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Let matrix $C_{n\times‎n}$ and equation $X^{2}=C$ be given, i want to find matrix $X$.

For $n=2$, $X$ is obtained by solving a system of equations;

$$\left\{ \begin{array}{l} x_{11}^2 + {x_{12}}{x_{21}} = {c_{11}}\\ {x_{11}}{x_{12}} + {x_{12}}{x_{22}} = {c_{12}}\\ {x_{21}}{x_{11}} + {x_{22}}{x_{21}} = {c_{21}}\\ {x_{21}}{x_{12}} + x_{22}^2 = {c_{22}} \end{array} \right. $$

but does analytical method (without solve a system of equations) for solving such equations exist?

Thanks.

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1  
Diagonalization is a possible method. –  Seirios Jan 13 '13 at 17:59
    
@M.Sina firstly you should ask yourself is such a matrix $X$ always exists. I doubt it. Diagonalization won't help if $C$ isn't diagonalizable. –  Git Gud Jan 13 '13 at 18:01
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I second @GitGud's comment. What structure do you have on $C$? For example, $C=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ does not have a square root. –  copper.hat Jan 13 '13 at 18:01
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@M.Sina In this case, note that $\det(X^2)=(\det X)^2\geq 0$. So there is no solution for every $C$ such that $\det C<0$. –  1015 Jan 13 '13 at 18:31
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For $n=2$ , someone has done a complete analysis in tw.knowledge.yahoo.com/question/question?qid=1010020803771. –  doraemonpaul May 17 '13 at 0:46

2 Answers 2

If $C$ is hermitian positive definite, then there exists an unitary matrix $U$ such that $C=U^*DU$, where $D=\text{diag}(\lambda _1, \dots, \lambda _n)$ with $\lambda _1, \dots, \lambda _n$ being non-negative reals and also $\lambda _1, \dots, \lambda _n$ are the eigenvalues of $C$.

Define $\sqrt D = \text{diag}(\sqrt \lambda _1, \dots,\sqrt \lambda _n)$ and let $X=U^*\sqrt DU$.

Then $$\begin{align} X^2&=(U^*\sqrt DU)(U^*\sqrt DU)^*\\ &=(U^*\sqrt DU)(U^*\sqrt D^*(U^*)^*)\\ &=(U^*\sqrt DU)(U^*\sqrt DU)\\ &=U^*\sqrt DI\sqrt DU\\ &=U^*DU=C. \end{align}$$

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No, an "analytical method (without solve a system of equations)" for solving such equations does not exist. This is because any matrix equation is by definition a system of equations.

However, considering when a matrix is already diagonalized, finding the solution is very simple. Read ahead for some considerations of the non-diagonalizable case.

Every matrix may be orthogonaly reduced to triangular. Take then the equation $Y^2 = R$ where $R$ is triangular and $C=QRQ^*$ where $QQ^*=I$. Then the solution for $X$ may be found

$$X^2=C=QRQ^* = QY^2Q^*$$ So that $$ X = QYQ^*$$

So it comes down to finding for a triangular matrix. Lets look at the $3 \times 3$ case. $$\pmatrix{d_0 & 0 & 0 \\ a_1 & d_1 & 0 \\ a_2 & b_2 & d_2}\pmatrix{d_0 & 0 & 0 \\ a_1 & d_1 & 0 \\ a_2 & b_2 & d_2} = \pmatrix{d_0^2 & 0 & 0 \\ a_1(d_0 + d_1) & d_1^2 & 0 \\ a_2(d_0 + d_2) + a_1b_2 & b_2(d_1 + d_2) & d_2^2}$$

This would not necessarily give a result " analytical method (without solve a system of equations)" but it does show that the square root does exist, and there are $2^n$ different solutions, where $n$ is the number of non-zero eigenvalues.

EDIT: I will rollback if this is wrong as it is ripped from a text, but...

Functions of Jordan Blocks

For a $k \times k$ Jordan block $J_*$ with eigenvalue $\lambda$, and for a function $f(z)$ such that $f(\lambda), f'(\lambda), \dots, f^{(k-1)}(\lambda)$ exist, $f(J_*)$ is defined to be $$f(J_*) = f\pmatrix{\lambda & 1 \\ & \ddots & \ddots \\ & & \ddots & 1 \\ &&& \lambda}=\pmatrix{f(\lambda) & f'(\lambda) & \frac{f''(\lambda)}{2!} & \cdots & \frac{f^{(k-1)}(\lambda)}{(k-1)!}\\ & f(\lambda)& f'(\lambda)& \ddots & \vdots\\ & & \ddots & \ddots & \frac{f''(\lambda)}{2!}\\ & & & f(\lambda) & f'(\lambda)\\ & & & & f(\lambda)\\}$$

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@copper though I suppose your example shows the problem with this when $d_i=0$ for one or more $d_i$. $2^n=0$ in that case though, doesn't it, so no roots exist... –  adam W Jan 13 '13 at 18:18
    
Why did you assume the square root of a triangular matrix is also triangular? –  ACARCHAU Jan 13 '13 at 18:21
    
I assumed that the square of a triangular is also triangular. –  adam W Jan 13 '13 at 18:23
    
@ACARCHAU you are correct, the square root need not be triangular. See jspecter's nice answer to my question. –  adam W Jan 14 '13 at 2:42

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