<s>[INST] Prove the existence of complex matrix that satisfies the following conditions. B is a 2 by 2 complex matrix. $B=0$ or $B^2 \neq 0$. Prove that a complex matrix C exists such that $C^2=B$.
I cannot understand what this problem wants me to do. "$B=0$ or $B^2 \neq 0$". Should I deal with the two cases? It would be great if somebody can give me some hints or solutions.
 [/INST] Ok, if $B=0$ then take $C=0$ and you are done, so $B^2 \neq 0$.
Since $B$ is complex matrix it has Jordan normal form. There exists invertable matrix $S$ such that $B=SXS^{-1}$ where X has one of 2 possible forms:
$X=\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} $ or $X=\begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix} $.
In the first case $\sqrt{X}=\begin{pmatrix} \sqrt{\lambda_1} & 0 \\ 0 & \sqrt{\lambda_2} \end{pmatrix}$ and $C = S \sqrt{X}S^{-1}$.
In the second case $\lambda \neq 0$, otherwise $B^2=SXS^{-1}SXS^{-1}=SX^2S^{-1}=0$. 
So $\sqrt{X}=\begin{pmatrix} \sqrt{\lambda} & \frac{1}{2 \sqrt{\lambda}} \\ 0 & \sqrt{\lambda} \end{pmatrix}$ and $C = S \sqrt{X}S^{-1}$.
And, of course, $C^2 = S \sqrt{X}S^{-1} S \sqrt{X}S^{-1} =SXS^{-1}=B$.
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