Definitions

# 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.

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

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 = [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 ^*.

So


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

$Phi : C^\left\{n times n\right\} rightarrow C^\left\{m times m\right\}$

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

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

where knm. Let {Ep q} ⊂ Cn × n be the n2 elementary matrix units. The positive matrix

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

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

$\left\{p_i, v_i \right\} ,$

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

$sum_j a_j ^* a_j = 1.$

Let

$p_i = a_i ^* a_i,$

and vi be the normalized ai. We see that

$\left\{p_i, v_i \right\} ,$

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 $D = V^\left\{-1\right\} A V$ 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, $A = V D V^\left\{-1\right\}$, and hence the square root of A is

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

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

begin\left\{align\right\}
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 ()