tag:blogger.com,1999:blog-4897507116249235354.post5552881263691556198..comments2010-01-04T15:00:33.403-08:00Comments on Thinking about Theoretical Computer Science: Lattice Based Cryptography - Getting Inside 2Akashhttp://www.blogger.com/profile/06989706264146599229noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-4897507116249235354.post-12916007238031185162010-01-04T15:00:33.403-08:002010-01-04T15:00:33.403-08:00I would just like to mention that the readers migh...I would just like to mention that the readers might want to look up the "fundamental parallelopiped" criterion for a set of n linearly independent vectors to qualify as a basis for a lattice. That too is one of the reasons behind hoping that two equivalent bases must have equal determinants.Shantanamhttps://www.blogger.com/profile/03513861297651643507noreply@blogger.comtag:blogger.com,1999:blog-4897507116249235354.post-74657423400197477352009-12-29T10:13:57.264-08:002009-12-29T10:13:57.264-08:00Nice one watchmath (may i have your name please)
...Nice one watchmath (may i have your name please)<br /><br />And thanks a lot for enabling us write blogs with mathematical symbols.<br /><br />Really great work (and a nice proof!)Akashhttps://www.blogger.com/profile/06989706264146599229noreply@blogger.comtag:blogger.com,1999:blog-4897507116249235354.post-33941868502785743582009-12-29T06:16:05.706-08:002009-12-29T06:16:05.706-08:00Great post. Keep it coming :).
Lets try to prove t...Great post. Keep it coming :).<br />Lets try to prove the statement above.<br />$(\Leftarrow)$ <br />Let $x$ be an integral vector. Take $y=Ux$. This $y$ also an integral vector (since $U$ is unimodular). It follows that $B_2x=B_1y$. NOw for $z=U^{-1}x$ ($z$ is clearly integral) we also have $B_1x=B_2z$. Hence $B_1$ and $B_2$ are equivalent.<br /><br />$(\Rightarrow)$<br />Suppose $B_1,B_2$ are equivalent. Then there are integer vectors $x_i$ such that<br />\[B_2e_i=B_1x_i\]<br />But $B_2e_i$ is exactly the $i$th column of $b_2$. Hence<br />\[B_2=B_1X\]<br />where $X$ is the matrix such that its ith column is $x_i$.<br />By similar argument there is an integer matrix $Y$ such that $B_1=B_2Y$. It follows that<br />\[B_1=B_1XY\]<br />Hence $\det(XY)=1$ which implies that $\det(X)=\pm 1$. So $X$ is unimodular.watchmathhttps://www.blogger.com/profile/13969830482192487872noreply@blogger.com