Square root of a matrix
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In mathematics, the square root of a matrix extends the notion of square root from numbers to matrices.
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[edit] Square roots of positive operators
In linear algebra and operator theory, given a bounded positive semidefinite operator T on a complex Hilbert space, B is a square root of T if T = B*B. According to the spectral theorem, the continuous functional calculus can be applied to obtain an operator T½ such that T½ is itself positive and (T½)2 = T. The operator T½ is the unique positive square root of T.
A bounded positive operator on a complex Hilbert space is self adjoint. So T = (T½)* T½. Conversely, it is trivially true that every operator of the form B*B is positive. Therefore, an operator T is positive if and only if T = B*B for some B (equivalently, T = CC* for some C).
The Cholesky factorization is a particular example of square root.
[edit] Unitary freedom of square roots
If T is a n × n positive matrix, all square roots of T are related by unitary transformations. More precisely, if T = AA* = BB*, then there exists an unitary U s.t. A = BU. We now verify this claim.
Take B = T½ to be the unique positive square root of T. It suffices to show that, for any square root A of T, A = UB. Let {ai}1 ≤ i ≤ n, and {bi}1 ≤ i ≤ n be the set of column vectors of A and B, respectively. By the construction of the square root, {bi}1 ≤ i ≤ n is the set of, not necessarily normalized, eigenvectors of a Hermitian matrix, therefore an orthogonal basis for Cn. Notice we include the eigenvectors corresponding to eigenvalue 0 when appropriate. The argument below hinges on the fact that {bi}1 ≤ i ≤ n is linearly independent and spans Cn.
The equality AA* = BB* can be rewritten as
By completeness of {bi}, for all j, there exists n scalars, by appending zeros if necessary, {ui j}1 ≤ i ≤ n s.t.
i.e.
We need to show the matrix U = [ui j] is unitary. Compute directly
So
By assumption,
Now because {bi} is a basis of Cn, the set {bi b*k} is a basis for n × n matrices. For linear spaces in general, the expression of an element in terms of a given basis is unique. Therefore
In other words, the n column vectors of U is an orthonormal set and the claim is proved.
The argument extends to the case where A and B are not necessarily square, provided one retains the assumption {bi} is linearly independent. In that case, the existence of the rectangular matrix U = [ui j] follows from the relation
rather than the completeness of {bi}. The conclusion UU* = I still holds. In general, B is n × m for some m ≤ n. A is n × k where m ≤ k ≤ n. U is a m × k partial isometry. (By a partial isometry, we mean a rectangular matrix U with UU* = I.)
[edit] Some applications
The unitary freedom of square roots has applications in linear algebra.
[edit] Kraus operators
By Choi's result, a linear map
is completely positive if and only if it is of the form
where k ≤ nm. Let {Ep q} ⊂ Cn × n be the n2 elementary matrix units. The positive matrix
is called the Choi matrix of Φ. The Kraus operators correspond to the, not necessarily square, square roots of MΦ: For any square root B of MΦ, one can obtain a family of Kraus operators Vi by undoing the Vec operation to each column bi of B. Thus all sets of Kraus operators are related by partial isometries.
[edit] Mixed ensembles
In quantum physics, a density matrix for a n-level quantum system is a n × n complex matrix ρ that is positive semidefinite with trace 1. If ρ can be expressed as
where ∑ pi = 1, the set
is said to be an ensemble that describes the mixed state ρ. Notice {vi} is not required to be orthogonal. Different ensembles describing the state ρ are related by unitary operators, via the square roots of ρ. For instance, suppose
The trace 1 condition means
Let
and vi be the normalized ai. We see that
gives the mixed state ρ.
[edit] Operators on Hilbert space
In general, if A, B are closed and densely defined operators on a Hilbert space H, and A*A = B*B, then A = UB where U is a partial isometry.
[edit] Computing the matrix square root
One can also define the square root of a square matrix A that is not necessarily positive-definite. A matrix B is said to be a square root of A if the matrix product B · B is equal to A.
[edit] By diagonalization
The square root of a diagonal matrix D is formed by taking the square root of all the entries on the diagonal. This suggests the following methods for general matrices:
An n × n matrix A is diagonalizable if there is a matrix V such that D = V − 1AV is a diagonal matrix. This happens if and only if A has n eigenvectors which constitute a basis for Cn; in this case, V can be chosen to be the matrix with the n eigenvectors as columns.
Now, A = VDV − 1, and hence the square root of A is
This approach works only for diagonalizable matrices. For non-diagonalizable matrices one can calculate the Jordan normal form followed by a series expansion, similar to the approach described in logarithm of a matrix.
[edit] Denman–Beavers square root iteration
Another way to find the square root of a matrix A is the Denman–Beavers square root iteration. Let Y0 = A and Z0 = I, where I is the identity matrix. The iteration is defined by
The matrix Yk converges quadratically to the square root A1/2, while Zk converges to its inverse, A−1/2 (Denman & Beavers 1976; Cheng et al. 2001)
[edit] See also
[edit] References
- Cheng, Sheung Hun; Higham, Nicholas J.; Kenney, Charles S. & Laub, Alan J. (2001), “Approximating the Logarithm of a Matrix to Specified Accuracy”, SIAM Journal on Matrix Analysis and Applications 22 (4): 1112–1125, doi:10.1137/S0895479899364015, <http://www.eeweb.ee.ucla.edu/publications/journalAlanLaubajlaub_simax22(4)_2001.pdf>
- Denman, Eugene D. & Beavers, Alex N. (1976), “The matrix sign function and computations in systems”, Applied Mathematics and Computation 2 (1): 63–94, DOI 10.1016/0096-3003(76)90020-5