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Square root of a diagonalizable matrix $A$ can be otained as $$B=\sqrt{A}=D \Lambda D^{-1} ~~~~(1),$$ where $$\Lambda=\begin{pmatrix} \lambda_1^{1/2} & 0 \\ 0 & \lambda_2^{1/2} \end{pmatrix}.$$ Here, $\lambda_{1,2}$ are the eigenvalues and $D$ is the diagonalizing matrix od $A$. Recently, in a question:

Solving a matrix equation of four unknowns

Square root of $$P=\begin{pmatrix} 4 & 5 \\ 3 & 6 \end{pmatrix},$$ was required such that $P=Q^2$. The eigenvalues of $P$ are $9,1$, the method (1) and also the the command `MatrixPower[P, 1/2]' of Mathematica gives $$Q=\frac{1}{4}\begin{pmatrix} 7 & 5 \\ 3 & 9 \end{pmatrix}.~~~~(2)$$ However a brute force method of construction $Q$ from four unknowns $a,b,c,d$ and solutions of the equations thereby along with the conditions that $Trace (P)=Trace (Q^2)$ and $\det(P)=\det(Q^2)$ yields a matrix $$Q= \frac{1}{2} \begin{pmatrix} 1 & 5 \\ 3 & 3 \end{pmatrix}.~~~~(3)$$ Can one explain why does the method in (1) and the Mathematica igonres (3)? How to resolve this ?

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  • $\begingroup$ The matrix square root isn't unique, as your example clearly illustrates (it is true that a symmetric and positive definite matrix has a unique symmetric and positive definite square root, but that breaks when the matrices aren't symmetric.) What is your actual question here? $\endgroup$ – Brian Borchers Jun 22 '19 at 3:56
  • $\begingroup$ @Brain Borchers, Thanks, the question is how does one get (3) from the method (1).? $\endgroup$ – Z Ahmed Jun 22 '19 at 4:08
  • $\begingroup$ To underscore @BrianBorchers comment, how many matrices $A$ are there such that $A^2=I$? $\endgroup$ – amd Jun 22 '19 at 4:39
  • $\begingroup$ @BrianBrochers You may also see dan_filea's solution. $\endgroup$ – Z Ahmed Jun 22 '19 at 6:50
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Let us diagonalize explicitly. $$ \begin{bmatrix} 4&5\\3&6 \end{bmatrix} = \begin{bmatrix} 5 & 5 \\ 5 & -3 \end{bmatrix} \begin{bmatrix} 9&\\&1 \end{bmatrix} \begin{bmatrix} 5 & 5 \\ 5 & -3 \end{bmatrix}^{-1} \ , $$ A square root of a slightly more general shape is then: $$ Q= \begin{bmatrix} 5 & 5 \\ 5 & -3 \end{bmatrix} \begin{bmatrix} \pm 3&\\&\pm 1 \end{bmatrix} \begin{bmatrix} 5 & 5 \\ 5 & -3 \end{bmatrix} ^{-1} \ . $$ Then the four matrices obtained in this way are: $$ \left[\begin{array}{rr} -\frac{7}{4} & -\frac{5}{4} \\ -\frac{3}{4} & -\frac{9}{4} \end{array}\right] \ ,\ \left[\begin{array}{rr} -\frac{1}{2} & -\frac{5}{2} \\ -\frac{3}{2} & -\frac{3}{2} \end{array}\right] \ ,\ \left[\begin{array}{rr} \frac{1}{2} & \frac{5}{2} \\ \frac{3}{2} & \frac{3}{2} \end{array}\right] \ ,\ \left[\begin{array}{rr} \frac{7}{4} & \frac{5}{4} \\ \frac{3}{4} & \frac{9}{4} \end{array}\right] \ . $$

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  • $\begingroup$ #dan_fulea, Yes, this answers it perfectly well. You also add some more educative remarks in this. one:. $\endgroup$ – Z Ahmed Jun 22 '19 at 4:16

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