I am reading a book about C*-algebra. And I meet with a problem.
Recall the range projection of an operator $a\in B(H)$ is the projection on the closure of $\{a(\eta):\eta\in H\}$(Here, $H$ is a hilbert space). Let $M$ be a von Neumann algebra and $a\in M$, *then the author says $(a^{\ast}a)^{1/2}$ and $a$ have the same range projecti*on. Why?
With the adjoint in the place you put it, the assertion is false: consider $M=B(H)$, $a\in M$ the unilateral shift. Then the range projection of $a$ is $I-E_{11}$, while the range projection of $(a^*a)^{1/2}=I$ is $I$.
But it is true that $a^*$ and $(a^*a)^{1/2}$ have the same range projection. The fastest way to see it is probably to use the polar decomposition: you have $a=v(a^*a)^{1/2}$ for some partial isometry $v$. Then, taking adjoints, $a^*=(a^*a)^{1/2}v$. This shows that any vector in the range of $a^*$ is also in the range of $(a^*a)^{1/2}$. Conversely, $(a^*a)^{1/2}$ is a limit of polynomials in $a^*a$, so any vector in its range is in the closure of the range of $a^*$.