Why if $\rho(I_{\mathfrak{X}} - YA)<1$ then $YA$ is invertible on the $R(YA)$?

Question

I am reading an article where I am stucked at one point. Below is my problem.

Given that $\mathfrak{X}$ and $\mathfrak{Y}$ are Banach spaces.

$A:\mathfrak{X} \to \mathfrak{Y}$ and $Y:\mathfrak{Y}\to \mathfrak{X}$ are linear bounded operators.

At one step it is written that if $\rho(I_{\mathfrak{X}} - YA)<1$ then $YA$ is invertible on the $R(YA)$, where $\rho$ stands for the spectral radious and $R(A)$ denotes the range space of the operator $A$. I am not able to understand why operator $YA$ is invertible on the $R(YA)$? Why not on its whole domain $\mathfrak{X}$ ?

Could anyone help me to clear my doubt. I would be very much thankful.

Thanks

Answer 1

If $T$ is a linear operator on a vector space, $I - T$ is not invertible iff $1$ is in the spectrum of $T$, and that implies the spectral radius of $T$ is at least $1$. Take $T = I - YA$. So if $\rho(I-YA) < 1$, $YA$ is invertible on $\mathfrak X$, and this is $R(YA)$ when $YA$ is invertible.