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We consider this matrix $$M=\left(\begin{array}{cc}0&-1\\1&0\end{array}\right).$$

I want to find a matrix $B$ such that $$BB=M.$$

Thank you for your help.

José Carlos Santos
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Student
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  • what do you mean by square root of a matrix? – Vinyl_cape_jawa Oct 23 '17 at 08:13
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    Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 5xum Oct 23 '17 at 08:13
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    Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. – 5xum Oct 23 '17 at 08:13

1 Answers1

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Since $M$ is the matrix of a rotation of $\frac\pi2$ radians around the origin, just take the matrix of a $\frac\pi4$ radians around the origin:$$\begin{pmatrix}\frac1{\sqrt2}&-\frac1{\sqrt2}\\\frac1{\sqrt2}&\frac1{\sqrt2}\end{pmatrix}.$$Or you can take the matrix of a $\frac{5\pi}4$ radians around the origin:$$\begin{pmatrix}-\frac1{\sqrt2}&\frac1{\sqrt2}\\-\frac1{\sqrt2}&-\frac1{\sqrt2}\end{pmatrix},$$which is, of course, the symmetric of the previous one. If $B.B=M$, then, obviously, $(-B).(-B)=M$ too. What made you think that the answer was unique?

José Carlos Santos
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