Is this configured as unity gain? Seems like there would be equal signal at the cathode or anode.
If you're interested, here's the logic behind Sluckey's confirmation of (slightly less than) unity gain:
1. Any signal input to the tube results in a plate current change. Current though a resistance causes a voltage drop across the resistance which is proportional to both the amount of current and resistance. Both the plate and cathode circuits have equal resistances, so any signal current will cause an equal voltage output appearing at the plate and cathode.
2. The tube doesn't know absolute voltage values with respect to input signal, it feels only a difference of voltage between its cathode and grid. The cathode output voltage due to signal current is in-phase with the grid input voltage, so as grid voltage increases cathode voltage also increases leaving little net voltage change between grid and cathode. So if there is a signal current causing a big voltage change at the cathode, there must be an even bigger grid voltage change because the cathode voltage offsets the net grid-to-cathode voltage.
3. All this means for a (for example) 40v output at the plate there is also a 40v output at the cathode due to equal resistances, and there must be a more-than-40v signal at the grid to not be overwhelmed by the cathode voltage. Therefore, there is a gain slightly less than unity to each output.
Extra side-note: 4. Your circuit is fixed-biased. If it was cathode biased instead, the input impedance due to the bootstrapping effect would be very much more than the value of the resistor between grid and cathode load. That's because the "ground side" of the grid resistor has a voltage change with output signal. That allows the use of a very much smaller coupling cap and a maximizing of the gain of the stage before this split-load inverter.