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So I tried to solve this matrix and I got that X has no solutions: $$\begin{pmatrix} i &-i \\ 0 & 1 \end{pmatrix}X^2\begin{pmatrix} -i &i \\ 1& 0 \end{pmatrix}=-\begin{pmatrix} 1 &i \\ 0 & -i \end{pmatrix}$$ Can you tell me if I am right or not? Ok so I did multiply that with the inverses of those matrices and I got that $X=\begin{pmatrix} a&b \\ c &d \end{pmatrix}=>X^2=\begin{pmatrix} a^2+bc &ab+bd \\ ac+cd& bc+d^2 \end{pmatrix}=\begin{pmatrix} 1+i& 2i-1\\ 1& i \end{pmatrix}=A=>AX=XA=>b=c(2i-1);a=c+d$

if we substitute what we got in $X^2=A$ and we do some operations and solve that system we get that $-3c^4-2c^2+1+i(8c^4-4c^2)=0 $ which is 0 if the real part and the imaginary part are 0 so we solve the system $-3c^4-2c^2+1=0$ and $8c^4-4c^2=0$ which has no solution. Am I right?

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  • $\begingroup$ How 'bout you tell us the approach you used, and show some details? Then we can help you better understand what you did wrong (or right). $\endgroup$ – John Hughes Jan 5 '17 at 17:26
  • $\begingroup$ @JohnHughes I edited the description with what I did. $\endgroup$ – Ghost Jan 6 '17 at 14:52
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Write the equation as $AX^2 B = C$, then $X^2 = A^{-1} C B^{-1} = D=\begin{bmatrix} 1+i & -1+2i \\ 1 & i\end{bmatrix}$.

$D$ has distinct eigenvalues, hence is diagonalisable, hence it has a (not unique) square root $X$ such that $X^2 = D$.

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