# LostTech.TensorFlow : API Documentation

Type LinearOperatorCirculant

Namespace tensorflow.linalg

Interfaces ILinearOperatorCirculant

`LinearOperator` acting like a circulant matrix.

This operator acts like a circulant matrix `A` with shape `[B1,...,Bb, N, N]` for some `b >= 0`. The first `b` indices index a batch member. For every batch index `(i1,...,ib)`, `A[i1,...,ib, : :]` is an `N x N` matrix. This matrix `A` is not materialized, but for purposes of broadcasting this shape will be relevant.

#### Description in terms of circulant matrices

Circulant means the entries of `A` are generated by a single vector, the convolution kernel `h`: `A_{mn} := h_{m-n mod N}`. With `h = [w, x, y, z]`,

``` A = |w z y x| |x w z y| |y x w z| |z y x w| ```

This means that the result of matrix multiplication `v = Au` has `Lth` column given circular convolution between `h` with the `Lth` column of `u`.

See http://ee.stanford.edu/~gray/toeplitz.pdf

#### Description in terms of the frequency spectrum

There is an equivalent description in terms of the [batch] spectrum `H` and Fourier transforms. Here we consider `A.shape = [N, N]` and ignore batch dimensions. Define the discrete Fourier transform (DFT) and its inverse by

``` DFT[ h[n] ] = H[k] := sum_{n = 0}^{N - 1} h_n e^{-i 2pi k n / N} IDFT[ H[k] ] = h[n] = N^{-1} sum_{k = 0}^{N - 1} H_k e^{i 2pi k n / N} ```

From these definitions, we see that

``` H[0] = sum_{n = 0}^{N - 1} h_n H[1] = "the first positive frequency" H[N - 1] = "the first negative frequency" ```

Loosely speaking, with `*` element-wise multiplication, matrix multiplication is equal to the action of a Fourier multiplier: `A u = IDFT[ H * DFT[u] ]`. Precisely speaking, given `[N, R]` matrix `u`, let `DFT[u]` be the `[N, R]` matrix with `rth` column equal to the DFT of the `rth` column of `u`. Define the `IDFT` similarly. Matrix multiplication may be expressed columnwise:

```(A u)_r = IDFT[ H * (DFT[u])_r ]```

#### Operator properties deduced from the spectrum.

Letting `U` be the `kth` Euclidean basis vector, and `U = IDFT[u]`. The above formulas show that`A U = H_k * U`. We conclude that the elements of `H` are the eigenvalues of this operator. Therefore

* This operator is positive definite if and only if `Real{H} > 0`.

A general property of Fourier transforms is the correspondence between Hermitian functions and real valued transforms.

Suppose `H.shape = [B1,...,Bb, N]`. We say that `H` is a Hermitian spectrum if, with `%` meaning modulus division,

```H[..., n % N] = ComplexConjugate[ H[..., (-n) % N] ]```

* This operator corresponds to a real matrix if and only if `H` is Hermitian. * This operator is self-adjoint if and only if `H` is real.

See e.g. "Discrete-Time Signal Processing", Oppenheim and Schafer.

#### Example of a self-adjoint positive definite operator #### Example of defining in terms of a real convolution kernel #### Example of Hermitian spectrum #### Example of forcing real `dtype` when spectrum is Hermitian #### Performance

Suppose `operator` is a `LinearOperatorCirculant` of shape `[N, N]`, and `x.shape = [N, R]`. Then

* `operator.matmul(x)` is `O(R*N*Log[N])` * `operator.solve(x)` is `O(R*N*Log[N])` * `operator.determinant()` involves a size `N` `reduce_prod`.

If instead `operator` and `x` have shape `[B1,...,Bb, N, N]` and `[B1,...,Bb, N, R]`, every operation increases in complexity by `B1*...*Bb`.

#### Matrix property hints

This `LinearOperator` is initialized with boolean flags of the form `is_X`, for `X = non_singular, self_adjoint, positive_definite, square`. These have the following meaning:

* If `is_X == True`, callers should expect the operator to have the property `X`. This is a promise that should be fulfilled, but is *not* a runtime assert. For example, finite floating point precision may result in these promises being violated. * If `is_X == False`, callers should expect the operator to not have `X`. * If `is_X == None` (the default), callers should have no expectation either way.
Show Example
```# spectrum is real ==> operator is self-adjoint
# spectrum is positive ==> operator is positive definite
spectrum = [6., 4, 2]  operator = LinearOperatorCirculant(spectrum)  # IFFT[spectrum]
operator.convolution_kernel()
==> [4 + 0j, 1 + 0.58j, 1 - 0.58j]  operator.to_dense()
==> [[4 + 0.0j, 1 - 0.6j, 1 + 0.6j],
[1 + 0.6j, 4 + 0.0j, 1 - 0.6j],
[1 - 0.6j, 1 + 0.6j, 4 + 0.0j]] ```

### Public instance methods

Returns the adjoint of the current `LinearOperator`.

Given `A` representing this `LinearOperator`, return `A*`. Note that calling `self.adjoint()` and `self.H` are equivalent.
##### Parameters
`string` name
A name for this `Op`.
##### Returns
`object`
`LinearOperator` which represents the adjoint of this `LinearOperator`.

Returns the adjoint of the current `LinearOperator`.

Given `A` representing this `LinearOperator`, return `A*`. Note that calling `self.adjoint()` and `self.H` are equivalent.
##### Parameters
`ImplicitContainer<T>` name
A name for this `Op`.
##### Returns
`object`
`LinearOperator` which represents the adjoint of this `LinearOperator`.

#### Tensorsolvevec(IndexedSlices rhs, bool adjoint, string name)

Solve single equation with best effort: `A X = rhs`.

The returned `Tensor` will be close to an exact solution if `A` is well conditioned. Otherwise closeness will vary. See class docstring for details.

Examples:
##### Parameters
`IndexedSlices` rhs
`Tensor` with same `dtype` as this operator. `rhs` is treated like a [batch] vector meaning for every set of leading dimensions, the last dimension defines a vector. See class docstring for definition of compatibility regarding batch dimensions.
`bool` adjoint
Python `bool`. If `True`, solve the system involving the adjoint of this `LinearOperator`: `A^H X = rhs`.
`string` name
A name scope to use for ops added by this method.
##### Returns
`Tensor`
`Tensor` with shape `[...,N]` and same `dtype` as `rhs`.
Show Example
```# Make an operator acting like batch matrix A.  Assume A.shape = [..., M, N]
operator = LinearOperator(...)
operator.shape = [..., M, N]  # Solve one linear system for every member of the batch.
RHS =... # shape [..., M]  X = operator.solvevec(RHS)
# X is the solution to the linear system
# sum_j A[..., :, j] X[..., j] = RHS[..., :]  operator.matvec(X)
==> RHS ```

#### Tensorsolvevec(IGraphNodeBase rhs, bool adjoint, string name)

Solve single equation with best effort: `A X = rhs`.

The returned `Tensor` will be close to an exact solution if `A` is well conditioned. Otherwise closeness will vary. See class docstring for details.

Examples:
##### Parameters
`IGraphNodeBase` rhs
`Tensor` with same `dtype` as this operator. `rhs` is treated like a [batch] vector meaning for every set of leading dimensions, the last dimension defines a vector. See class docstring for definition of compatibility regarding batch dimensions.
`bool` adjoint
Python `bool`. If `True`, solve the system involving the adjoint of this `LinearOperator`: `A^H X = rhs`.
`string` name
A name scope to use for ops added by this method.
##### Returns
`Tensor`
`Tensor` with shape `[...,N]` and same `dtype` as `rhs`.
Show Example
```# Make an operator acting like batch matrix A.  Assume A.shape = [..., M, N]
operator = LinearOperator(...)
operator.shape = [..., M, N]  # Solve one linear system for every member of the batch.
RHS =... # shape [..., M]  X = operator.solvevec(RHS)
# X is the solution to the linear system
# sum_j A[..., :, j] X[..., j] = RHS[..., :]  operator.matvec(X)
==> RHS ```

#### Tensorsolvevec(PythonClassContainer rhs, bool adjoint, string name)

Solve single equation with best effort: `A X = rhs`.

The returned `Tensor` will be close to an exact solution if `A` is well conditioned. Otherwise closeness will vary. See class docstring for details.

Examples:
##### Parameters
`PythonClassContainer` rhs
`Tensor` with same `dtype` as this operator. `rhs` is treated like a [batch] vector meaning for every set of leading dimensions, the last dimension defines a vector. See class docstring for definition of compatibility regarding batch dimensions.
`bool` adjoint
Python `bool`. If `True`, solve the system involving the adjoint of this `LinearOperator`: `A^H X = rhs`.
`string` name
A name scope to use for ops added by this method.
##### Returns
`Tensor`
`Tensor` with shape `[...,N]` and same `dtype` as `rhs`.
Show Example
```# Make an operator acting like batch matrix A.  Assume A.shape = [..., M, N]
operator = LinearOperator(...)
operator.shape = [..., M, N]  # Solve one linear system for every member of the batch.
RHS =... # shape [..., M]  X = operator.solvevec(RHS)
# X is the solution to the linear system
# sum_j A[..., :, j] X[..., j] = RHS[..., :]  operator.matvec(X)
==> RHS ```

#### objectsolvevec_dyn(object rhs, ImplicitContainer<T> adjoint, ImplicitContainer<T> name)

Solve single equation with best effort: `A X = rhs`.

The returned `Tensor` will be close to an exact solution if `A` is well conditioned. Otherwise closeness will vary. See class docstring for details.

Examples:
##### Parameters
`object` rhs
`Tensor` with same `dtype` as this operator. `rhs` is treated like a [batch] vector meaning for every set of leading dimensions, the last dimension defines a vector. See class docstring for definition of compatibility regarding batch dimensions.
`ImplicitContainer<T>` adjoint
Python `bool`. If `True`, solve the system involving the adjoint of this `LinearOperator`: `A^H X = rhs`.
`ImplicitContainer<T>` name
A name scope to use for ops added by this method.
##### Returns
`object`
`Tensor` with shape `[...,N]` and same `dtype` as `rhs`.
Show Example
```# Make an operator acting like batch matrix A.  Assume A.shape = [..., M, N]
operator = LinearOperator(...)
operator.shape = [..., M, N]  # Solve one linear system for every member of the batch.
RHS =... # shape [..., M]  X = operator.solvevec(RHS)
# X is the solution to the linear system
# sum_j A[..., :, j] X[..., j] = RHS[..., :]  operator.matvec(X)
==> RHS ```