I have a bounded invertible operator $F$ which matrix in the standard orthonormal basis is upper triangular with units on the main diagonal. Let $F_t=tF+(1-t)E$, where $E$ is identity matrix. I want to prove (if it is correct), that in case of $F$ acting in infinite dimensional Hilbert space every $F_t$ is invertible. To show that it is enough to prove that there exists $\gamma>0$ such that for every $N$ and for every "finite dimensional" vector $x=(x_1,x_2,\dotsc,x_N,0,0\dotsc)$ stands $$ \|F_tx\|\geq \gamma\|x\| $$ Notice that $\gamma$ can depend on $t$.
I have proved that for 3-dimensional case $\|F_t^{-1}\|\leq\sqrt{6}\|F\|^2$ for every $t$. I am now trying to show that in infinite dimensional situation such $\gamma$ should exist, because if it doesn't, it would mean that $F_t$ shrinks some vectors too much, but this would somehow contradict the 3-dimensional case. I haven't been successfull at all.
If I find an inequality for $\|F_t^{-1}\|$ that uses $\|F\|$ and $\|F^{-1}\|$ and is indepenent of dimension, this will prove the fact, but all I got is dependent of dimension.
Which ideas can I use?
EDIT: The $\|\cdot\|$ denotes the specral norm, i.e. $\sup_{\|x\|=1}\|Fx\|$