Suppose $\tau$ is an operator on a finite dimensional complex inner product space. I'm read the following,
If $\rho$ is the unique positive square root of the positive operatore $\tau^*\tau$, then $$ \|\rho v\|^2=\langle \rho v,\rho v\rangle=\langle \rho^2 v,v\rangle=\langle \tau^*\tau,v\rangle=\|\tau v\|^2. $$ Define $\nu$ on $im(\rho)$ by $\nu(\rho v)=\tau v$ for $v\in V$. The above equation shows $\rho x=\rho y$ implies $\tau x=\tau y$ so the above definition is well defined.
I don't follow how it shows $\nu$ is well defined. Can anyone clarify?
Suppose $\rho v_1=\rho v_2$. Then you need $\nu$ to be defined on it in a unique way. But for $\rho v_1$ you are defining $\nu$ as $\tau v_1$ and for $\rho v_2$ as $\tau v_2$. You need then that $\tau v_1=\tau v_2$.
Notice that if $\rho(v_1-v_2)=0$ then, by the norm equalities above $\tau(v_1-v_2)=0$.