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This is the main entry point.

lineax.linear_solve(operator: AbstractLinearOperator, vector: PyTree[ArrayLike], solver: AbstractLinearSolver = AutoLinearSolver(well_posed=True), *, options: Optional[dict[str, Any]] = None, state: PyTree[typing.Any] = sentinel, throw: bool = True) -> Solution ¤

Solves a linear system.

Given an operator represented as a matrix \(A\), and a vector \(b\): if the operator is square and nonsingular (so that the problem is well-posed), then this returns the usual solution \(x\) to \(Ax = b\), defined as \(A^{-1}b\).

If the operator is overdetermined, then this either returns the least-squares solution \(\min_x \| Ax - b \|_2\), or throws an error. (Depending on the choice of solver.)

If the operator is underdetermined, then this either returns the minimum-norm solution \(\min_x \|x\|_2 \text{ subject to } Ax = b\), or throws an error. (Depending on the choice of solver.)


This function is equivalent to either numpy.linalg.solve, or to its generalisation numpy.linalg.lstsq, depending on the choice of solver.

The default solver is lineax.AutoLinearSolver(well_posed=True). This automatically selects a solver depending on the structure (e.g. triangular) of your problem, and will throw an error if your system is overdetermined or underdetermined.

Use lineax.AutoLinearSolver(well_posed=False) if your system is known to be overdetermined or underdetermined (although handling this case implies greater computational cost).


These three kinds of solution to a linear system are collectively known as the "pseudoinverse solution" to a linear system. That is, given our matrix \(A\), let \(A^\dagger\) denote the Moore--Penrose pseudoinverse of \(A\). Then the usual/least-squares/minimum-norm solution are all equal to \(A^\dagger b\).


  • operator: a linear operator. This is the '\(A\)' in '\(Ax = b\)'.

    Most frequently this operator is simply represented as a JAX matrix (i.e. a rank-2 JAX array), but any lineax.AbstractLinearOperator is supported.

    Note that if it is a matrix, then it should be passed as an lineax.MatrixLinearOperator, e.g.

    matrix = jax.random.normal(key, (5, 5))  # JAX array of shape (5, 5)
    operator = lx.MatrixLinearOperator(matrix)  # Wrap into a linear operator
    solution = lx.linear_solve(operator, ...)
    rather than being passed directly.

  • vector: the vector to solve against. This is the '\(b\)' in '\(Ax = b\)'.

  • solver: the solver to use. Should be any lineax.AbstractLinearSolver. The default is lineax.AutoLinearSolver which behaves as discussed above.

    If the operator is overdetermined or underdetermined , then passing lineax.SVD is typical.

  • options: Individual solvers may accept additional runtime arguments; for example lineax.CG allows for specifying a preconditioner. See each individual solver's documentation for more details. Keyword only argument.

  • state: If performing multiple linear solves with the same operator, then it is possible to save re-use some computation between these solves, and to pass the result of any intermediate computation in as this argument. See lineax.AbstractLinearSolver.init for more details. Keyword only argument.

  • throw: How to report any failures. (E.g. an iterative solver running out of steps, or a well-posed-only solver being run with a singular operator.)

    If True then a failure will raise an error. Note that errors are only reliably raised on CPUs. If on GPUs then the error may only be printed to stderr, whilst on TPUs then the behaviour is undefined.

    If False then the returned solution object will have a result field indicating whether any failures occured. (See lineax.Solution.)

    Keyword only argument.


An lineax.Solution object containing the solution to the linear system.