I have the following matrix: $$\begin{pmatrix} 0 & -7 & 1 \\ 0 & 4 & 0 \\ -2 & 1 & 3 \\ \end{pmatrix}$$ I already found the eigenvalues which are $$\lambda_1=1$$ $$\lambda_2=2$$ $$\lambda_3=4$$ They are distinct so the matrix is clearly diagonalizable. I know a matrix is similar to its diagonal matrix so we can write this as follows: $$A= PDP^{-1}$$ We also have that : $$A^n=PD^nP^{-1}$$ Now is it possible to find $B$ such that $B^2=A$ ? I have been trying to use the fact that $A$ is similiar to its diagonal, but i do not know if i have to use that or not.



Try to compute $(PD^\frac12P^{-1})^2$

  • $\begingroup$ Ohhh nice i can see it now :D. Does it work for a matrix with real coefficients too? $\endgroup$ – TheNicouU Mar 23 '18 at 5:20
  • $\begingroup$ I thought your current matrix has real coefficients? $\endgroup$ – Siong Thye Goh Mar 23 '18 at 5:24
  • $\begingroup$ Yes it has, sorry that was a silly question! Thanks for your help :) $\endgroup$ – TheNicouU Mar 23 '18 at 5:30
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    $\begingroup$ I see, it need not be the case, for example, let $A=-1$, unless you want to work with complex number, it doesn't exist for $n=2$. $\endgroup$ – Siong Thye Goh Mar 23 '18 at 5:30
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    $\begingroup$ sure, I was referring to the scalar $-1$ having no real square root. $\endgroup$ – Siong Thye Goh Mar 23 '18 at 16:30

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