tf.linalg.lstsq

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Solves one or more linear least-squares problems.

tf.linalg.lstsq(
    matrix, rhs, l2_regularizer=0.0, fast=True, name=None
)

matrix is a tensor of shape [..., M, N] whose inner-most 2 dimensions form M-by-N matrices. Rhs is a tensor of shape [..., M, K] whose inner-most 2 dimensions form M-by-K matrices. The computed output is a Tensor of shape [..., N, K] whose inner-most 2 dimensions form M-by-K matrices that solve the equations matrix[..., :, :] * output[..., :, :] = rhs[..., :, :] in the least squares sense.

Below we will use the following notation for each pair of matrix and right-hand sides in the batch:

matrix=$A \in \Re^{m \times n}$, rhs=$B \in \Re^{m \times k}$, output=$X \in \Re^{n \times k}$, l2_regularizer=$\lambda$.

If fast is True, then the solution is computed by solving the normal equations using Cholesky decomposition. Specifically, if $m \ge n$ then $X = (A^T A + \lambda I)^{-1} A^T B$, which solves the least-squares problem $X = \mathrm{argmin}_{Z \in \Re^{n \times k} } ||A Z - B||_F^2 + \lambda ||Z||_F^2$. If $m \lt n$ then output is computed as $X = A^T (A A^T + \lambda I)^{-1} B$, which (for $\lambda = 0$) is the minimum-norm solution to the under-determined linear system, i.e. $X = \mathrm{argmin}_{Z \in \Re^{n \times k} } ||Z||_F^2$, subject to $A Z = B$. Notice that the fast path is only numerically stable when $A$ is numerically full rank and has a condition number $\mathrm{cond}(A) \lt \frac{1}{\sqrt{\epsilon_{mach} } }$ or$\lambda$ is sufficiently large.

If fast is False an algorithm based on the numerically robust complete orthogonal decomposition is used. This computes the minimum-norm least-squares solution, even when $A$ is rank deficient. This path is typically 6-7 times slower than the fast path. If fast is False then l2_regularizer is ignored.