Square root of a matrix

From Wikipedia, the free encyclopedia

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

1 Square roots of positive operators
2 Numerical analysis
- 2.1 Numerical method
- 2.2 Diagonalizable matrices
3 See also
4 References

[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 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.

[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 = 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.)

[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 {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.

[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 ∑ 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.

[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 A^1/2:

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, A^1/2 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

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.

Retrieved from "http://en.wikipedia.org../../../s/q/u/Square_root_of_a_matrix.html"

Category: Matrix theory