Read the equation
Multiplying one factor by a nonzero scalar and dividing the other by that scalar cancels in their outer product. The individual vectors can therefore move in magnitude while the effective weight update remains identical. This is why an operator-based distance avoids a basic ambiguity in factor-only measurements.
Terms & definitions
| Term | Meaning in this source |
|---|---|
| lambda_ | Any nonzero rescaling scalar; c in another version |
| a_t | Input-side factor |
| b_t | Output-side factor |
How it works
Read the definition from its right-hand expression to the quantity on the left. Its surrounding source records the intended input, state, or parameter context. Where the source leaves an operation unspecified, this page preserves that gap.
Assumptions & limits
The scalar must be nonzero.
Original context
Read the surrounding source passage
Here st is the effective scaling coefficient from the adapter configuration, at is the input-side direction, and bt the output-side direction. Standard and rank-stabilized scaling differ in general; at rank one their rank denominators coincide. Configuration values, per-layer patterns, orientation conventions, and optional methods such as DoRA must be checked rather than assumed [1,4]. Comparing bt alone is useful within a fixed parameterization, but the factorization is not unique. For any nonzero scalar c: cbtat/cT=btatT. In particular, reversing the signs of both factors changes the apparent direction of each while leaving the effective matrix unchanged. A factor-level rotation therefore deserves a corresponding effective-update check. This does not assert that the published rotation was a gauge artifact; that question requires the actual tensors. 6.2 Low-memory exact formulas The Frobenius inner product between two rank-one updates can be computed without constructing the dense matrices:
Exact original notation
\left(c{b}_{t}\right){\left({a}_{t}/c\right)}^{T}={b}_{t}{a}_{t}^{T}.Related builds
Sources & provenance
\left(c{b}_{t}\right){\left({a}_{t}/c\right)}^{T}={b}_{t}{a}_{t}^{T}.
\left(c{b}_{t}\right){\left({a}_{t}/c\right)}^{T}={b}_{t}{a}_{t}^{T}.