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square root of a matrix: Information from Answers.com

️Wed Jul 01 2015

In mathematics, the square root of a matrix extends the notion of square root from numbers to matrices.

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 T^½ such that T^½ is itself positive and (T^½)² = T. The operator T^½ is the unique positive square root of T.

A bounded positive operator on a complex Hilbert space is self adjoint. So T = (T^½)* T^½. 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.

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 = T^½ to be the unique positive square root of T. It suffices to show that, for any square root A of T, A = UB. Let {a_i}_{1 ≤ i ≤ n}, and {b_i}_{1 ≤ i ≤ n} be the set of column vectors of A and B, respectively. By the construction of the square root, {b_i}_{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 {b_i}_{1 ≤ i ≤ n} is linearly independent and spans Cⁿ.

The equality AA* = BB* can be rewritten as

$\sum_j a_j a_j ^* = \sum_i b_i b_i ^*.$

By completeness of {b_i}, for all j, there exists n scalars, by appending zeros if necessary, {u_{i j}}_{1 ≤ i ≤ n} s.t.

$a_j = \sum _i u_{ij} b_i \,$

i.e.

$\begin{bmatrix} a_1 & \cdots & a_n \end{bmatrix} = \begin{bmatrix} b_1 & \cdots & b_n \end{bmatrix} \begin{bmatrix} u_{11} & \cdots & u_{1n} \\ \vdots & \ddots & \vdots \\ u_{n1} & \cdots & u_{nn} \end{bmatrix} .$

We need to show the matrix U = [u_{i j}] is unitary. Compute directly

$a_j a_j ^* = \sum _i u_{ij} b_i \cdot \sum _k {\bar u_{kj} } b_k ^* = \sum _{i k} u_{ij} {\bar u_{kj} } b_i b_k ^*.$

$\sum_j a_j a_j ^* = \sum_j \sum _{i k} u_{ij} {\bar u_{kj} } b_i b_k ^* = \sum_{i k} \sum _{j } u_{ij} {\bar u_{kj} } b_i b_k ^* = \sum_{i k} \alpha_{i k} b_i b_k ^*.$

By assumption,

$\sum_j a_j a_j ^* = \sum_{i k} \alpha_{i k} b_i b_k ^* = \sum_i b_i b_i^*.$

Now because {b_i} is a basis of Cⁿ, the set {b_i 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

$\alpha_{i k} = \sum _{j } u_{ij} {\bar u_{kj} } = \delta _{i k}.$

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 {b_i} is linearly independent. In that case, the existence of the rectangular matrix U = [u_i
j] follows from the relation

$\sum_j a_j a_j ^* = \sum_i b_i b_i ^*,$

rather than the completeness of {b_i}. 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.)

Some applications

The unitary freedom of square roots has applications in linear algebra.

Kraus operators

By Choi's result, a linear map

Φ:C^n×n→C^m×m

is completely positive if and only if it is of the form

$\Phi(A) = \sum_i ^k V_i A V_i^*$

where k ≤ nm. Let {E_{p q}} ⊂ C^{n × n} be the n² elementary matrix units. The positive matrix

$M_{\Phi} = ( \Phi (E_{pq}) )_{pq} \in C^{nm \times nm}$

is called the Choi matrix of Φ. The Kraus operators correspond to the, not necessarily square, square roots of M_Φ: For any square root B of M_Φ, one can obtain a family of Kraus operators V_i by undoing the Vec operation to each column b_i of B. Thus all sets of Kraus operators are related by partial isometries.

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

$\rho = \sum_i p_i v_i v_i^*$

where ∑ p_i = 1, the set

$\{p_i, v_i \} \,$

is said to be an ensemble that describes the mixed state ρ. Notice {v_i} 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

$\rho = \sum_j a_j a_j^*.$

The trace 1 condition means

$\sum_j a_j ^* a_j = 1.$

Let

$p_i = a_i ^* a_i,$

and v_i be the normalized a_i. We see that

$\{p_i, v_i \} \,$

gives the mixed state ρ.

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.

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.

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 ^{- 1}AV 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

$A^{1/2} = V D^{1/2} V^{-1}. \,$

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.

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

Failed to parse (unknown function\begin): \begin{align} Y_{k+1} &= \tfrac12 (Y_k + Z_k^{-1}), \\ Z_{k+1} &= \tfrac12 (Z_k + Y_k^{-1}). \end{align}

Both Y_k and Z_k converge to the square root A^{1 / 2}.

References

Sheung Hun Cheng, Nicholas J. Higham, Charles S. Kenney, and Alan J. Laub, Approximating the Logarithm of a Matrix to Specified Accuracy, SIAM Journal on Matrix Analysis and Applications, vol. 22 (2001), no. 4, pp. 1112–1125.

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