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

\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 ≤ 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 = [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 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

Main article: Choi's theorem on completely positive maps

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 {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.

[edit] Mixed ensembles

Main article: Density matrix

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.

Let

p_i = a_i ^* a_i,

and vi be the normalized ai. We see that

\{p_i, v_i \} \,

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] Numerical analysis

One can also define the square root of a n × n, not necessarily positive, matrix A. A matrix B is said to be a square root of A if the matrix product B B is equal to A.

[edit] Numerical method

Given a square matrix A, one way to find its square root is the Denman-Beavers square root iteration, described below:

Given matrix A, and the goal being to find A1/2:
Let Y0 = A and Z0 = I, where I is the identity matrix.
Iterate:
Y_{k+1} = (Y_k + Z_k^{-1})/2
Z_{k+1} = (Z_k + Y_k^{-1})/2,
then
\lim_{k\rightarrow \infty} Y_k = A^{1/2}.

[edit] Diagonalizable matrices

A n × n matrix A is diagonalizable if and only if the eigenvectors of A constitute a basis for Cn.

When A is diagonalizable, A1/2 can be computed in a natural way:

1. Find the matrix V such that the columns of V are eigenvectors of A.
2. Find the inverse V−1 of V.
3. Let
A' = V^{-1} \cdot A \cdot V.
The matrix A' will be a diagonal matrix whose diagonal elements are eigenvalues of A. (This is the diagonalization of A.)
4. Replace each diagonal element of A' by its square root to obtain
\sqrt{A'} ,
then
\sqrt{A} = V \cdot \sqrt{A'} \cdot V^{-1}.

On a graphing calculator, this method is more efficient than the previous one.

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] See also

[edit] References

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