groove of nail matrix

Square root of a matrix

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 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.

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 ≤ in, and {bi}1 ≤ in be the set of column vectors of A and B, respectively. By the construction of the square root, {bi}1 ≤ in 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 ≤ in 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 {bi}, for all j, there exists n scalars, by appending zeros if necessary, {ui j}1 ≤ in s.t.

a_j = sum _i u_{ij} b_i ,


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 = [ui 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 {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

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

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

rather than the completeness of {bi}. The conclusion UU* = I still holds. In general, B is n × m for some mn. A is n × k where mkn. 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

Phi : C^{n times n} rightarrow C^{m times m}

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

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

where knm. Let {Ep q} ⊂ Cn × 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 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.

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 ∑ pi = 1, the set

{p_i, v_i } ,

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

rho = sum_j a_j a_j^*.

The trace 1 condition means

sum_j a_j ^* a_j = 1.


p_i = a_i ^* a_i,

and vi be the normalized ai. 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} A V 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 = V D V^{-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

Y_{k+1} &= tfrac12 (Y_k + Z_k^{-1}), Z_{k+1} &= tfrac12 (Z_k + Y_k^{-1}). end{align} The matrix Y_k converges quadratically to the square root A1/2, while Z_k converges to its inverse, A−1/2 ()

See also


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