Linear Algebra MA237 Section 1Fall 2009 Dr. ByrneHomework for Section 3.1 Due 10/09I. Using the definition of linear transformation.Verify whether or not the following transformations T: R2 R2 are linear. C) 112121xxxxTT: 2  2 so we must have u,v  2. Let 21uuu and21vvv.Check property 1:11)(2121uuuuTuT11)(2121vvvvTvT11)(22112211vuvuvuvuTvuT221111)()(22112121vuvuvvuuvTuTWe have )()()( vuTvTuT because 112222112211vuvuvuvu. This implies the function is not linear, there is no need to check the second property.F)1121xxTT: 2  2 so we must have u,v  2. Let 21uuu and21vvv.Check property 1:11)(21uuTuT11)(21vvTvT11)(2211vuvuTvuT221111)()( vTuTWe have )()()( vuTvTuT because1122. This implies the function is not linear, there is no need to check the second