Square root of a matrix

The square root of a matrix, or matrix root, is a linear algebra term that generalizes the concept of the square root of a real number . A square root of a square matrix is a matrix that, when multiplied by itself, results in the output matrix . A unique square root can be defined for symmetric positive semidefinite matrices. In general, however, a square root does not need to exist, nor does it need to be unique if it does exist.

Square root of a positive semi-definite matrix

definition

For a symmetric positive semidefinite matrix , a symmetric positive semidefinite matrix is ​​called the square root or, for short, the root of if ${\displaystyle A\in \mathbb {R} ^{n\times n}}$ ${\displaystyle B\in \mathbb {R} ^{n\times n}}$ ${\displaystyle A,}$

${\displaystyle B^{2}=B\cdot B=A}$

applies. The square root of is uniquely determined and is denoted by. ${\displaystyle A}$ ${\displaystyle A^{\frac {1}{2}}}$

presentation

The square root of is given as follows. According to the spectral theorem, there is an orthogonal matrix ${\displaystyle A}$

${\displaystyle Q=(v_{1}\mid \cdots \mid v_{n})\in \mathbb {R} ^{n\times n}}$

with pairwise orthonormal eigenvectors of as columns and a diagonal matrix ${\displaystyle v_{1},\ldots ,v_{n}\in \mathbb {R} ^{n}}$ ${\displaystyle A}$

${\displaystyle D=\operatorname {diag} (d_{1},\ldots ,d_{n})\in \mathbb {R} ^{n\times n}}$

with the eigenvalues associated with these eigenvectors on the diagonal , so that ${\displaystyle d_{1},\ldots ,d_{n}\in \mathbb {R} }$

${\displaystyle A=QDQ^{T}}$

applies. The square root of then results in ${\displaystyle A}$

${\displaystyle A^{\frac {1}{2}}=QD^{\frac {1}{2}}Q^{T},}$

where the diagonal matrix

${\displaystyle D^{\frac {1}{2}}={\text{diag}}\left({\sqrt {d_{1}}},\ldots ,{\sqrt {d_{n}}}\right)\in \mathbb {R} ^{n\times n}}$

has the square roots of the eigenvalues ​​of on the diagonal. Since the eigenvalues ​​of a positive semidefinite matrix are always real and nonnegative, their square roots can also be chosen to be real and nonnegative. ${\displaystyle A}$ ${\displaystyle A}$

example

The matrix

${\displaystyle A={\begin{pmatrix}5&4\\4&5\end{pmatrix}}}$

has the eigenvalues and and form the corresponding orthonormal basis from eigenvectors. So it applies ${\displaystyle \lambda _{1}=9}$ ${\displaystyle \lambda _{2}=1.}$ ${\displaystyle v_{1}={\tfrac {1}{\sqrt {2}}}{\begin{pmatrix}1\\1\end{pmatrix}}}$ ${\displaystyle v_{2}={\tfrac {1}{\sqrt {2}}}{\begin{pmatrix}1\\-1\end{pmatrix}}}$

${\displaystyle A={\frac {1}{\sqrt {2}}}{\begin{pmatrix}1&1\\1&-1\end{pmatrix}}{\begin{pmatrix}9&0\\0&1\end{pmatrix}}{\frac {1}{\sqrt {2}}}{\begin{pmatrix}1&1\\1&-1\end{pmatrix}}}$

summarized to

${\displaystyle A={\frac {1}{2}}{\begin{pmatrix}1&1\\1&-1\end{pmatrix}}{\begin{pmatrix}9&0\\0&1\end{pmatrix}}{\begin{pmatrix}1&1\\1&-1\end{pmatrix}}}$

and thus

${\displaystyle A^{\frac {1}{2}}={\frac {1}{2}}{\begin{pmatrix}1&1\\1&-1\end{pmatrix}}{\begin{pmatrix}3&0\\0&1\end{pmatrix}}{\begin{pmatrix}1&1\\1&-1\end{pmatrix}}={\begin{pmatrix}2&1\\1&2\end{pmatrix}}.}$

properties

The square of the matrix is the matrix ${\displaystyle A^{\frac {1}{2}}}$ ${\displaystyle A\colon }$

${\displaystyle A^{\frac {1}{2}}A^{\frac {1}{2}}=QD^{\frac {1}{2}}Q^{T}QD^{\frac {1}{2}}Q^{T}=QDQ^{T}=A}$

The matrix is symmetrical: ${\displaystyle A^{\frac {1}{2}}}$

${\displaystyle \left(A^{\frac {1}{2}}\right)^{T}=\left(QD^{\frac {1}{2}}Q^{T}\right)^{T}=QD^{\frac {1}{2}}Q^{T}=A^{\frac {1}{2}}}$

The matrix is positive semidefinite (the displacement property of the standard scalar product is used ): ${\displaystyle A^{\frac {1}{2}}}$

${\displaystyle \left\langle x,A^{\frac {1}{2}}x\right\rangle =\left\langle x,QD^{\frac {1}{2}}Q^{T}x\right\rangle =\left\langle D^{\frac {1}{4}}Q^{T}x,D^{\frac {1}{4}}Q^{T}x\right\rangle \geq 0}$

for all where . If positive is definite, then positive is definite. ${\displaystyle x\in \mathbb {R} ^{n},}$ ${\displaystyle D^{\frac {1}{4}}={\text{diag}}\left({\sqrt[{4}]{d_{1}}},\ldots ,{\sqrt[{4}]{d_{n}}}\right)}$ ${\displaystyle A}$ ${\displaystyle A^{\frac {1}{2}}}$

Square roots of arbitrary matrices

definition

The root of a square matrix is any matrix that, when multiplied by itself, results in: ${\displaystyle A}$ ${\displaystyle B,}$ ${\displaystyle A}$

${\displaystyle B{\text{ ist Wurzel von }}A\quad \Leftrightarrow \quad B^{2}=BB=A}$

There are also sources in which a root of is mentioned when applies. ${\displaystyle B}$ ${\displaystyle A}$ ${\displaystyle A=B\cdot B^{T}}$

A root of is also written . In this notation, however, it is unclear which root is meant, since several can exist. ${\displaystyle A}$ ${\displaystyle A^{\frac {1}{2}}.}$

Number of existing roots

As with the square root of real or complex numbers, the square root of a matrix is ​​generally ambiguous. Is about a root from then too ${\displaystyle A^{\frac {1}{2}}}$ ${\displaystyle A,}$ ${\displaystyle -A^{\frac {1}{2}}.}$

Unlike the root of a complex number, matrices can have more than two roots.

For example, matrices, whose characteristic polynomial breaks down into different linear factors in pairs, have up to different roots. ${\displaystyle n\times n}$ ${\displaystyle 2^{n}}$

There are even matrices with an infinite number of roots. The identity matrix , for example, has a root for every complex number . ${\displaystyle {\bigl (}{\begin{smallmatrix}1&0\\0&1\end{smallmatrix}}{\bigr )}}$ ${\displaystyle {\bigl (}{\begin{smallmatrix}1&z\\0&-1\end{smallmatrix}}{\bigr )}}$ ${\displaystyle z}$

There are also matrices for which there is no root at all: An example is ${\displaystyle A={\bigl (}{\begin{smallmatrix}0&1\\0&0\end{smallmatrix}}{\bigr )}.}$

Geometric interpretation of roots

Looking at the array as a linear transformation , that is, as a mapping between vector spaces through which a vector , a vector is assigned, then a root is a transformation that is twice in succession must perform in order to convert. ${\displaystyle A}$ ${\displaystyle {\vec {v}}}$ ${\displaystyle {\vec {v}}'=A{\vec {v}}}$ ${\displaystyle A^{\frac {1}{2}}}$ ${\displaystyle {\vec {v}}}$ ${\displaystyle {\vec {v}}'}$

Example:

${\displaystyle A}$be the two-dimensional rotation matrix with the angle ${\displaystyle \alpha \colon }$

${\displaystyle A={\begin{pmatrix}\cos \alpha &-\sin \alpha \\\sin \alpha &\cos \alpha \end{pmatrix}}}$

Then every rotation matrix belonging to an angle with an integer is a square root of For with the first multiplication of a vector with a rotation by half the angle and with the second multiplication again. ${\displaystyle (t\cdot 360^{\circ }+\alpha )/2}$ ${\displaystyle t}$ ${\displaystyle A.}$ ${\displaystyle t=0}$ ${\displaystyle {\vec {v}}}$ ${\displaystyle A^{\frac {1}{2}}}$ ${\displaystyle \alpha /2}$

Calculating a Root

One can easily determine the roots of a matrix of size if it is a diagonal matrix or can at least be converted into a diagonal form (see diagonalization ). ${\displaystyle A}$ ${\displaystyle n\times n}$ ${\displaystyle A}$

Case 1: diagonal matrix

If the diagonal entries of a diagonal matrix are different in pairs, then all the roots of the diagonal matrix can be determined simply by determining a root from each entry on the main diagonal . If you denote the diagonal entries of as usual with , you get the roots of the matrices ${\displaystyle A}$ ${\displaystyle a_{11},\dotsc ,a_{nn}}$ ${\displaystyle A}$

${\displaystyle {\begin{pmatrix}\pm {\sqrt {a_{11}}}&0&\cdots &0\\0&\pm {\sqrt {a_{22}}}&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\0&0&\dots &\pm {\sqrt {a_{nn}}}\end{pmatrix}}}$

For each of the diagonal elements, the sign can be chosen arbitrarily, so that one obtains different roots in pairs, if all diagonal entries are different from zero. If a diagonal entry is zero, then one obtains roots that are different in pairs. Since the matrix can also have negative values ​​on the diagonal, the roots can also contain complex numbers. ${\displaystyle n}$ ${\displaystyle 2^{n}}$ ${\displaystyle 2^{n-1}}$ ${\displaystyle A}$

It should be noted that there can be further roots if the diagonal entries are not different in pairs. However, these are then not diagonal matrices. For example, the identity matrix has an infinite number of roots, as already explained above . Diagonal matrices with negative diagonal entries can in this case also have real roots. For example:

${\displaystyle {\begin{pmatrix}0&1\\-1&0\end{pmatrix}}^{2}={\begin{pmatrix}-1&0\\0&-1\end{pmatrix}}}$

Case 2: Diagonalizable matrix

If the matrix can be diagonalized, the roots of can be found in the following way : ${\displaystyle A}$ ${\displaystyle A}$

First one determines an invertible matrix and a diagonal matrix so that . The matrix then has eigenvectors of the matrix as columns and the matrix has the associated eigenvalues as diagonal entries . ${\displaystyle T}$ ${\displaystyle D}$ ${\displaystyle A=TDT^{-1}}$ ${\displaystyle T}$ ${\displaystyle A}$ ${\displaystyle D}$

If now is a root of then is a root of the matrix , because the following applies: ${\displaystyle D^{\frac {1}{2}}}$ ${\displaystyle D,}$ ${\displaystyle A^{\frac {1}{2}}:=TD^{\frac {1}{2}}T^{-1}}$ ${\displaystyle A}$

${\displaystyle (TD^{\frac {1}{2}}T^{-1})^{2}=TD^{\frac {1}{2}}T^{-1}TD^{\frac {1}{2}}T^{-1}=TD^{\frac {1}{2}}D^{\frac {1}{2}}T^{-1}=TDT^{-1}=A}$

Since is a diagonal matrix, one obtains possible roots as in case 1. Here, too, it should be noted that some eigenvalues ​​of the diagonal matrix can be negative, the roots of which are then complex. If the matrix pairwise distinct eigenvalues, are also obtained as in case 1 or different solutions. ${\displaystyle D}$ ${\displaystyle n}$ ${\displaystyle 2^{n}}$ ${\displaystyle 2^{n-1}}$

Case 3: Non-diagonalizable matrix

If the matrix cannot be diagonalized, the method shown cannot be used to calculate a root. However, this does not mean that it does not have a root: For example, the shear matrix cannot be diagonalized, but it does have the root ${\displaystyle A}$ ${\displaystyle A}$ ${\displaystyle A={\bigl (}{\begin{smallmatrix}1&2\\0&1\end{smallmatrix}}{\bigr )}}$ ${\displaystyle {\bigl (}{\begin{smallmatrix}1&1\\0&1\end{smallmatrix}}{\bigr )}.}$

If we allow complex numbers in arithmetic, every matrix can be transformed into Jordanian normal form , even if it cannot be diagonalized. ${\displaystyle A}$

One determines matrices their inverse and with where has the following block diagonal form: ${\displaystyle Q,}$ ${\displaystyle Q^{-1}}$ ${\displaystyle J}$ ${\displaystyle J=Q^{-1}AQ,}$ ${\displaystyle J}$

${\displaystyle J={\begin{pmatrix}J_{1}&&0\\&\ddots &\\0&&J_{k}\end{pmatrix}}}$

They are Jordan blocks of shape ${\displaystyle J_{i}}$

${\displaystyle J_{i}={\begin{pmatrix}\lambda _{i}&1&&&0\\&\lambda _{i}&1&&\\&&\ddots {}&\ddots {}\\&&&\lambda _{i}&1\\0&&&&\lambda _{i}\end{pmatrix}}.}$

A square root of is calculated according to ${\displaystyle A}$

${\displaystyle A=A^{\frac {1}{2}}A^{\frac {1}{2}}=QJQ^{-1}=QJ^{\frac {1}{2}}J^{\frac {1}{2}}Q^{-1}=(QJ^{\frac {1}{2}}Q^{-1})(QJ^{\frac {1}{2}}Q^{-1})\Rightarrow A^{\frac {1}{2}}=QJ^{\frac {1}{2}}Q^{-1}.}$

The root of is to be taken from each Jordan block individually. ${\displaystyle J}$ ${\displaystyle J_{i}}$

If true, the power of a Jordan block is through ${\displaystyle \lambda _{i}\,\neq \,0}$ ${\displaystyle J_{i}^{\beta }}$ ${\displaystyle J_{i}}$

${\displaystyle J_{i}^{\beta }={\begin{pmatrix}\alpha _{i0}&\alpha _{i1}&\alpha _{i2}&\alpha _{i3}\\0&\alpha _{i0}&\alpha _{i1}&\alpha _{i2}\\0&0&\alpha _{i0}&\alpha _{i1}\\0&0&0&\alpha _{i0}\end{pmatrix}}}$

given with where the -th derivative of the power function is. This results explicitly and where the size of the Jordan block is denoted by (in the illustration ), the subdiagonals by ( is the diagonal) and the gamma function by . Set for the square root . ${\displaystyle \alpha _{ij}={\tfrac {1}{j!}}f^{(j)}(\lambda _{i}),}$ ${\displaystyle f^{(j)}}$ ${\displaystyle j}$ ${\displaystyle f(x)=x^{\beta }}$ ${\displaystyle \alpha _{ij}={\frac {\Gamma (\beta +1)\lambda _{i}^{\beta -j}}{\Gamma (j+1)\Gamma (\beta -j+1)}}}$ ${\displaystyle j=0,1,\dotsc ,(m_{i}-1),}$ ${\displaystyle J_{i}}$ ${\displaystyle m_{i}}$ ${\displaystyle m_{i}=4}$ ${\displaystyle j}$ ${\displaystyle j=0}$ ${\displaystyle \Gamma }$ ${\displaystyle \beta =1/2}$

For example, for ${\displaystyle m_{i}=2}$

${\displaystyle J_{i}^{\frac {1}{2}}={\begin{pmatrix}{\sqrt {\lambda _{i}}}&{\frac {1}{2{\sqrt {\lambda _{i}}}}}\\0&{\sqrt {\lambda _{i}}}\end{pmatrix}}.}$

If and at the same time , the root of the Jordan block does not exist. ${\displaystyle \lambda _{i}=0}$ ${\displaystyle m_{i}>1}$ ${\displaystyle J_{i}}$

Outside the Jordan blocks there are all zeros.

If so, the number has two roots, so you get two different roots for each Jordan block this way . Combination creates roots, where the number of Jordan blocks denotes. ${\displaystyle \lambda _{i}\neq 0,}$ ${\displaystyle \lambda _{i}}$ ${\displaystyle J_{i}}$ ${\displaystyle 2^{k}}$ ${\displaystyle k}$ ${\displaystyle J_{i}}$

This technique generally gives only some, and not all, of the square roots of a matrix.

See also

• Matrix potency
• Matrix exponential
• Matrix Riccati equation , also contains an iteration method for the square root

Web links

• Eric W. Weisstein : Matrix Square Root . In: MathWorld (English).
• rspuzio: Square root of positive definite matrix . In: PlanetMath . (English)

Individual evidence

1. ↑ ^{a b} Christian Kanzow: Numerics of linear systems of equations. Direct and iterative procedures. Springer-Verlag, Berlin / Heidelberg / New York 2005, pp. 13–15.