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I feel ridiculous for asking this, but I can't seem to find a clear answer.

Let

$$U = \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$$

Show that $U$, $-U$, $-I$ where $I$ is the $2 \times 2$ identity matrix, each is its own inverse and the product of any two of these is the third one.

Now, all I want to know is what does the '$-$' mean in front of the $U$ and $I$. I have tried to Google this, but to no avail. I have also gone through my textbook, as well as past papers but I cannot find a clear indication.

Any pointers as to the meaning of '$-$' will be appreciated.

Thanks

beep-boop
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DrDeanification
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    The negative, multiplication with $-1$. $$-\begin{pmatrix} a & b\ c & d\end{pmatrix} = \begin{pmatrix} -a & -b \ -c & -d\end{pmatrix}$$ – Daniel Fischer Sep 04 '14 at 17:58
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    It is the negative of the matrix, that is, $[-U]{ij} = - [U]{ij}$. – copper.hat Sep 04 '14 at 17:58
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    Because the set of square matrices forms a vector space, it is an abelian group with respect to addition. We want $A+(-A)=0$, the zero matrix. This forces $-A$ as above. – Dietrich Burde Sep 04 '14 at 18:59
  • This question is not ridiculous if you don't know the answer, but your way of using MathJax was silly and accordingly I've cleaned it up. See my edits. – Michael Hardy Sep 09 '14 at 17:52

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