Unformatted text preview: AB is thereFore singular, as X 6 = 0. 26. (i) Let B be a singular n × n matrix. Then BX = 0 For some non–zero column vector X . Then ( AB ) X = A ( BX ) = A 0 = 0 and hence AB is also singular. (ii) Suppose A is a singular n × n matrix. Then A t is also singular and hence by (i) so is B t A t = ( AB ) t . Consequently AB is also singular 29...
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 Fall '10
 Dreibelbis

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