Let $\mathscr{R}$ be a finite von Neumann algebra and for an $n\in \Bbb{N}$ consider the von Neumann algebra $M_n(\mathscr{R})$. If $E_1$ denotes the projection (in $M_n(\mathscr{R})$) whose $(1, 1)$-entry is equal to $I$ and all other entry is $0$. I want to prove that
$E_1$ is finite.
For this I assume a subprojection $F\le E_1$ and choose a partial isometry $V$ with initial projection $E_1$ and finial projection $F$. To prove: $F = E_1$.
I have proved earlier that such a $F$ has only non zero entry at $(1, 1)$ position and rest is zero. I'll call this $(1, 1)$-entry as $F_0$. So enough to prove that $F_0=I$. To do this the author gave a hint as follows: If $V$ is a partial isometry with initial projection $E_1$ and final projection $F$ then $E_1VE_1= V$
I cannot understand why this last statement is true? Since $E_1$ is the initial projection so $VE_1=V$...but from here I get $E_1VE_1 = E_1V=??$
Is it true in general??? Is it true that
if $V$ is a partial isometry with initial projection $E$ then $EVE=V$?
To your question in the end: That is not true in general. For example, if $\xi,\eta$ are orthonormal vectors, then $V=\langle\xi,\cdot\,\rangle\eta$ is a partial isometry with initial projection $E=\langle\xi,\cdot\,\rangle\xi$, but $EVE=0\neq V$.
To prove this claim in your specific situation, you have to use that $F\leq E_1$, which implies $E_1 F=F$. Thus $$ E_1VE_1=E_1 V=E_1FV=FV=V. $$