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In mathematics, the square root of a matrix extends the notion of square root from numbers to matrices.
## Square roots of positive operators

### Unitary freedom of square roots

### Some applications

#### Kraus operators

#### Mixed ensembles

### Operators on Hilbert space

## Computing the matrix square root

### By diagonalization

### Denman–Beavers square root iteration

^{1/2}, while $Z\_k$ converges to its inverse, A^{−1/2} ()
## See also

## References

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.

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 C^{n}. 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 C^{n}.

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.

- $$

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

- $$

So

- $$

By assumption,

- $$

Now because {b_{i}} is a basis of C^{n}, 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

- $$

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.)

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

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 k ≤ nm. Let {E_{p q}} ⊂ C^{n × n} be the n^{2} 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.

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.

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.

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.

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 C^{n}; 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.

Another way to find the square root of a matrix A is the Denman–Beavers square root iteration. Let Y_{0} = A and Z_{0} = I, where I is the identity matrix. The iteration is defined by

- $begin\{align\}$

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Last updated on Thursday August 07, 2008 at 08:20:30 PDT (GMT -0700)

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