numerical diagonalization of a unitary matrix - matrix

To numerically diagonalize a unitary matrix I use the LAPACK routine zgeev.
The problem is: In case of degeneracies the degenerate subspace is not orthonormalized, since the routine is for general matrices.
However, since in my case the matrices are unitary, the basis can be always orthonormalized. Is there a better solution than applying QR-algorithm afterwards to the degenerate subspace?

Short answer: Schur decomposition!
If a square matrix A is complex, then its Schur factorization is A=ZTZ*, where Z is unitary and T is upper triangular.
If A happens to be unitary, T must also be unitary. Since T is both unitary and triangular, it is diagonal (proof here,.or there)
Let's consider the vectors Z.e_i, where e_i are the vectors of the canonical basis. These vectors obviously form an orthonormal basis. Moreover, these vectors are eigenvectors of the matrix A.
Hence, the columns of the unitary matrix Z are eigenvectors of the unitary matrix A and form an orthonormal basis.
As a consequence, computing a Schur decomposition of a unitary matrix is equivalent to finding one of its orthogonal basis of eigenvectors.
ZGEESX computes the eigenvalues, the Schur form, and, optionally, the matrix of Schur vectors for GE matrices
The resulting T can also be tested to check that A is unitary.
Here is a piece of python code testing it, though scipy's scipy.linalg.schur makes use of Lapack's zgees for Schur decomposition. I used hpaulj's code to generate random unitary matrix as shown in How to create random orthonormal matrix in python numpy
import numpy as np
import scipy.linalg
#from hpaulj, https://stackoverflow.com/questions/38426349/how-to-create-random-orthonormal-matrix-in-python-numpy
def rvs(dim=3):
random_state = np.random
H = np.eye(dim)
D = np.ones((dim,))
for n in range(1, dim):
x = random_state.normal(size=(dim-n+1,))
D[n-1] = np.sign(x[0])
x[0] -= D[n-1]*np.sqrt((x*x).sum())
# Householder transformation
Hx = (np.eye(dim-n+1) - 2.*np.outer(x, x)/(x*x).sum())
mat = np.eye(dim)
mat[n-1:, n-1:] = Hx
H = np.dot(H, mat)
# Fix the last sign such that the determinant is 1
D[-1] = (-1)**(1-(dim % 2))*D.prod()
# Equivalent to np.dot(np.diag(D), H) but faster, apparently
H = (D*H.T).T
return H
n=42
A= rvs(n)
A = A.astype(complex)
T,Z=scipy.linalg.schur(A,output='complex',lwork=None,overwrite_a=False,sort=None,check_finite=True)
#print T
normT=np.linalg.norm(T,ord=None) #2-norm
eigenvalues=[]
for i in range(n):
eigenvalues.append(T[i,i])
T[i,i]=0.
normTu=np.linalg.norm(T,ord=None)
print 'must be very low if A is unitary: ',normTu/normT
#print Z
for i in range(n):
v=Z[:,i]
w=A.dot(v)-eigenvalues[i]*v
print i,'must be very low if column i of Z is eigenvector of A: ',np.linalg.norm(w,ord=None)/np.linalg.norm(v,ord=None)

Related

Hermitian matrix of h-chain using np.diag

I hope you are well.
Greatly appreciate any help provided.
I am attempting to create a NxN hermitian matrix which takes the molecular chain of H atoms using a function and numpy diagonal.
def H_chain(N, E, A):
y = np.ones(N,N)
x = np.arange(E).reshape((N,N))
z = np.arange(A).reshape((N,N))
g = np.diag(np.diag(y)*x)+np.diag(np.diag(y, k=-1)*-z+np.diag(np.diag(y,k=1)*-z))
return g
#The function should return something that looks like this
#[E,-A,0,0,0],
#[-A,E,-A,0,0]
#[0,-A,E,-A,0]
#[0,0,-A,E,-A]
#[0,0,0,-A,E]
Returns an N x N matrix representing the Hamiltonian of an N-atom chain molecule with
site-energies E and inter-site coupling rate A.
Ideally I should be summing three matrices together using the numpy diagonal function and the numpy ones to fill the diagonals aswell.
Much appreciate any help.
Thank you.
Kind regards,

Computing a single element of the adjugate or inverse of a symbolic binary matrix

I'm trying to get a single element of an adjugate A_adj of a matrix A, both of which need to be symbolic expressions, where the symbols x_i are binary and the matrix A is symmetric and sparse. Python's sympy works great for small problems:
from sympy import zeros, symbols
size = 4
A = zeros(size,size)
x_i = [x for x in symbols(f'x0:{size}')]
for i in range(size-1):
A[i,i] += 0.5*x_i[i]
A[i+1,i+1] += 0.5*x_i[i]
A[i,i+1] = A[i+1,i] = -0.3*(i+1)*x_i[i]
A_adj_0 = A[1:,1:].det()
A_adj_0
This calculates the first element A_adj_0 of the cofactor matrix (which is the corresponding minor) and correctly gives me 0.125x_0x_1x_2 - 0.28x_2x_2^2 - 0.055x_1^2x_2 - 0.28x_1x_2^2, which is the expression I need, but there are two issues:
This is completely unfeasible for larger matrices (I need this for sizes of ~100).
The x_i are binary variables (i.e. either 0 or 1) and there seems to be no way for sympy to simplify expressions of binary variables, i.e. simplifying polynomials x_i^n = x_i.
The first issue can be partly addressed by instead solving a linear equation system Ay = b, where b is set to the first basis vector [1, 0, 0, 0], such that y is the first column of the inverse of A. The first entry of y is the first element of the inverse of A:
b = zeros(size,1)
b[0] = 1
y = A.LUsolve(b)
s = {x_i[i]: 1 for i in range(size)}
print(y[0].subs(s) * A.subs(s).det())
print(A_adj_0.subs(s))
The problem here is that the expression for the first element of y is extremely complicated, even after using simplify() and so on. It would be a very simple expression with simplification of binary expressions as mentioned in point 2 above. It's a faster method, but still unfeasible for larger matrices.
This boils down to my actual question:
Is there an efficient way to compute a single element of the adjugate of a sparse and symmetric symbolic matrix, where the symbols are binary values?
I'm open to using other software as well.
Addendum 1:
It seems simplifying binary expressions in sympy is possible with a simple custom substitution which I wasn't aware of:
A_subs = A_adj_0
for i in range(size):
A_subs = A_subs.subs(x_i[i]*x_i[i], x_i[i])
A_subs
You should make sure to use Rational rather than floats in sympy so S(1)/2 or Rational(1, 2) rather than 0.5.
There is a new (undocumented and for the moment internal) implementation of matrices in sympy called DomainMatrix. It is likely to be a lot faster for a problem like this and always produces polynomial results in a fully expanded form. I expect that it will be much faster for this kind of problem but it still seems to be fairly slow for this because is is not sparse internally (yet - that will probably change in the next release) and it does not take advantage of the simplification from the symbols being binary-valued. It can be made to work over GF(2) but not with symbols that are assumed to be in GF(2) which is something different.
In case it is helpful though this is how you would use it in sympy 1.7.1:
from sympy import zeros, symbols, Rational
from sympy.polys.domainmatrix import DomainMatrix
size = 10
A = zeros(size,size)
x_i = [x for x in symbols(f'x0:{size}')]
for i in range(size-1):
A[i,i] += Rational(1, 2)*x_i[i]
A[i+1,i+1] += Rational(1, 2)*x_i[i]
A[i,i+1] = A[i+1,i] = -Rational(3, 10)*(i+1)*x_i[i]
# Convert to DomainMatrix:
dM = DomainMatrix.from_list_sympy(size-1, size-1, A[1:, 1:].tolist())
# Compute determinant and convert back to normal sympy expression:
# Could also use dM.det().as_expr() although it might be slower
A_adj_0 = dM.charpoly()[-1].as_expr()
# Reduce powers:
A_adj_0 = A_adj_0.replace(lambda e: e.is_Pow, lambda e: e.args[0])
print(A_adj_0)

vectorization of a single loop in matlab (multiplication and then addition)

I have a nX2 matrix A and a 3D matrix K. I would like to take element-wise multiplication specifying 2 indices in 3rd dimension of K designated by each row vector in A and take summation of them.
For instance of a simplified example when n=2,
A=[1 2;3 4];%2X2 matrix
K=unifrnd(0.1,0.1,2,2,4);%just random 3D matrix
L=zeros(2,2);%save result to here
for t=1:2
L=L+prod(K(:,:,A(t,:)),3);
end
Can I get rid of the for loop in this case?
How's this?
B = A.'; %'
L = squeeze(sum(prod(...
reshape(permute(K(:,:,B(:)),[3 1 2]),2,[],size(K,1),size(K,2)),...
1),...
2));
Although your test case is too simple, so I can't be entirely sure that it's correct.
The idea is that we first take all the indices in A, in column-major order, then reshape the elements of K such that the first two dimensions are of size [2, n], and the second two dimensions are the original 2 of K. We then take the product, then the sum along the necessary dimensions, ending up with a matrix that has to be squeezed to get a 2d matrix.
Using a bit more informative test case:
K = rand(2,3,4);
A = randi(4,4,2);
L = zeros(2,3);%save result to here
for t=1:size(A,1)
L = L+prod(K(:,:,A(t,:)),3);
end
B = A.'; %'
L2 = squeeze(sum(prod(reshape(permute(K(:,:,B(:)),[3 1 2]),2,[],size(K,1),size(K,2)),1),2));
Then
>> isequal(L,L2)
ans =
1
With some reshaping magic -
%// Get sizes
[m1,n1,r1] = size(K);
[m2,n2] = size(A);
%// Index into 3rd dim of K; perform reductions and reshape back
Lout = reshape(sum(prod(reshape(K(:,:,A'),[],n2,m2),2),3),m1,n1);
Explanation :
Index into the third dimension of K with a transposed version of A (transposed because we are using rows of A for indexing).
Perform the prod() and sum() operations.
Finally reshape back to a shape same as K but without the third dimension as that was removed in the earlier reduction steps.

Numpy Hermitian Matrix class

Are you aware of something like a hermitian matrix class in numpy? I'd like to optimize matrix calculations like
B = U * A * U.H
, where A (and thus, B) are hermitian. Without specification, all matrix elements of B are calculated. In fact, it should be able to save a factor of about 2 here. Do I miss something?
The method I need should take take the upper/lower triangle of A, the full matrix of U and return the upper/lower triangle of B.
I don't think there exists a method for your specific problem, but with a little thought you might be able to build an algorithm from the low-level BLAS routines that are wrapped in SciPy. For example, dgemm, dsymm, and dtrmm do general, symmetric, and triangular matrix products respectively. Here's an example of using them:
from scipy.linalg.blas import dgemm, dsymm, dtrmm
A = np.random.rand(10, 10)
B = np.random.rand(10, 10)
S = np.dot(A, A.T) # symmetric matrix
T = np.triu(S) # upper triangular matrix
# normal matrix-matrix product
assert np.allclose(dgemm(1, A, B), np.dot(A, B))
# symmetric mat-mat product using only upper-triangle
assert np.allclose(dsymm(1, T, B), np.dot(S, B))
# upper-triangular mat-mat product
assert np.allclose(dtrmm(1, T, B), np.dot(T, B))
There are many other low-level BLAS routines available; I find the NETLIB page to be a good resource to learn what they do. You may be able to cleverly use some combination of the available routines to efficiently solve the problem you have in mind.
Edit: it looks like there are LAPACK routines that quickly compute exactly what you want: dsytrd or zhetrd, but unfortunately these don't appear to be wrapped directly in scipy.linalg.lapack, though scipy does provide cython wrappers for them. Best of luck!
I needed tridiagonal reduction of a symmetric/Hermitian matrix A,
T = Q^H * A * Q
– presumably OP's underlying problem – and I've just submitted a pull request to SciPy for properly interfacing LAPACK's {s,d}sytrd (for real symmetric matrices) and {c,z}hetrd (for Hermitian matrices). All routines use either only the upper or the lower triangular part of the matrix.
Once this has been merged, it can be used like
import numpy as np
n = 3
A = np.zeros((n, n), dtype=dtype)
A[np.triu_indices_from(A)] = np.arange(1, 2*n+1, dtype=dtype)
# query lwork -- optional
lwork, info = sytrd_lwork(n)
assert info == 0
data, d, e, tau, info = sytrd(A, lwork=lwork)
assert info == 0
The vectors d and e now contain the main diagonal and the upper and lower diagonal, respectively.

Spark distributed matrix multiply and pseudo-inverse calculating

I am very new in Apache Spark Scala. Can you help me with some operations?
I have two distributed matrix H and Y in Spark Scala.
I want to compute the pseudo-inverse of H and then multiply H and Y.
How can I do this?
Here is an implementation for the inverse.
import org.apache.spark.mllib.linalg.{Vectors,Vector,Matrix,SingularValueDecomposition,DenseMatrix,DenseVector}
import org.apache.spark.mllib.linalg.distributed.RowMatrix
def computeInverse(X: RowMatrix): DenseMatrix = {
val nCoef = X.numCols.toInt
val svd = X.computeSVD(nCoef, computeU = true)
if (svd.s.size < nCoef) {
sys.error(s"RowMatrix.computeInverse called on singular matrix.")
}
// Create the inv diagonal matrix from S
val invS = DenseMatrix.diag(new DenseVector(svd.s.toArray.map(x => math.pow(x,-1))))
// U cannot be a RowMatrix
val U = new DenseMatrix(svd.U.numRows().toInt,svd.U.numCols().toInt,svd.U.rows.collect.flatMap(x => x.toArray))
// If you could make V distributed, then this may be better. However its alreadly local...so maybe this is fine.
val V = svd.V
// inv(X) = V*inv(S)*transpose(U) --- the U is already transposed.
(V.multiply(invS)).multiply(U)
}
To calculate the pseudo-inverse of non-square matrices you need to be able to calculate the transpose (easy) and the matrix inverse (others have supplied that functionality). There are two different calculations, depending on whether M has full column rank or full row rank.
Full column rank means that the columns of the matrix are linearly independent which requires that the the number of columns is less than or equal to the number of rows. (In pathological cases, an mxn matrix with m>=n might still not have full column rank, but we'll ignore that statistical impossibility. If it is a possibility in your case, the matrix inversion step below will fail.) For full column rank, the pseudo-inverse is
M^+ = (M^T M)^{-1} M^T
where M^T is the transpose of M. Matrix multiply M^T by M, then take the inverse, then matrix multiply by M^T again. (I'm assuming M has real number entries; if the entries are complex numbers, you also have to take complex conjugates.)
A quick check to make sure you have calculated the psuedo-inverse correctly is to check M^+ M. It should be the identity matrix (up to floating point error).
On the other hand, if M has full row rank, in other words M is mxn with m<=n, the pseudo-inverse is
M^+ = M^T (M M^T)^{-1}
To check whether you have the correct pseudo-inverse in this case, right multiply with the original matrix: M M^+. That should equal the identity matrix, up to floating point error.
Matrix multiplication is the easier one: there are several Matrix implementations with a multiply method in packages org.apache.spark.mllib.linalg and org.apache.spark.mllib.linalg.distributed. Pick whatever fits your needs most.
I have not seen (pseudo-)inverse anywhere in the Spark API. But RowMatrix is able to compute the singular value decomposition which can be used to calculate the inverse of a matrix. Here is a very naive implementation, inspired by How can we compute Pseudoinverse for any Matrix (warning: dimensions of the 2x2 matrix are hard-coded):
val m = new RowMatrix(sc.parallelize(Seq(Vectors.dense(4, 3), Vectors.dense(3, 2))))
val svd = m.computeSVD(2, true)
val v = svd.V
val sInvArray = svd.s.toArray.toList.map(x => 1.0 / x).toArray
val sInverse = new DenseMatrix(2, 2, Matrices.diag(Vectors.dense(sInvArray)).toArray)
val uArray = svd.U.rows.collect.toList.map(_.toArray.toList).flatten.toArray
val uTranspose = new DenseMatrix(2, 2, uArray) // already transposed because DenseMatrix is column-major
val inverse = v.multiply(sInverse).multiply(uTranspose)
// -1.9999999999998297 2.999999999999767
// 2.9999999999997637 -3.9999999999996767
Unfortunately, a lot of conversion from Matrix to Array and so forth is necessary. If you need a fully distributed implementation, try using DistributedMatrix instead of DenseMatrix. If not, maybe using Breeze is preferable here.

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