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Here is $3\times3$ matrix$$\begin{pmatrix} 1& -18& 0\\ 0 & 4& 0\\ -8& -13 & 9\end{pmatrix}$$ How can I find two different matrices so that $R^2=A$?

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  • $\begingroup$ Reduce to RREF first? $\endgroup$ – pie314271 Jan 19 '17 at 15:03
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    $\begingroup$ Diagonalise? If $R$ works so does $-R$. $\endgroup$ – JP McCarthy Jan 19 '17 at 15:04
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Hint

Finding a square root is easy for a diagonal matrix. If you can write $A$ as $PDP^{-1}$ with $D$ a diagonal matrix and if $S^2 = D$, then $R=PSP^{-1}$ is a square root of $A$.

Your matrix is diagonalizable and has eigenvalues $1$, $4$ and $9$.

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    $\begingroup$ Interestingly, the eigenvalues are perfect squares, so this is a well crafted exercise, and your solution is for sure interesting and...doable. I just got myself a nice exercise for my poor hardworking students...+1 $\endgroup$ – imranfat Jan 19 '17 at 15:12
  • $\begingroup$ Well they needn't be for this method to work but indeed, this probably isn't a coincidence... :-). $\endgroup$ – StackTD Jan 19 '17 at 15:14
  • $\begingroup$ It's a nice application of diagonalization which I hadn't thought off.... $\endgroup$ – imranfat Jan 19 '17 at 15:15
  • $\begingroup$ You're welcome @Ioanah. $\endgroup$ – StackTD Jan 20 '17 at 11:07

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