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$A=\left[ \begin{array}{ccc} 4 & 0 & 0 \\ 0 & 1 & i \\ 0 & -i & 1 \end{array} \right]$

Find a matrix $ B $ such that $B^*B$ =$A$ (Star means conjugate transpose of $B$)

I think that $A$ is hermitian, and so $A^* =A$ Also, We can edit this as $BB^*=A$. But I couldn't solve it completely.. Thank you

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First of all, you can't take the Hermitian of a product of matrices like that.

$$(B^*B)^*=B^*B\neq BB^* $$

You can examine the $4$ and $\left( \begin{array}{cc} 1 & i \\ -i & 1 \\ \end{array} \right)$ separately since the off diagonal elements are zero. I was able to simply guess which matrix would give $\left( \begin{array}{cc} 1 & i \\ -i & 1 \\ \end{array} \right)$ when multiplied with its conjugate transpose.

$$B=\left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & -i & 1 \\ \end{array} \right)$$

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  • $\begingroup$ I don't understand the part of finding B in the last step $\endgroup$ – Bener Gül May 24 '15 at 17:22
  • $\begingroup$ Neither did I, I just guessed it. $\endgroup$ – grdgfgr May 24 '15 at 17:28

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