Suppose that X and A are matrices and A is positive definite. (A is a positive definite matrix if x’Ax > 0 for every nonzero vector x.) The matrix X is called the square-root of A if X2 = A. Here, X2 is the multiplication of X by itself, that is X2 = X⋅X. Furthermore, the square-root of A is denoted by A1/2. How to find the square-root of a matrix? This article addresses the problem.

Let A be a k × k positive definite matrix with the following spectral decomposition.

where (\lambda_i,\vec{e}_i) are eigenvalue-eigenvector pairs of A with the properties that all eigenvectors have norm 1 and they are orthogonal to each other. By defining the matrix P= (\vec{e}_1 \quad \vec{e}_2 \quad \cdots \quad \vec{e}_k) and the matrix \Lambda = [\lambda_{ij}] where \lambda_{ij} = \begin{Bmatrix} 0 ; \: i \neq j \\ \lambda_i ; \: i = j \end{matrix}, the decomposition can also be expressed as:

A = P \Lambda P'

It can be proved that the matrix defined by A^{1/2}= \sum_{i=1}^k \sqrt{\lambda_i} \vec{e}_i \cdot \vec{e}_i \: ' is a solution to the matrix equation X2 = A. (In other words, A1/2⋅A1/2 = A.) Additionally, if we define the matrix \Lambda^{1/2}= [l_{ij}] where l_{ij} = \begin{Bmatrix} 0 ; \: i \neq j \\ \sqrt{\lambda_i} ; \: i = j \end{matrix}, then A1/2 can also be expressed as:

A^{1/2}=P \Lambda^{1/2}P'

Some of the properties of the square-root matrix are as follows.

  1. A1/2 is a symmetric matrix.
  2. The inverse of the matrix A1/2, which is (A^{1/2})^{-1} and denoted by A-1/2, can be determined by the following formula: A^{-1/2} =P \Lambda^{-1/2}P' where \Lambda^{-1/2}=[l_{ij}] and l_{ij} = \begin{Bmatrix} 0 ; \: i \neq j \\ \frac{1}{\sqrt{\lambda_i}} ; \: i = j \end{matrix}.
  3. The square of A-1/2 is the inverse of A. In other words, (A^{-1/2})^2 = A^{-1/2}\cdot A^{-1/2} = A^{-1}

 

Example
Given a positive definite matrix A = \begin{pmatrix}2 & -2 \\ -2 & 5 \end{pmatrix}, find A1/2, A-1/2, and A-1.

Answer
The eigenvalues ​​of A are λ1 = 6 and λ2 = 1. For λ1 = 6, we get \vec{e}_1 = \begin{pmatrix}1/\sqrt{5} \\ -2/\sqrt {5} \end{pmatrix} and for λ2 = 1 we have \vec{e}_2 = \begin{pmatrix}2/\sqrt{5} \\ 1/\sqrt{5} \end{pmatrix}. From \vec{e}_1, \: \vec{e}_2, λ1, and λ2 we get the matrices P = \begin{pmatrix}1/\sqrt{5} & 2 /\sqrt{5} \\ -2/\sqrt{5} & 1/\sqrt{5} \end{pmatrix} and \Lambda = \begin{pmatrix}6 & 0 \\ 0 & 1 \end{pmatrix}. As a consequence, the spectral decomposition of A is as follows.

To find A1/2, the first step is to calculate Λ1/2, that is \Lambda^{1/2} = \begin{pmatrix}\sqrt{6} & 0 \\ 0 & \sqrt{1} \end{pmatrix} = \begin{pmatrix}\sqrt{6} & 0\\ 0 & 1 \end{pmatrix}. This results in the following square-root matrix.

 

To determine A-1/2, the first step is to compute Λ-1/2, i.e. \Lambda^{-1/2} = \begin{pmatrix}\frac{1}{\sqrt{6}} & 0 \\ 0 & \frac{1}{\sqrt{1}} \end{pmatrix} = \begin{pmatrix}\frac{1}{\sqrt{6}} & 0\\ 0 & 1 \end{pmatrix}. Therefore, we obtain the square-root matrix:

To determine A-1, we can use the formula above: A-1/2 ⋅ A-1/2 = A-1

It follows that

 

Note about finding A-1

Indeed there are other ways to determine A-1, but in this article we have deliberately applied A^{-1/2} \cdot A^{-1/2} = A^{-1} just to demonstrate how to use the formula. Furthermore, if A has the spectral decomposition A = PΛP’, its inverse can also be expressed as A^{-1} = \sum_{i=1}^k \frac{1}{\lambda_i} \vec{ e}_i \cdot \vec{e}_i \: ' or A-1 = PΛ-1P’ where \Lambda^{-1}=[l_{ij}] and l_{ij} = \begin{Bmatrix} 0 ; \: i \neq j \\ \frac{1}{\lambda_i} ; \: i = j \end{matrix}.

If A-1 in the example above is determined using the formula A-1 = PΛ-1P’, then \Lambda^{-1} = \begin{pmatrix}\frac{1}{6} & 0 \\ 0 & \frac{1}{1} \end{pmatrix} = \begin{pmatrix}\frac{1}{6} & 0\\ 0 & 1 \end{pmatrix} and it follows that:

On

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