Why is this statement true for two equivalent projections in $B(H)$? In a book of operator theory it is stated that two projections $P$ and $Q$ in a von Neumann algebra $A$ are equivalent if there exist $V$ in $A$ that $V^*V=P$ and $VV^*=Q$. 
After this definition, it states immediately that if $A=B(H)$ then $P$ and $Q$ are equivalent iff $\dim(\operatorname{Im} P)=\dim(\operatorname{Im}Q)$. 
I cannot prove this, and I don't understand that why this is true for $B(H)$ but not for every von Neumann algebra.
Thanks for your answers. 
 A: *

*If $A=B(H)$, then $P$ and $Q$ are literally the orthogonal projections onto the closed subspaces $PH$ and $QH$, respectively, of $H$. Can you think of $V$ as a map between the Hilbert spaces $PH$ and $QH$? If so, what kind of map is it? On the other hand, if $PH$ and $QH$ have the same dimension, you can pick orthonormal bases for $PH$ and $QH$ of the same cardinality; can you see how this lets you define $V$?

*Let $A$ be a general von Neumann algebra, which is, therefore, a weakly closed $\ast$-subalgebra of $B(H)$ for some Hilbert space $H$. If projections $P$ and $Q$ in $A$ are equivalent in $A$, then they are, in particular, equivalent in $B(H)$, so you can apply 1. without a problem. On the other hand, suppose that projections $P$ and $Q$ in $A$ are equivalent in $B(H)$, so that $P=V^\ast V$ and $Q = VV^\ast$ for some $V \in B(H)$. What guarantee do you have that $V$ can be chosen to be an element of $A$? As a very elementary example, think of the projections
$$
 P = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad Q = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}
$$
in the von Neumann algebra $A := \mathbb{C} \oplus \mathbb{C}$ of diagonal matrices in $B(\mathbb{C}^2) = M_2(\mathbb{C})$. Can you actually find $V \in A$ such that $P = V^\ast V$ and $Q = VV^\ast$?

