T$ is the smallest non-negative real number $c$ such that 
$$\|Tv\|\leq c\|v\|,$$
for all $v\in V$. Since this is a norm, for all scalar $\lambda$, you have $\|\lambda T\|=\vert\lambda\vert\|T\|$ (where $\|T\|$ is the norm defined above). Now, going back to your problem, you are assuming that
$$\|T\|\leq 1.$$
But for any operator $T$, by picking $\lambda$ suitably, you can make $\|\lambda T\|\leq 1$. Therefore (and this is what Daniel Fischer mentioned above), if you take a non-normal $T$, you  can make $\lambda T$ into a counterexample.
A: For the second question the answer is yes. I'll suppose that $T$ is diagonalizable (i.e. no Jordan blocks). If $T$ is not unitary then there are eigenvectors $u,v$ with eigenvalues $a,b$, $a\neq b$ ($|a|=|b|=1$) such that $(u,v)\neq 0$. If $c$ is a complecx number then $\|u+cv\|^2=\|u\|^2+|c|^2\|v\|^2+2Re\, c(v,u)$ and $\|T(u+cv)\|^2=\|u\|^2+|c|^2\|v\|^2+2Re\, a^{-1}bc(v,u)$. If you choose $c$ so that $c(v,u)$ is real and negative (e.g. $c=-(u,v)$) then the inequality (for the vector $u+cv$) fails. I'll leave the non-diagonalizable case as an excercise.
