tag:blogger.com,1999:blog-5244458674117363159.post4426633945681875552..comments2023-09-24T02:39:05.162-07:00Comments on Daily GRE: Math GRE - #15Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-5244458674117363159.post-79082937205272499672011-08-16T07:37:50.135-07:002011-08-16T07:37:50.135-07:00@8002959260108264927.0
Yup! That was exactly the w...@8002959260108264927.0<br />Yup! That was exactly the way I thought about it too!Paul Liuhttps://www.blogger.com/profile/16809371907394009052noreply@blogger.comtag:blogger.com,1999:blog-5244458674117363159.post-80029592601082649272011-08-16T03:34:44.590-07:002011-08-16T03:34:44.590-07:00This one almost got me. For some reason I thought ...This one almost got me. For some reason I thought 1 wasn't true, but knew that 2 & 3 were, so I did the same thing you did - only the last one has both 2 & 3, so it must be right.<br /><br />After thinking about it, we have<br /><br />\[A-2I=PBP^{-1}-2PIP^{-1}=P(B-2I)P^{-1}\]<br /><br />so $A-2I$ and $B-2I$ are similar by the same matrix $P$.Kevinhttps://www.blogger.com/profile/13910448998912161723noreply@blogger.com