What does LoRA's α/r ratio control, and why is α = 2r a common default?

**LoRA** freezes the big pretrained weight and learns a small low-rank correction on the side. The correction is a product of two skinny matrices, B times A, and it gets added back to the frozen weight. Before that correction is added, it gets multiplied by a single number: α divided by r. Here r is the rank, the width of the two skinny matrices, and α is a knob you pick in the config. This number is not learned during training. You set it once and it stays fixed. Think of α/r as a volume control on the correction. A bigger value means the learned update is applied more strongly each step; a smaller value means it is applied more gently. In practice it behaves a lot like an extra learning-rate multiplier sitting on top of the adapter. Here is the catch. If you raise r to give the adapter more capacity but leave α alone, the ratio α/r shrinks, so the update silently gets weaker. People often see the bigger adapter score lower and wrongly conclude rank hurt them. That is why recipes tie α to r, like α equals two times r. The ratio stays constant, so you can change rank freely and compare fairly.

In the update, the scalar α/r multiplies the entire low-rank product before it is added to the frozen base weight. The canonical form is: $$W = W_0 + \frac{\alpha}{r}\, B A$$ That scalar is a fixed hyperparameter, not a learned value. It behaves like an effective learning-rate multiplier on the adapter. A larger ratio applies the learned update more strongly each step; a smaller ratio applies it more gently. The practical consequence is the whole reason the convention exists. Because the ratio multiplies the update, changing r while holding α fixed rescales the update magnitude. Doubling r from 16 to 32 with α at 16 halves the effective scale. The bigger adapter then looks worse, when really the optimiser just took smaller steps. Setting α proportional to r, such as α = 2r, holds the ratio constant as rank varies. Now you can sweep r in a search without retuning the learning rate. Any measured difference is attributable to capacity, not to step size. That rank-invariance under sweeps is why most recipes ship this default.