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$A= \begin{pmatrix} 1 & 2+3i \\ 2-3i & -1 \end{pmatrix}$
What is matrix $B$ such that $B^2=A$?

Its eigenvalues are $\sqrt{14}, -\sqrt{14}$
and I tried to use formula $B=U\sqrt{\lambda}U^{H}$ where $U$ is unitary matrix.
But then $\sqrt{ }$ of $-\sqrt{14}$ is not possible.

How can I solve this?

  • 0
    In addition to the other answers, you can consider taking the square root of A using the algorithm here [ http://en.wikipedia.org/wiki/Square_root_of_a_2_by_2_matrix ]2012-12-12

2 Answers 2

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The characteristic polynomial of $A$ is $\lambda^2 - 14$, so $A^2 - 14 I = 0$. Thus $(a A + bI)^2 = (14 a^2 + b^2) I + 2 a b A$. To make this equal to $A$, we want $2ab = 1$ and $14 a^2 + b^2 = 0$. Thus $b = 1/(2a)$ and $14 a^2 = -1/(4a^2)$, or $a^4 = -1/56$. Take $a$ to be any of the four fourth roots of $-1/56$, and you have a solution.

  • 0
    I see, you did that by applying the teorem of Cayley-Hamilton2017-06-30
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You can always use the "brute force" approach: let $B=\pmatrix{r&s\cr t&u\cr}$ calculate $B^2$, set it equal to $A$, and solve the resulting (nonlinear) system of four equations in four unknowns.