Two sided ideal of a von Neumann algebra is linear span of its positive elements

Question

Let $H$ be a Hilbert space and $M$ be a von Neumann algebra acting on the Hilbert space $H$. Let $I$ be a two-sided ideal in $M$. Then $I$ is linear span of its positive elements.
My attempt: Since $I$ is a two-sided ideal in $M$, then $I=I^*$. Now let $x \in I$. Then we can write $x=\dfrac{x+x^*}{2}+i\dfrac{x-x^*}{2i}$, that is, $x=y+iz$ where $y,z$ are two self-adjoint elements of $I$. Now let $y \in I$ be such that $y=y^*$. Then if we show that $y$ can be written as linear span of positive elements of $I$, then we are done. But I am unable to show that. Please help me here. Thank you very much.

Answer 1

This follows easily from the polar decomposition.

Since $M$ is a von Neumann algebra, you can write $y=v|y|$ with $v\in M$ a partial isometry. The partial isometry $v$ in the polar decomposition satisfies that $v^*v$ is the range projection of $y^*$. Then $$ |y|=v^*y\in I. $$ Now, since $y$ is selfadjoint and $t\leq |t|$ for all $t\in\mathbb R$, $$ y=\frac12\,\big[(|y|+y)-(|y|-y)\big] $$ is a difference of two positive elements in $I$.